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Find the dates on which more than one revisions were made.
CREATE TABLE Catalogs (date_of_latest_revision VARCHAR)
SELECT date_of_latest_revision FROM Catalogs GROUP BY date_of_latest_revision HAVING COUNT(*) > 1
How many products are there in the records?
CREATE TABLE catalog_contents (Id VARCHAR)
SELECT COUNT(*) FROM catalog_contents
Name all the products with next entry ID greater than 8.
CREATE TABLE catalog_contents (catalog_entry_name VARCHAR, next_entry_id INTEGER)
SELECT catalog_entry_name FROM catalog_contents WHERE next_entry_id > 8
How many aircrafts do we have?
CREATE TABLE Aircraft (Id VARCHAR)
SELECT COUNT(*) FROM Aircraft
Show name and distance for all aircrafts.
CREATE TABLE Aircraft (name VARCHAR, distance VARCHAR)
SELECT name, distance FