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Find the name of services that have been used for more than 2 times in first notification of loss.
CREATE TABLE services (service_name VARCHAR, service_id VARCHAR); CREATE TABLE first_notification_of_loss (service_id VARCHAR)
SELECT t2.service_name FROM first_notification_of_loss AS t1 JOIN services AS t2 ON t1.service_id = t2.service_id GROUP BY t1.service_id HAVING COUNT(*) > 2
### Context: CREATE TABLE services (service_name VARCHAR, service_id VARCHAR); CREATE TABLE first_notification_of_loss (service_id VARCHAR) ### Question: Find the name of services that have been used for more than 2 times in first notification of loss. ### Answer: SELECT t2.service_name FROM first_notification_of_loss AS t1 JOIN services AS t2 ON t1.service_id = t2.service_id GROUP BY t1.service_id HAVING COUNT(*) > 2
What is the effective date of the claim that has the largest amount of total settlement?
CREATE TABLE settlements (claim_id VARCHAR, settlement_amount INTEGER); CREATE TABLE claims (Effective_Date VARCHAR, claim_id VARCHAR)
SELECT t1.Effective_Date FROM claims AS t1 JOIN settlements AS t2 ON t1.claim_id = t2.claim_id GROUP BY t1.claim_id ORDER BY SUM(t2.settlement_amount) DESC LIMIT 1
### Context: CREATE TABLE settlements (claim_id VARCHAR, settlement_amount INTEGER); CREATE TABLE claims (Effective_Date VARCHAR, claim_id VARCHAR) ### Question: What is the effective date of the claim that has the largest amount of total settlement? ### Answer: SELECT t1.Effective_Date FROM claims AS t1 JOIN settlements AS t2 ON t1.claim_id = t2.claim_id GROUP BY t1.claim_id ORDER BY SUM(t2.settlement_amount) DESC LIMIT 1
How many policies are listed for the customer named "Dayana Robel"?
CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR); CREATE TABLE customers_policies (customer_id VARCHAR)
SELECT COUNT(*) FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id WHERE t1.customer_name = "Dayana Robel"
### Context: CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR); CREATE TABLE customers_policies (customer_id VARCHAR) ### Question: How many policies are listed for the customer named "Dayana Robel"? ### Answer: SELECT COUNT(*) FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id WHERE t1.customer_name = "Dayana Robel"
What is the name of the customer who has the most policies listed?
CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE customers_policies (customer_id VARCHAR)
SELECT t1.customer_name FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id GROUP BY t1.customer_name ORDER BY COUNT(*) DESC LIMIT 1
### Context: CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE customers_policies (customer_id VARCHAR) ### Question: What is the name of the customer who has the most policies listed? ### Answer: SELECT t1.customer_name FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id GROUP BY t1.customer_name ORDER BY COUNT(*) DESC LIMIT 1
What are all the policy types of the customer named "Dayana Robel"?
CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR); CREATE TABLE available_policies (policy_type_code VARCHAR, policy_id VARCHAR); CREATE TABLE customers_policies (customer_id VARCHAR, policy_id VARCHAR)
SELECT DISTINCT t3.policy_type_code FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id JOIN available_policies AS t3 ON t2.policy_id = t3.policy_id WHERE t1.customer_name = "Dayana Robel"
### Context: CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR); CREATE TABLE available_policies (policy_type_code VARCHAR, policy_id VARCHAR); CREATE TABLE customers_policies (customer_id VARCHAR, policy_id VARCHAR) ### Question: What are all the policy types of the customer named "Dayana Robel"? ### Answer: SELECT DISTINCT t3.policy_type_code FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id JOIN available_policies AS t3 ON t2.policy_id = t3.policy_id WHERE t1.customer_name = "Dayana Robel"
What are all the policy types of the customer that has the most policies listed?
CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR); CREATE TABLE available_policies (policy_type_code VARCHAR, policy_id VARCHAR); CREATE TABLE customers_policies (customer_id VARCHAR, policy_id VARCHAR)
SELECT DISTINCT t3.policy_type_code FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id JOIN available_policies AS t3 ON t2.policy_id = t3.policy_id WHERE t1.customer_name = (SELECT t1.customer_name FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id GROUP BY t1.customer_name ORDER BY COUNT(*) DESC LIMIT 1)
### Context: CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR); CREATE TABLE available_policies (policy_type_code VARCHAR, policy_id VARCHAR); CREATE TABLE customers_policies (customer_id VARCHAR, policy_id VARCHAR) ### Question: What are all the policy types of the customer that has the most policies listed? ### Answer: SELECT DISTINCT t3.policy_type_code FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id JOIN available_policies AS t3 ON t2.policy_id = t3.policy_id WHERE t1.customer_name = (SELECT t1.customer_name FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id GROUP BY t1.customer_name ORDER BY COUNT(*) DESC LIMIT 1)
List all the services in the alphabetical order.
CREATE TABLE services (service_name VARCHAR)
SELECT service_name FROM services ORDER BY service_name
### Context: CREATE TABLE services (service_name VARCHAR) ### Question: List all the services in the alphabetical order. ### Answer: SELECT service_name FROM services ORDER BY service_name
How many services are there?
CREATE TABLE services (Id VARCHAR)
SELECT COUNT(*) FROM services
### Context: CREATE TABLE services (Id VARCHAR) ### Question: How many services are there? ### Answer: SELECT COUNT(*) FROM services
Find the names of users who do not have a first notification of loss record.
CREATE TABLE first_notification_of_loss (customer_id VARCHAR); CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE customers (customer_name VARCHAR)
SELECT customer_name FROM customers EXCEPT SELECT t1.customer_name FROM customers AS t1 JOIN first_notification_of_loss AS t2 ON t1.customer_id = t2.customer_id
### Context: CREATE TABLE first_notification_of_loss (customer_id VARCHAR); CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE customers (customer_name VARCHAR) ### Question: Find the names of users who do not have a first notification of loss record. ### Answer: SELECT customer_name FROM customers EXCEPT SELECT t1.customer_name FROM customers AS t1 JOIN first_notification_of_loss AS t2 ON t1.customer_id = t2.customer_id
Find the names of customers who have used either the service "Close a policy" or the service "Upgrade a policy".
CREATE TABLE first_notification_of_loss (customer_id VARCHAR, service_id VARCHAR); CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE services (service_id VARCHAR, service_name VARCHAR)
SELECT t1.customer_name FROM customers AS t1 JOIN first_notification_of_loss AS t2 ON t1.customer_id = t2.customer_id JOIN services AS t3 ON t2.service_id = t3.service_id WHERE t3.service_name = "Close a policy" OR t3.service_name = "Upgrade a policy"
### Context: CREATE TABLE first_notification_of_loss (customer_id VARCHAR, service_id VARCHAR); CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE services (service_id VARCHAR, service_name VARCHAR) ### Question: Find the names of customers who have used either the service "Close a policy" or the service "Upgrade a policy". ### Answer: SELECT t1.customer_name FROM customers AS t1 JOIN first_notification_of_loss AS t2 ON t1.customer_id = t2.customer_id JOIN services AS t3 ON t2.service_id = t3.service_id WHERE t3.service_name = "Close a policy" OR t3.service_name = "Upgrade a policy"
Find the names of customers who have used both the service "Close a policy" and the service "New policy application".
CREATE TABLE first_notification_of_loss (customer_id VARCHAR, service_id VARCHAR); CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE services (service_id VARCHAR, service_name VARCHAR)
SELECT t1.customer_name FROM customers AS t1 JOIN first_notification_of_loss AS t2 ON t1.customer_id = t2.customer_id JOIN services AS t3 ON t2.service_id = t3.service_id WHERE t3.service_name = "Close a policy" INTERSECT SELECT t1.customer_name FROM customers AS t1 JOIN first_notification_of_loss AS t2 ON t1.customer_id = t2.customer_id JOIN services AS t3 ON t2.service_id = t3.service_id WHERE t3.service_name = "New policy application"
### Context: CREATE TABLE first_notification_of_loss (customer_id VARCHAR, service_id VARCHAR); CREATE TABLE customers (customer_name VARCHAR, customer_id VARCHAR); CREATE TABLE services (service_id VARCHAR, service_name VARCHAR) ### Question: Find the names of customers who have used both the service "Close a policy" and the service "New policy application". ### Answer: SELECT t1.customer_name FROM customers AS t1 JOIN first_notification_of_loss AS t2 ON t1.customer_id = t2.customer_id JOIN services AS t3 ON t2.service_id = t3.service_id WHERE t3.service_name = "Close a policy" INTERSECT SELECT t1.customer_name FROM customers AS t1 JOIN first_notification_of_loss AS t2 ON t1.customer_id = t2.customer_id JOIN services AS t3 ON t2.service_id = t3.service_id WHERE t3.service_name = "New policy application"
Find the IDs of customers whose name contains "Diana".
CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR)
SELECT customer_id FROM customers WHERE customer_name LIKE "%Diana%"
### Context: CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR) ### Question: Find the IDs of customers whose name contains "Diana". ### Answer: SELECT customer_id FROM customers WHERE customer_name LIKE "%Diana%"
What are the maximum and minimum settlement amount on record?
CREATE TABLE settlements (settlement_amount INTEGER)
SELECT MAX(settlement_amount), MIN(settlement_amount) FROM settlements
### Context: CREATE TABLE settlements (settlement_amount INTEGER) ### Question: What are the maximum and minimum settlement amount on record? ### Answer: SELECT MAX(settlement_amount), MIN(settlement_amount) FROM settlements
List all the customers in increasing order of IDs.
CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR)
SELECT customer_id, customer_name FROM customers ORDER BY customer_id
### Context: CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR) ### Question: List all the customers in increasing order of IDs. ### Answer: SELECT customer_id, customer_name FROM customers ORDER BY customer_id
Retrieve the open and close dates of all the policies associated with the customer whose name contains "Diana"
CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR); CREATE TABLE customers_policies (date_opened VARCHAR, date_closed VARCHAR, customer_id VARCHAR)
SELECT t2.date_opened, t2.date_closed FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id WHERE t1.customer_name LIKE "%Diana%"
### Context: CREATE TABLE customers (customer_id VARCHAR, customer_name VARCHAR); CREATE TABLE customers_policies (date_opened VARCHAR, date_closed VARCHAR, customer_id VARCHAR) ### Question: Retrieve the open and close dates of all the policies associated with the customer whose name contains "Diana" ### Answer: SELECT t2.date_opened, t2.date_closed FROM customers AS t1 JOIN customers_policies AS t2 ON t1.customer_id = t2.customer_id WHERE t1.customer_name LIKE "%Diana%"
How many kinds of enzymes are there?
CREATE TABLE enzyme (Id VARCHAR)
SELECT COUNT(*) FROM enzyme
### Context: CREATE TABLE enzyme (Id VARCHAR) ### Question: How many kinds of enzymes are there? ### Answer: SELECT COUNT(*) FROM enzyme
List the name of enzymes in descending lexicographical order.
CREATE TABLE enzyme (name VARCHAR)
SELECT name FROM enzyme ORDER BY name DESC
### Context: CREATE TABLE enzyme (name VARCHAR) ### Question: List the name of enzymes in descending lexicographical order. ### Answer: SELECT name FROM enzyme ORDER BY name DESC
List the names and the locations that the enzymes can make an effect.
CREATE TABLE enzyme (name VARCHAR, LOCATION VARCHAR)
SELECT name, LOCATION FROM enzyme
### Context: CREATE TABLE enzyme (name VARCHAR, LOCATION VARCHAR) ### Question: List the names and the locations that the enzymes can make an effect. ### Answer: SELECT name, LOCATION FROM enzyme
What is the maximum Online Mendelian Inheritance in Man (OMIM) value of the enzymes?
CREATE TABLE enzyme (OMIM INTEGER)
SELECT MAX(OMIM) FROM enzyme
### Context: CREATE TABLE enzyme (OMIM INTEGER) ### Question: What is the maximum Online Mendelian Inheritance in Man (OMIM) value of the enzymes? ### Answer: SELECT MAX(OMIM) FROM enzyme
What is the product, chromosome and porphyria related to the enzymes which take effect at the location 'Cytosol'?
CREATE TABLE enzyme (product VARCHAR, chromosome VARCHAR, porphyria VARCHAR, LOCATION VARCHAR)
SELECT product, chromosome, porphyria FROM enzyme WHERE LOCATION = 'Cytosol'
### Context: CREATE TABLE enzyme (product VARCHAR, chromosome VARCHAR, porphyria VARCHAR, LOCATION VARCHAR) ### Question: What is the product, chromosome and porphyria related to the enzymes which take effect at the location 'Cytosol'? ### Answer: SELECT product, chromosome, porphyria FROM enzyme WHERE LOCATION = 'Cytosol'
What are the names of enzymes who does not produce 'Heme'?
CREATE TABLE enzyme (name VARCHAR, product VARCHAR)
SELECT name FROM enzyme WHERE product <> 'Heme'
### Context: CREATE TABLE enzyme (name VARCHAR, product VARCHAR) ### Question: What are the names of enzymes who does not produce 'Heme'? ### Answer: SELECT name FROM enzyme WHERE product <> 'Heme'
What are the names and trade names of the medicines which has 'Yes' value in the FDA record?
CREATE TABLE medicine (name VARCHAR, trade_name VARCHAR, FDA_approved VARCHAR)
SELECT name, trade_name FROM medicine WHERE FDA_approved = 'Yes'
### Context: CREATE TABLE medicine (name VARCHAR, trade_name VARCHAR, FDA_approved VARCHAR) ### Question: What are the names and trade names of the medicines which has 'Yes' value in the FDA record? ### Answer: SELECT name, trade_name FROM medicine WHERE FDA_approved = 'Yes'
What are the names of enzymes in the medicine named 'Amisulpride' that can serve as an 'inhibitor'?
CREATE TABLE medicine (id VARCHAR, name VARCHAR); CREATE TABLE medicine_enzyme_interaction (enzyme_id VARCHAR, medicine_id VARCHAR, interaction_type VARCHAR); CREATE TABLE enzyme (name VARCHAR, id VARCHAR)
SELECT T1.name FROM enzyme AS T1 JOIN medicine_enzyme_interaction AS T2 ON T1.id = T2.enzyme_id JOIN medicine AS T3 ON T2.medicine_id = T3.id WHERE T3.name = 'Amisulpride' AND T2.interaction_type = 'inhibitor'
### Context: CREATE TABLE medicine (id VARCHAR, name VARCHAR); CREATE TABLE medicine_enzyme_interaction (enzyme_id VARCHAR, medicine_id VARCHAR, interaction_type VARCHAR); CREATE TABLE enzyme (name VARCHAR, id VARCHAR) ### Question: What are the names of enzymes in the medicine named 'Amisulpride' that can serve as an 'inhibitor'? ### Answer: SELECT T1.name FROM enzyme AS T1 JOIN medicine_enzyme_interaction AS T2 ON T1.id = T2.enzyme_id JOIN medicine AS T3 ON T2.medicine_id = T3.id WHERE T3.name = 'Amisulpride' AND T2.interaction_type = 'inhibitor'
What are the ids and names of the medicine that can interact with two or more enzymes?
CREATE TABLE medicine_enzyme_interaction (medicine_id VARCHAR); CREATE TABLE medicine (id VARCHAR, Name VARCHAR)
SELECT T1.id, T1.Name FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id GROUP BY T1.id HAVING COUNT(*) >= 2
### Context: CREATE TABLE medicine_enzyme_interaction (medicine_id VARCHAR); CREATE TABLE medicine (id VARCHAR, Name VARCHAR) ### Question: What are the ids and names of the medicine that can interact with two or more enzymes? ### Answer: SELECT T1.id, T1.Name FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id GROUP BY T1.id HAVING COUNT(*) >= 2
What are the ids, names and FDA approval status of medicines in descending order of the number of enzymes that it can interact with.
CREATE TABLE medicine_enzyme_interaction (medicine_id VARCHAR); CREATE TABLE medicine (id VARCHAR, Name VARCHAR, FDA_approved VARCHAR)
SELECT T1.id, T1.Name, T1.FDA_approved FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id GROUP BY T1.id ORDER BY COUNT(*) DESC
### Context: CREATE TABLE medicine_enzyme_interaction (medicine_id VARCHAR); CREATE TABLE medicine (id VARCHAR, Name VARCHAR, FDA_approved VARCHAR) ### Question: What are the ids, names and FDA approval status of medicines in descending order of the number of enzymes that it can interact with. ### Answer: SELECT T1.id, T1.Name, T1.FDA_approved FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id GROUP BY T1.id ORDER BY COUNT(*) DESC
What is the id and name of the enzyme with most number of medicines that can interact as 'activator'?
CREATE TABLE medicine_enzyme_interaction (enzyme_id VARCHAR, interaction_type VARCHAR); CREATE TABLE enzyme (id VARCHAR, name VARCHAR)
SELECT T1.id, T1.name FROM enzyme AS T1 JOIN medicine_enzyme_interaction AS T2 ON T1.id = T2.enzyme_id WHERE T2.interaction_type = 'activitor' GROUP BY T1.id ORDER BY COUNT(*) DESC LIMIT 1
### Context: CREATE TABLE medicine_enzyme_interaction (enzyme_id VARCHAR, interaction_type VARCHAR); CREATE TABLE enzyme (id VARCHAR, name VARCHAR) ### Question: What is the id and name of the enzyme with most number of medicines that can interact as 'activator'? ### Answer: SELECT T1.id, T1.name FROM enzyme AS T1 JOIN medicine_enzyme_interaction AS T2 ON T1.id = T2.enzyme_id WHERE T2.interaction_type = 'activitor' GROUP BY T1.id ORDER BY COUNT(*) DESC LIMIT 1
What is the interaction type of the enzyme named 'ALA synthase' and the medicine named 'Aripiprazole'?
CREATE TABLE enzyme (id VARCHAR, name VARCHAR); CREATE TABLE medicine (id VARCHAR, name VARCHAR); CREATE TABLE medicine_enzyme_interaction (interaction_type VARCHAR, medicine_id VARCHAR, enzyme_id VARCHAR)
SELECT T1.interaction_type FROM medicine_enzyme_interaction AS T1 JOIN medicine AS T2 ON T1.medicine_id = T2.id JOIN enzyme AS T3 ON T1.enzyme_id = T3.id WHERE T3.name = 'ALA synthase' AND T2.name = 'Aripiprazole'
### Context: CREATE TABLE enzyme (id VARCHAR, name VARCHAR); CREATE TABLE medicine (id VARCHAR, name VARCHAR); CREATE TABLE medicine_enzyme_interaction (interaction_type VARCHAR, medicine_id VARCHAR, enzyme_id VARCHAR) ### Question: What is the interaction type of the enzyme named 'ALA synthase' and the medicine named 'Aripiprazole'? ### Answer: SELECT T1.interaction_type FROM medicine_enzyme_interaction AS T1 JOIN medicine AS T2 ON T1.medicine_id = T2.id JOIN enzyme AS T3 ON T1.enzyme_id = T3.id WHERE T3.name = 'ALA synthase' AND T2.name = 'Aripiprazole'
What is the most common interaction type between enzymes and medicine? And how many are there?
CREATE TABLE medicine_enzyme_interaction (interaction_type VARCHAR)
SELECT interaction_type, COUNT(*) FROM medicine_enzyme_interaction GROUP BY interaction_type ORDER BY COUNT(*) DESC LIMIT 1
### Context: CREATE TABLE medicine_enzyme_interaction (interaction_type VARCHAR) ### Question: What is the most common interaction type between enzymes and medicine? And how many are there? ### Answer: SELECT interaction_type, COUNT(*) FROM medicine_enzyme_interaction GROUP BY interaction_type ORDER BY COUNT(*) DESC LIMIT 1
How many medicines have the FDA approval status 'No' ?
CREATE TABLE medicine (FDA_approved VARCHAR)
SELECT COUNT(*) FROM medicine WHERE FDA_approved = 'No'
### Context: CREATE TABLE medicine (FDA_approved VARCHAR) ### Question: How many medicines have the FDA approval status 'No' ? ### Answer: SELECT COUNT(*) FROM medicine WHERE FDA_approved = 'No'
How many enzymes do not have any interactions?
CREATE TABLE medicine_enzyme_interaction (id VARCHAR, enzyme_id VARCHAR); CREATE TABLE enzyme (id VARCHAR, enzyme_id VARCHAR)
SELECT COUNT(*) FROM enzyme WHERE NOT id IN (SELECT enzyme_id FROM medicine_enzyme_interaction)
### Context: CREATE TABLE medicine_enzyme_interaction (id VARCHAR, enzyme_id VARCHAR); CREATE TABLE enzyme (id VARCHAR, enzyme_id VARCHAR) ### Question: How many enzymes do not have any interactions? ### Answer: SELECT COUNT(*) FROM enzyme WHERE NOT id IN (SELECT enzyme_id FROM medicine_enzyme_interaction)
What is the id and trade name of the medicines can interact with at least 3 enzymes?
CREATE TABLE medicine (id VARCHAR, trade_name VARCHAR); CREATE TABLE medicine_enzyme_interaction (medicine_id VARCHAR)
SELECT T1.id, T1.trade_name FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id GROUP BY T1.id HAVING COUNT(*) >= 3
### Context: CREATE TABLE medicine (id VARCHAR, trade_name VARCHAR); CREATE TABLE medicine_enzyme_interaction (medicine_id VARCHAR) ### Question: What is the id and trade name of the medicines can interact with at least 3 enzymes? ### Answer: SELECT T1.id, T1.trade_name FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id GROUP BY T1.id HAVING COUNT(*) >= 3
What are the distinct name, location and products of the enzymes which has any 'inhibitor' interaction?
CREATE TABLE enzyme (name VARCHAR, location VARCHAR, product VARCHAR, id VARCHAR); CREATE TABLE medicine_enzyme_interaction (enzyme_id VARCHAR, interaction_type VARCHAR)
SELECT DISTINCT T1.name, T1.location, T1.product FROM enzyme AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.enzyme_id = T1.id WHERE T2.interaction_type = 'inhibitor'
### Context: CREATE TABLE enzyme (name VARCHAR, location VARCHAR, product VARCHAR, id VARCHAR); CREATE TABLE medicine_enzyme_interaction (enzyme_id VARCHAR, interaction_type VARCHAR) ### Question: What are the distinct name, location and products of the enzymes which has any 'inhibitor' interaction? ### Answer: SELECT DISTINCT T1.name, T1.location, T1.product FROM enzyme AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.enzyme_id = T1.id WHERE T2.interaction_type = 'inhibitor'
List the medicine name and trade name which can both interact as 'inhibitor' and 'activitor' with enzymes.
CREATE TABLE medicine (name VARCHAR, trade_name VARCHAR, id VARCHAR); CREATE TABLE medicine_enzyme_interaction (medicine_id VARCHAR)
SELECT T1.name, T1.trade_name FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id WHERE interaction_type = 'inhibitor' INTERSECT SELECT T1.name, T1.trade_name FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id WHERE interaction_type = 'activitor'
### Context: CREATE TABLE medicine (name VARCHAR, trade_name VARCHAR, id VARCHAR); CREATE TABLE medicine_enzyme_interaction (medicine_id VARCHAR) ### Question: List the medicine name and trade name which can both interact as 'inhibitor' and 'activitor' with enzymes. ### Answer: SELECT T1.name, T1.trade_name FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id WHERE interaction_type = 'inhibitor' INTERSECT SELECT T1.name, T1.trade_name FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id WHERE interaction_type = 'activitor'
Show the medicine names and trade names that cannot interact with the enzyme with product 'Heme'.
CREATE TABLE medicine (name VARCHAR, trade_name VARCHAR); CREATE TABLE medicine_enzyme_interaction (medicine_id VARCHAR, enzyme_id VARCHAR); CREATE TABLE medicine (name VARCHAR, trade_name VARCHAR, id VARCHAR); CREATE TABLE enzyme (id VARCHAR, product VARCHAR)
SELECT name, trade_name FROM medicine EXCEPT SELECT T1.name, T1.trade_name FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id JOIN enzyme AS T3 ON T3.id = T2.enzyme_id WHERE T3.product = 'Protoporphyrinogen IX'
### Context: CREATE TABLE medicine (name VARCHAR, trade_name VARCHAR); CREATE TABLE medicine_enzyme_interaction (medicine_id VARCHAR, enzyme_id VARCHAR); CREATE TABLE medicine (name VARCHAR, trade_name VARCHAR, id VARCHAR); CREATE TABLE enzyme (id VARCHAR, product VARCHAR) ### Question: Show the medicine names and trade names that cannot interact with the enzyme with product 'Heme'. ### Answer: SELECT name, trade_name FROM medicine EXCEPT SELECT T1.name, T1.trade_name FROM medicine AS T1 JOIN medicine_enzyme_interaction AS T2 ON T2.medicine_id = T1.id JOIN enzyme AS T3 ON T3.id = T2.enzyme_id WHERE T3.product = 'Protoporphyrinogen IX'
How many distinct FDA approval statuses are there for the medicines?
CREATE TABLE medicine (FDA_approved VARCHAR)
SELECT COUNT(DISTINCT FDA_approved) FROM medicine
### Context: CREATE TABLE medicine (FDA_approved VARCHAR) ### Question: How many distinct FDA approval statuses are there for the medicines? ### Answer: SELECT COUNT(DISTINCT FDA_approved) FROM medicine
Which enzyme names have the substring "ALA"?
CREATE TABLE enzyme (name VARCHAR)
SELECT name FROM enzyme WHERE name LIKE "%ALA%"
### Context: CREATE TABLE enzyme (name VARCHAR) ### Question: Which enzyme names have the substring "ALA"? ### Answer: SELECT name FROM enzyme WHERE name LIKE "%ALA%"
find the number of medicines offered by each trade.
CREATE TABLE medicine (trade_name VARCHAR)
SELECT trade_name, COUNT(*) FROM medicine GROUP BY trade_name
### Context: CREATE TABLE medicine (trade_name VARCHAR) ### Question: find the number of medicines offered by each trade. ### Answer: SELECT trade_name, COUNT(*) FROM medicine GROUP BY trade_name
List all schools and their nicknames in the order of founded year.
CREATE TABLE university (school VARCHAR, nickname VARCHAR, founded VARCHAR)
SELECT school, nickname FROM university ORDER BY founded
### Context: CREATE TABLE university (school VARCHAR, nickname VARCHAR, founded VARCHAR) ### Question: List all schools and their nicknames in the order of founded year. ### Answer: SELECT school, nickname FROM university ORDER BY founded
List all public schools and their locations.
CREATE TABLE university (school VARCHAR, LOCATION VARCHAR, affiliation VARCHAR)
SELECT school, LOCATION FROM university WHERE affiliation = 'Public'
### Context: CREATE TABLE university (school VARCHAR, LOCATION VARCHAR, affiliation VARCHAR) ### Question: List all public schools and their locations. ### Answer: SELECT school, LOCATION FROM university WHERE affiliation = 'Public'
When was the school with the largest enrollment founded?
CREATE TABLE university (founded VARCHAR, enrollment VARCHAR)
SELECT founded FROM university ORDER BY enrollment DESC LIMIT 1
### Context: CREATE TABLE university (founded VARCHAR, enrollment VARCHAR) ### Question: When was the school with the largest enrollment founded? ### Answer: SELECT founded FROM university ORDER BY enrollment DESC LIMIT 1
Find the founded year of the newest non public school.
CREATE TABLE university (founded VARCHAR, affiliation VARCHAR)
SELECT founded FROM university WHERE affiliation <> 'Public' ORDER BY founded DESC LIMIT 1
### Context: CREATE TABLE university (founded VARCHAR, affiliation VARCHAR) ### Question: Find the founded year of the newest non public school. ### Answer: SELECT founded FROM university WHERE affiliation <> 'Public' ORDER BY founded DESC LIMIT 1
How many schools are in the basketball match?
CREATE TABLE basketball_match (school_id VARCHAR)
SELECT COUNT(DISTINCT school_id) FROM basketball_match
### Context: CREATE TABLE basketball_match (school_id VARCHAR) ### Question: How many schools are in the basketball match? ### Answer: SELECT COUNT(DISTINCT school_id) FROM basketball_match
What is the highest acc percent score in the competition?
CREATE TABLE basketball_match (acc_percent VARCHAR)
SELECT acc_percent FROM basketball_match ORDER BY acc_percent DESC LIMIT 1
### Context: CREATE TABLE basketball_match (acc_percent VARCHAR) ### Question: What is the highest acc percent score in the competition? ### Answer: SELECT acc_percent FROM basketball_match ORDER BY acc_percent DESC LIMIT 1
What is the primary conference of the school that has the lowest acc percent score in the competition?
CREATE TABLE basketball_match (school_id VARCHAR, acc_percent VARCHAR); CREATE TABLE university (Primary_conference VARCHAR, school_id VARCHAR)
SELECT t1.Primary_conference FROM university AS t1 JOIN basketball_match AS t2 ON t1.school_id = t2.school_id ORDER BY t2.acc_percent LIMIT 1
### Context: CREATE TABLE basketball_match (school_id VARCHAR, acc_percent VARCHAR); CREATE TABLE university (Primary_conference VARCHAR, school_id VARCHAR) ### Question: What is the primary conference of the school that has the lowest acc percent score in the competition? ### Answer: SELECT t1.Primary_conference FROM university AS t1 JOIN basketball_match AS t2 ON t1.school_id = t2.school_id ORDER BY t2.acc_percent LIMIT 1
What is the team name and acc regular season score of the school that was founded for the longest time?
CREATE TABLE university (school_id VARCHAR, founded VARCHAR); CREATE TABLE basketball_match (team_name VARCHAR, ACC_Regular_Season VARCHAR, school_id VARCHAR)
SELECT t2.team_name, t2.ACC_Regular_Season FROM university AS t1 JOIN basketball_match AS t2 ON t1.school_id = t2.school_id ORDER BY t1.founded LIMIT 1
### Context: CREATE TABLE university (school_id VARCHAR, founded VARCHAR); CREATE TABLE basketball_match (team_name VARCHAR, ACC_Regular_Season VARCHAR, school_id VARCHAR) ### Question: What is the team name and acc regular season score of the school that was founded for the longest time? ### Answer: SELECT t2.team_name, t2.ACC_Regular_Season FROM university AS t1 JOIN basketball_match AS t2 ON t1.school_id = t2.school_id ORDER BY t1.founded LIMIT 1
Find the location and all games score of the school that has Clemson as its team name.
CREATE TABLE basketball_match (All_Games VARCHAR, school_id VARCHAR); CREATE TABLE university (location VARCHAR, school_id VARCHAR)
SELECT t2.All_Games, t1.location FROM university AS t1 JOIN basketball_match AS t2 ON t1.school_id = t2.school_id WHERE team_name = 'Clemson'
### Context: CREATE TABLE basketball_match (All_Games VARCHAR, school_id VARCHAR); CREATE TABLE university (location VARCHAR, school_id VARCHAR) ### Question: Find the location and all games score of the school that has Clemson as its team name. ### Answer: SELECT t2.All_Games, t1.location FROM university AS t1 JOIN basketball_match AS t2 ON t1.school_id = t2.school_id WHERE team_name = 'Clemson'
What are the average enrollment size of the universities that are founded before 1850?
CREATE TABLE university (enrollment INTEGER, founded INTEGER)
SELECT AVG(enrollment) FROM university WHERE founded < 1850
### Context: CREATE TABLE university (enrollment INTEGER, founded INTEGER) ### Question: What are the average enrollment size of the universities that are founded before 1850? ### Answer: SELECT AVG(enrollment) FROM university WHERE founded < 1850
Show the enrollment and primary_conference of the oldest college.
CREATE TABLE university (enrollment VARCHAR, primary_conference VARCHAR, founded VARCHAR)
SELECT enrollment, primary_conference FROM university ORDER BY founded LIMIT 1
### Context: CREATE TABLE university (enrollment VARCHAR, primary_conference VARCHAR, founded VARCHAR) ### Question: Show the enrollment and primary_conference of the oldest college. ### Answer: SELECT enrollment, primary_conference FROM university ORDER BY founded LIMIT 1
What is the total and minimum enrollment of all schools?
CREATE TABLE university (enrollment INTEGER)
SELECT SUM(enrollment), MIN(enrollment) FROM university
### Context: CREATE TABLE university (enrollment INTEGER) ### Question: What is the total and minimum enrollment of all schools? ### Answer: SELECT SUM(enrollment), MIN(enrollment) FROM university
Find the total student enrollment for different affiliation type schools.
CREATE TABLE university (affiliation VARCHAR, enrollment INTEGER)
SELECT SUM(enrollment), affiliation FROM university GROUP BY affiliation
### Context: CREATE TABLE university (affiliation VARCHAR, enrollment INTEGER) ### Question: Find the total student enrollment for different affiliation type schools. ### Answer: SELECT SUM(enrollment), affiliation FROM university GROUP BY affiliation
How many schools do not participate in the basketball match?
CREATE TABLE university (school_id VARCHAR); CREATE TABLE basketball_match (school_id VARCHAR)
SELECT COUNT(*) FROM university WHERE NOT school_id IN (SELECT school_id FROM basketball_match)
### Context: CREATE TABLE university (school_id VARCHAR); CREATE TABLE basketball_match (school_id VARCHAR) ### Question: How many schools do not participate in the basketball match? ### Answer: SELECT COUNT(*) FROM university WHERE NOT school_id IN (SELECT school_id FROM basketball_match)
Find the schools that were either founded after 1850 or public.
CREATE TABLE university (school VARCHAR, founded VARCHAR, affiliation VARCHAR)
SELECT school FROM university WHERE founded > 1850 OR affiliation = 'Public'
### Context: CREATE TABLE university (school VARCHAR, founded VARCHAR, affiliation VARCHAR) ### Question: Find the schools that were either founded after 1850 or public. ### Answer: SELECT school FROM university WHERE founded > 1850 OR affiliation = 'Public'
Find how many different affiliation types there are.
CREATE TABLE university (affiliation VARCHAR)
SELECT COUNT(DISTINCT affiliation) FROM university
### Context: CREATE TABLE university (affiliation VARCHAR) ### Question: Find how many different affiliation types there are. ### Answer: SELECT COUNT(DISTINCT affiliation) FROM university
Find how many school locations have the word 'NY'.
CREATE TABLE university (LOCATION VARCHAR)
SELECT COUNT(*) FROM university WHERE LOCATION LIKE "%NY%"
### Context: CREATE TABLE university (LOCATION VARCHAR) ### Question: Find how many school locations have the word 'NY'. ### Answer: SELECT COUNT(*) FROM university WHERE LOCATION LIKE "%NY%"
Find the team names of the universities whose enrollments are smaller than the average enrollment size.
CREATE TABLE university (school_id VARCHAR); CREATE TABLE basketball_match (team_name VARCHAR, school_id VARCHAR); CREATE TABLE university (enrollment INTEGER)
SELECT t2.team_name FROM university AS t1 JOIN basketball_match AS t2 ON t1.school_id = t2.school_id WHERE enrollment < (SELECT AVG(enrollment) FROM university)
### Context: CREATE TABLE university (school_id VARCHAR); CREATE TABLE basketball_match (team_name VARCHAR, school_id VARCHAR); CREATE TABLE university (enrollment INTEGER) ### Question: Find the team names of the universities whose enrollments are smaller than the average enrollment size. ### Answer: SELECT t2.team_name FROM university AS t1 JOIN basketball_match AS t2 ON t1.school_id = t2.school_id WHERE enrollment < (SELECT AVG(enrollment) FROM university)
Find the number of universities that have over a 20000 enrollment size for each affiliation type.
CREATE TABLE university (affiliation VARCHAR, enrollment INTEGER)
SELECT COUNT(*), affiliation FROM university WHERE enrollment > 20000 GROUP BY affiliation
### Context: CREATE TABLE university (affiliation VARCHAR, enrollment INTEGER) ### Question: Find the number of universities that have over a 20000 enrollment size for each affiliation type. ### Answer: SELECT COUNT(*), affiliation FROM university WHERE enrollment > 20000 GROUP BY affiliation
Find the total number of students enrolled in the colleges that were founded after the year of 1850 for each affiliation type.
CREATE TABLE university (affiliation VARCHAR, Enrollment INTEGER, founded INTEGER)
SELECT SUM(Enrollment), affiliation FROM university WHERE founded > 1850 GROUP BY affiliation
### Context: CREATE TABLE university (affiliation VARCHAR, Enrollment INTEGER, founded INTEGER) ### Question: Find the total number of students enrolled in the colleges that were founded after the year of 1850 for each affiliation type. ### Answer: SELECT SUM(Enrollment), affiliation FROM university WHERE founded > 1850 GROUP BY affiliation
What is the maximum enrollment across all schools?
CREATE TABLE university (Enrollment INTEGER)
SELECT MAX(Enrollment) FROM university
### Context: CREATE TABLE university (Enrollment INTEGER) ### Question: What is the maximum enrollment across all schools? ### Answer: SELECT MAX(Enrollment) FROM university
List all information regarding the basketball match.
CREATE TABLE basketball_match (Id VARCHAR)
SELECT * FROM basketball_match
### Context: CREATE TABLE basketball_match (Id VARCHAR) ### Question: List all information regarding the basketball match. ### Answer: SELECT * FROM basketball_match
List names of all teams in the basketball competition, ordered by all home scores in descending order.
CREATE TABLE basketball_match (team_name VARCHAR, All_Home VARCHAR)
SELECT team_name FROM basketball_match ORDER BY All_Home DESC
### Context: CREATE TABLE basketball_match (team_name VARCHAR, All_Home VARCHAR) ### Question: List names of all teams in the basketball competition, ordered by all home scores in descending order. ### Answer: SELECT team_name FROM basketball_match ORDER BY All_Home DESC
the names of models that launched between 2002 and 2004.
CREATE TABLE chip_model (Model_name VARCHAR, Launch_year INTEGER)
SELECT Model_name FROM chip_model WHERE Launch_year BETWEEN 2002 AND 2004
### Context: CREATE TABLE chip_model (Model_name VARCHAR, Launch_year INTEGER) ### Question: the names of models that launched between 2002 and 2004. ### Answer: SELECT Model_name FROM chip_model WHERE Launch_year BETWEEN 2002 AND 2004
Which model has the least amount of RAM? List the model name and the amount of RAM.
CREATE TABLE chip_model (Model_name VARCHAR, RAM_MiB VARCHAR)
SELECT Model_name, RAM_MiB FROM chip_model ORDER BY RAM_MiB LIMIT 1
### Context: CREATE TABLE chip_model (Model_name VARCHAR, RAM_MiB VARCHAR) ### Question: Which model has the least amount of RAM? List the model name and the amount of RAM. ### Answer: SELECT Model_name, RAM_MiB FROM chip_model ORDER BY RAM_MiB LIMIT 1
What are the chip model and screen mode of the phone with hardware model name "LG-P760"?
CREATE TABLE phone (chip_model VARCHAR, screen_mode VARCHAR, Hardware_Model_name VARCHAR)
SELECT chip_model, screen_mode FROM phone WHERE Hardware_Model_name = "LG-P760"
### Context: CREATE TABLE phone (chip_model VARCHAR, screen_mode VARCHAR, Hardware_Model_name VARCHAR) ### Question: What are the chip model and screen mode of the phone with hardware model name "LG-P760"? ### Answer: SELECT chip_model, screen_mode FROM phone WHERE Hardware_Model_name = "LG-P760"
How many phone hardware models are produced by the company named "Nokia Corporation"?
CREATE TABLE phone (Company_name VARCHAR)
SELECT COUNT(*) FROM phone WHERE Company_name = "Nokia Corporation"
### Context: CREATE TABLE phone (Company_name VARCHAR) ### Question: How many phone hardware models are produced by the company named "Nokia Corporation"? ### Answer: SELECT COUNT(*) FROM phone WHERE Company_name = "Nokia Corporation"
What is maximum and minimum RAM size of phone produced by company named "Nokia Corporation"?
CREATE TABLE phone (chip_model VARCHAR, Company_name VARCHAR); CREATE TABLE chip_model (RAM_MiB INTEGER, Model_name VARCHAR)
SELECT MAX(T1.RAM_MiB), MIN(T1.RAM_MiB) FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model WHERE T2.Company_name = "Nokia Corporation"
### Context: CREATE TABLE phone (chip_model VARCHAR, Company_name VARCHAR); CREATE TABLE chip_model (RAM_MiB INTEGER, Model_name VARCHAR) ### Question: What is maximum and minimum RAM size of phone produced by company named "Nokia Corporation"? ### Answer: SELECT MAX(T1.RAM_MiB), MIN(T1.RAM_MiB) FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model WHERE T2.Company_name = "Nokia Corporation"
What is the average ROM size of phones produced by the company named "Nokia Corporation"?
CREATE TABLE phone (chip_model VARCHAR, Company_name VARCHAR); CREATE TABLE chip_model (ROM_MiB INTEGER, Model_name VARCHAR)
SELECT AVG(T1.ROM_MiB) FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model WHERE T2.Company_name = "Nokia Corporation"
### Context: CREATE TABLE phone (chip_model VARCHAR, Company_name VARCHAR); CREATE TABLE chip_model (ROM_MiB INTEGER, Model_name VARCHAR) ### Question: What is the average ROM size of phones produced by the company named "Nokia Corporation"? ### Answer: SELECT AVG(T1.ROM_MiB) FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model WHERE T2.Company_name = "Nokia Corporation"
List the hardware model name and company name for all the phones that were launched in year 2002 or have RAM size greater than 32.
CREATE TABLE chip_model (Model_name VARCHAR, Launch_year VARCHAR, RAM_MiB VARCHAR); CREATE TABLE phone (Hardware_Model_name VARCHAR, Company_name VARCHAR, chip_model VARCHAR)
SELECT T2.Hardware_Model_name, T2.Company_name FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model WHERE T1.Launch_year = 2002 OR T1.RAM_MiB > 32
### Context: CREATE TABLE chip_model (Model_name VARCHAR, Launch_year VARCHAR, RAM_MiB VARCHAR); CREATE TABLE phone (Hardware_Model_name VARCHAR, Company_name VARCHAR, chip_model VARCHAR) ### Question: List the hardware model name and company name for all the phones that were launched in year 2002 or have RAM size greater than 32. ### Answer: SELECT T2.Hardware_Model_name, T2.Company_name FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model WHERE T1.Launch_year = 2002 OR T1.RAM_MiB > 32
Find all phones that have word 'Full' in their accreditation types. List the Hardware Model name and Company name.
CREATE TABLE phone (Hardware_Model_name VARCHAR, Company_name VARCHAR, Accreditation_type VARCHAR)
SELECT Hardware_Model_name, Company_name FROM phone WHERE Accreditation_type LIKE 'Full'
### Context: CREATE TABLE phone (Hardware_Model_name VARCHAR, Company_name VARCHAR, Accreditation_type VARCHAR) ### Question: Find all phones that have word 'Full' in their accreditation types. List the Hardware Model name and Company name. ### Answer: SELECT Hardware_Model_name, Company_name FROM phone WHERE Accreditation_type LIKE 'Full'
Find the Char cells, Pixels and Hardware colours for the screen of the phone whose hardware model name is "LG-P760".
CREATE TABLE phone (screen_mode VARCHAR, Hardware_Model_name VARCHAR); CREATE TABLE screen_mode (Char_cells VARCHAR, Pixels VARCHAR, Hardware_colours VARCHAR, Graphics_mode VARCHAR)
SELECT T1.Char_cells, T1.Pixels, T1.Hardware_colours FROM screen_mode AS T1 JOIN phone AS T2 ON T1.Graphics_mode = T2.screen_mode WHERE T2.Hardware_Model_name = "LG-P760"
### Context: CREATE TABLE phone (screen_mode VARCHAR, Hardware_Model_name VARCHAR); CREATE TABLE screen_mode (Char_cells VARCHAR, Pixels VARCHAR, Hardware_colours VARCHAR, Graphics_mode VARCHAR) ### Question: Find the Char cells, Pixels and Hardware colours for the screen of the phone whose hardware model name is "LG-P760". ### Answer: SELECT T1.Char_cells, T1.Pixels, T1.Hardware_colours FROM screen_mode AS T1 JOIN phone AS T2 ON T1.Graphics_mode = T2.screen_mode WHERE T2.Hardware_Model_name = "LG-P760"
List the hardware model name and company name for the phone whose screen mode type is "Graphics."
CREATE TABLE phone (Hardware_Model_name VARCHAR, Company_name VARCHAR, screen_mode VARCHAR); CREATE TABLE screen_mode (Graphics_mode VARCHAR, Type VARCHAR)
SELECT T2.Hardware_Model_name, T2.Company_name FROM screen_mode AS T1 JOIN phone AS T2 ON T1.Graphics_mode = T2.screen_mode WHERE T1.Type = "Graphics"
### Context: CREATE TABLE phone (Hardware_Model_name VARCHAR, Company_name VARCHAR, screen_mode VARCHAR); CREATE TABLE screen_mode (Graphics_mode VARCHAR, Type VARCHAR) ### Question: List the hardware model name and company name for the phone whose screen mode type is "Graphics." ### Answer: SELECT T2.Hardware_Model_name, T2.Company_name FROM screen_mode AS T1 JOIN phone AS T2 ON T1.Graphics_mode = T2.screen_mode WHERE T1.Type = "Graphics"
Find the name of the company that has the least number of phone models. List the company name and the number of phone model produced by that company.
CREATE TABLE phone (Company_name VARCHAR)
SELECT Company_name, COUNT(*) FROM phone GROUP BY Company_name ORDER BY COUNT(*) LIMIT 1
### Context: CREATE TABLE phone (Company_name VARCHAR) ### Question: Find the name of the company that has the least number of phone models. List the company name and the number of phone model produced by that company. ### Answer: SELECT Company_name, COUNT(*) FROM phone GROUP BY Company_name ORDER BY COUNT(*) LIMIT 1
List the name of the company that produced more than one phone model.
CREATE TABLE phone (Company_name VARCHAR)
SELECT Company_name FROM phone GROUP BY Company_name HAVING COUNT(*) > 1
### Context: CREATE TABLE phone (Company_name VARCHAR) ### Question: List the name of the company that produced more than one phone model. ### Answer: SELECT Company_name FROM phone GROUP BY Company_name HAVING COUNT(*) > 1
List the maximum, minimum and average number of used kb in screen mode.
CREATE TABLE screen_mode (used_kb INTEGER)
SELECT MAX(used_kb), MIN(used_kb), AVG(used_kb) FROM screen_mode
### Context: CREATE TABLE screen_mode (used_kb INTEGER) ### Question: List the maximum, minimum and average number of used kb in screen mode. ### Answer: SELECT MAX(used_kb), MIN(used_kb), AVG(used_kb) FROM screen_mode
List the name of the phone model launched in year 2002 and with the highest RAM size.
CREATE TABLE chip_model (Model_name VARCHAR, Launch_year VARCHAR, RAM_MiB VARCHAR); CREATE TABLE phone (Hardware_Model_name VARCHAR, chip_model VARCHAR)
SELECT T2.Hardware_Model_name FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model WHERE T1.Launch_year = 2002 ORDER BY T1.RAM_MiB DESC LIMIT 1
### Context: CREATE TABLE chip_model (Model_name VARCHAR, Launch_year VARCHAR, RAM_MiB VARCHAR); CREATE TABLE phone (Hardware_Model_name VARCHAR, chip_model VARCHAR) ### Question: List the name of the phone model launched in year 2002 and with the highest RAM size. ### Answer: SELECT T2.Hardware_Model_name FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model WHERE T1.Launch_year = 2002 ORDER BY T1.RAM_MiB DESC LIMIT 1
What are the wifi and screen mode type of the hardware model named "LG-P760"?
CREATE TABLE phone (chip_model VARCHAR, screen_mode VARCHAR, Hardware_Model_name VARCHAR); CREATE TABLE chip_model (WiFi VARCHAR, Model_name VARCHAR); CREATE TABLE screen_mode (Type VARCHAR, Graphics_mode VARCHAR)
SELECT T1.WiFi, T3.Type FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model JOIN screen_mode AS T3 ON T2.screen_mode = T3.Graphics_mode WHERE T2.Hardware_Model_name = "LG-P760"
### Context: CREATE TABLE phone (chip_model VARCHAR, screen_mode VARCHAR, Hardware_Model_name VARCHAR); CREATE TABLE chip_model (WiFi VARCHAR, Model_name VARCHAR); CREATE TABLE screen_mode (Type VARCHAR, Graphics_mode VARCHAR) ### Question: What are the wifi and screen mode type of the hardware model named "LG-P760"? ### Answer: SELECT T1.WiFi, T3.Type FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model JOIN screen_mode AS T3 ON T2.screen_mode = T3.Graphics_mode WHERE T2.Hardware_Model_name = "LG-P760"
List the hardware model name for the phones that have screen mode type "Text" or RAM size greater than 32.
CREATE TABLE chip_model (Model_name VARCHAR, RAM_MiB VARCHAR); CREATE TABLE phone (Hardware_Model_name VARCHAR, chip_model VARCHAR, screen_mode VARCHAR); CREATE TABLE screen_mode (Graphics_mode VARCHAR, Type VARCHAR)
SELECT T2.Hardware_Model_name FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model JOIN screen_mode AS T3 ON T2.screen_mode = T3.Graphics_mode WHERE T3.Type = "Text" OR T1.RAM_MiB > 32
### Context: CREATE TABLE chip_model (Model_name VARCHAR, RAM_MiB VARCHAR); CREATE TABLE phone (Hardware_Model_name VARCHAR, chip_model VARCHAR, screen_mode VARCHAR); CREATE TABLE screen_mode (Graphics_mode VARCHAR, Type VARCHAR) ### Question: List the hardware model name for the phones that have screen mode type "Text" or RAM size greater than 32. ### Answer: SELECT T2.Hardware_Model_name FROM chip_model AS T1 JOIN phone AS T2 ON T1.Model_name = T2.chip_model JOIN screen_mode AS T3 ON T2.screen_mode = T3.Graphics_mode WHERE T3.Type = "Text" OR T1.RAM_MiB > 32
List the hardware model name for the phones that were produced by "Nokia Corporation" or whose screen mode type is "Graphics."
CREATE TABLE phone (Hardware_Model_name VARCHAR, screen_mode VARCHAR); CREATE TABLE screen_mode (Graphics_mode VARCHAR, Type VARCHAR)
SELECT DISTINCT T2.Hardware_Model_name FROM screen_mode AS T1 JOIN phone AS T2 ON T1.Graphics_mode = T2.screen_mode WHERE T1.Type = "Graphics" OR t2.Company_name = "Nokia Corporation"
### Context: CREATE TABLE phone (Hardware_Model_name VARCHAR, screen_mode VARCHAR); CREATE TABLE screen_mode (Graphics_mode VARCHAR, Type VARCHAR) ### Question: List the hardware model name for the phones that were produced by "Nokia Corporation" or whose screen mode type is "Graphics." ### Answer: SELECT DISTINCT T2.Hardware_Model_name FROM screen_mode AS T1 JOIN phone AS T2 ON T1.Graphics_mode = T2.screen_mode WHERE T1.Type = "Graphics" OR t2.Company_name = "Nokia Corporation"
List the hardware model name for the phons that were produced by "Nokia Corporation" but whose screen mode type is not Text.
CREATE TABLE phone (Hardware_Model_name VARCHAR, screen_mode VARCHAR); CREATE TABLE screen_mode (Graphics_mode VARCHAR, Type VARCHAR)
SELECT DISTINCT T2.Hardware_Model_name FROM screen_mode AS T1 JOIN phone AS T2 ON T1.Graphics_mode = T2.screen_mode WHERE t2.Company_name = "Nokia Corporation" AND T1.Type <> "Text"
### Context: CREATE TABLE phone (Hardware_Model_name VARCHAR, screen_mode VARCHAR); CREATE TABLE screen_mode (Graphics_mode VARCHAR, Type VARCHAR) ### Question: List the hardware model name for the phons that were produced by "Nokia Corporation" but whose screen mode type is not Text. ### Answer: SELECT DISTINCT T2.Hardware_Model_name FROM screen_mode AS T1 JOIN phone AS T2 ON T1.Graphics_mode = T2.screen_mode WHERE t2.Company_name = "Nokia Corporation" AND T1.Type <> "Text"
List the phone hardware model and company name for the phones whose screen usage in kb is between 10 and 15.
CREATE TABLE phone (Hardware_Model_name VARCHAR, Company_name VARCHAR, screen_mode VARCHAR); CREATE TABLE screen_mode (Graphics_mode VARCHAR, used_kb INTEGER)
SELECT DISTINCT T2.Hardware_Model_name, T2.Company_name FROM screen_mode AS T1 JOIN phone AS T2 ON T1.Graphics_mode = T2.screen_mode WHERE T1.used_kb BETWEEN 10 AND 15
### Context: CREATE TABLE phone (Hardware_Model_name VARCHAR, Company_name VARCHAR, screen_mode VARCHAR); CREATE TABLE screen_mode (Graphics_mode VARCHAR, used_kb INTEGER) ### Question: List the phone hardware model and company name for the phones whose screen usage in kb is between 10 and 15. ### Answer: SELECT DISTINCT T2.Hardware_Model_name, T2.Company_name FROM screen_mode AS T1 JOIN phone AS T2 ON T1.Graphics_mode = T2.screen_mode WHERE T1.used_kb BETWEEN 10 AND 15
Find the number of phones for each accreditation type.
CREATE TABLE phone (Accreditation_type VARCHAR)
SELECT Accreditation_type, COUNT(*) FROM phone GROUP BY Accreditation_type
### Context: CREATE TABLE phone (Accreditation_type VARCHAR) ### Question: Find the number of phones for each accreditation type. ### Answer: SELECT Accreditation_type, COUNT(*) FROM phone GROUP BY Accreditation_type
Find the accreditation level that more than 3 phones use.
CREATE TABLE phone (Accreditation_level VARCHAR)
SELECT Accreditation_level FROM phone GROUP BY Accreditation_level HAVING COUNT(*) > 3
### Context: CREATE TABLE phone (Accreditation_level VARCHAR) ### Question: Find the accreditation level that more than 3 phones use. ### Answer: SELECT Accreditation_level FROM phone GROUP BY Accreditation_level HAVING COUNT(*) > 3
Find the details for all chip models.
CREATE TABLE chip_model (Id VARCHAR)
SELECT * FROM chip_model
### Context: CREATE TABLE chip_model (Id VARCHAR) ### Question: Find the details for all chip models. ### Answer: SELECT * FROM chip_model
How many models do not have the wifi function?
CREATE TABLE chip_model (wifi VARCHAR)
SELECT COUNT(*) FROM chip_model WHERE wifi = 'No'
### Context: CREATE TABLE chip_model (wifi VARCHAR) ### Question: How many models do not have the wifi function? ### Answer: SELECT COUNT(*) FROM chip_model WHERE wifi = 'No'
List all the model names sorted by their launch year.
CREATE TABLE chip_model (model_name VARCHAR, launch_year VARCHAR)
SELECT model_name FROM chip_model ORDER BY launch_year
### Context: CREATE TABLE chip_model (model_name VARCHAR, launch_year VARCHAR) ### Question: List all the model names sorted by their launch year. ### Answer: SELECT model_name FROM chip_model ORDER BY launch_year
Find the average ram mib size of the chip models that are never used by any phone.
CREATE TABLE chip_model (RAM_MiB INTEGER, model_name VARCHAR, chip_model VARCHAR); CREATE TABLE phone (RAM_MiB INTEGER, model_name VARCHAR, chip_model VARCHAR)
SELECT AVG(RAM_MiB) FROM chip_model WHERE NOT model_name IN (SELECT chip_model FROM phone)
### Context: CREATE TABLE chip_model (RAM_MiB INTEGER, model_name VARCHAR, chip_model VARCHAR); CREATE TABLE phone (RAM_MiB INTEGER, model_name VARCHAR, chip_model VARCHAR) ### Question: Find the average ram mib size of the chip models that are never used by any phone. ### Answer: SELECT AVG(RAM_MiB) FROM chip_model WHERE NOT model_name IN (SELECT chip_model FROM phone)
Find the names of the chip models that are not used by any phone with full accreditation type.
CREATE TABLE chip_model (model_name VARCHAR, chip_model VARCHAR, Accreditation_type VARCHAR); CREATE TABLE phone (model_name VARCHAR, chip_model VARCHAR, Accreditation_type VARCHAR)
SELECT model_name FROM chip_model EXCEPT SELECT chip_model FROM phone WHERE Accreditation_type = 'Full'
### Context: CREATE TABLE chip_model (model_name VARCHAR, chip_model VARCHAR, Accreditation_type VARCHAR); CREATE TABLE phone (model_name VARCHAR, chip_model VARCHAR, Accreditation_type VARCHAR) ### Question: Find the names of the chip models that are not used by any phone with full accreditation type. ### Answer: SELECT model_name FROM chip_model EXCEPT SELECT chip_model FROM phone WHERE Accreditation_type = 'Full'
Find the pixels of the screen modes that are used by both phones with full accreditation types and phones with Provisional accreditation types.
CREATE TABLE phone (screen_mode VARCHAR, Accreditation_type VARCHAR); CREATE TABLE screen_mode (pixels VARCHAR, Graphics_mode VARCHAR)
SELECT t1.pixels FROM screen_mode AS t1 JOIN phone AS t2 ON t1.Graphics_mode = t2.screen_mode WHERE t2.Accreditation_type = 'Provisional' INTERSECT SELECT t1.pixels FROM screen_mode AS t1 JOIN phone AS t2 ON t1.Graphics_mode = t2.screen_mode WHERE t2.Accreditation_type = 'Full'
### Context: CREATE TABLE phone (screen_mode VARCHAR, Accreditation_type VARCHAR); CREATE TABLE screen_mode (pixels VARCHAR, Graphics_mode VARCHAR) ### Question: Find the pixels of the screen modes that are used by both phones with full accreditation types and phones with Provisional accreditation types. ### Answer: SELECT t1.pixels FROM screen_mode AS t1 JOIN phone AS t2 ON t1.Graphics_mode = t2.screen_mode WHERE t2.Accreditation_type = 'Provisional' INTERSECT SELECT t1.pixels FROM screen_mode AS t1 JOIN phone AS t2 ON t1.Graphics_mode = t2.screen_mode WHERE t2.Accreditation_type = 'Full'
How many countries are there in total?
CREATE TABLE country (Id VARCHAR)
SELECT COUNT(*) FROM country
### Context: CREATE TABLE country (Id VARCHAR) ### Question: How many countries are there in total? ### Answer: SELECT COUNT(*) FROM country
Show the country name and capital of all countries.
CREATE TABLE country (Country_name VARCHAR, Capital VARCHAR)
SELECT Country_name, Capital FROM country
### Context: CREATE TABLE country (Country_name VARCHAR, Capital VARCHAR) ### Question: Show the country name and capital of all countries. ### Answer: SELECT Country_name, Capital FROM country
Show all official native languages that contain the word "English".
CREATE TABLE country (Official_native_language VARCHAR)
SELECT Official_native_language FROM country WHERE Official_native_language LIKE "%English%"
### Context: CREATE TABLE country (Official_native_language VARCHAR) ### Question: Show all official native languages that contain the word "English". ### Answer: SELECT Official_native_language FROM country WHERE Official_native_language LIKE "%English%"
Show all distinct positions of matches.
CREATE TABLE match_season (POSITION VARCHAR)
SELECT DISTINCT POSITION FROM match_season
### Context: CREATE TABLE match_season (POSITION VARCHAR) ### Question: Show all distinct positions of matches. ### Answer: SELECT DISTINCT POSITION FROM match_season
Show the players from college UCLA.
CREATE TABLE match_season (Player VARCHAR, College VARCHAR)
SELECT Player FROM match_season WHERE College = "UCLA"
### Context: CREATE TABLE match_season (Player VARCHAR, College VARCHAR) ### Question: Show the players from college UCLA. ### Answer: SELECT Player FROM match_season WHERE College = "UCLA"
Show the distinct position of players from college UCLA or Duke.
CREATE TABLE match_season (POSITION VARCHAR, College VARCHAR)
SELECT DISTINCT POSITION FROM match_season WHERE College = "UCLA" OR College = "Duke"
### Context: CREATE TABLE match_season (POSITION VARCHAR, College VARCHAR) ### Question: Show the distinct position of players from college UCLA or Duke. ### Answer: SELECT DISTINCT POSITION FROM match_season WHERE College = "UCLA" OR College = "Duke"
Show the draft pick numbers and draft classes of players whose positions are defenders.
CREATE TABLE match_season (Draft_Pick_Number VARCHAR, Draft_Class VARCHAR, POSITION VARCHAR)
SELECT Draft_Pick_Number, Draft_Class FROM match_season WHERE POSITION = "Defender"
### Context: CREATE TABLE match_season (Draft_Pick_Number VARCHAR, Draft_Class VARCHAR, POSITION VARCHAR) ### Question: Show the draft pick numbers and draft classes of players whose positions are defenders. ### Answer: SELECT Draft_Pick_Number, Draft_Class FROM match_season WHERE POSITION = "Defender"
How many distinct teams are involved in match seasons?
CREATE TABLE match_season (Team VARCHAR)
SELECT COUNT(DISTINCT Team) FROM match_season
### Context: CREATE TABLE match_season (Team VARCHAR) ### Question: How many distinct teams are involved in match seasons? ### Answer: SELECT COUNT(DISTINCT Team) FROM match_season
Show the players and the years played.
CREATE TABLE player (Player VARCHAR, Years_Played VARCHAR)
SELECT Player, Years_Played FROM player
### Context: CREATE TABLE player (Player VARCHAR, Years_Played VARCHAR) ### Question: Show the players and the years played. ### Answer: SELECT Player, Years_Played FROM player
Show all team names.
CREATE TABLE Team (Name VARCHAR)
SELECT Name FROM Team
### Context: CREATE TABLE Team (Name VARCHAR) ### Question: Show all team names. ### Answer: SELECT Name FROM Team
Show the season, the player, and the name of the country that player belongs to.
CREATE TABLE match_season (Season VARCHAR, Player VARCHAR, Country VARCHAR); CREATE TABLE country (Country_name VARCHAR, Country_id VARCHAR)
SELECT T2.Season, T2.Player, T1.Country_name FROM country AS T1 JOIN match_season AS T2 ON T1.Country_id = T2.Country
### Context: CREATE TABLE match_season (Season VARCHAR, Player VARCHAR, Country VARCHAR); CREATE TABLE country (Country_name VARCHAR, Country_id VARCHAR) ### Question: Show the season, the player, and the name of the country that player belongs to. ### Answer: SELECT T2.Season, T2.Player, T1.Country_name FROM country AS T1 JOIN match_season AS T2 ON T1.Country_id = T2.Country
Which players are from Indonesia?
CREATE TABLE country (Country_id VARCHAR, Country_name VARCHAR); CREATE TABLE match_season (Player VARCHAR, Country VARCHAR)
SELECT T2.Player FROM country AS T1 JOIN match_season AS T2 ON T1.Country_id = T2.Country WHERE T1.Country_name = "Indonesia"
### Context: CREATE TABLE country (Country_id VARCHAR, Country_name VARCHAR); CREATE TABLE match_season (Player VARCHAR, Country VARCHAR) ### Question: Which players are from Indonesia? ### Answer: SELECT T2.Player FROM country AS T1 JOIN match_season AS T2 ON T1.Country_id = T2.Country WHERE T1.Country_name = "Indonesia"
What are the distinct positions of the players from a country whose capital is Dublin?
CREATE TABLE country (Country_id VARCHAR, Capital VARCHAR); CREATE TABLE match_season (Position VARCHAR, Country VARCHAR)
SELECT DISTINCT T2.Position FROM country AS T1 JOIN match_season AS T2 ON T1.Country_id = T2.Country WHERE T1.Capital = "Dublin"
### Context: CREATE TABLE country (Country_id VARCHAR, Capital VARCHAR); CREATE TABLE match_season (Position VARCHAR, Country VARCHAR) ### Question: What are the distinct positions of the players from a country whose capital is Dublin? ### Answer: SELECT DISTINCT T2.Position FROM country AS T1 JOIN match_season AS T2 ON T1.Country_id = T2.Country WHERE T1.Capital = "Dublin"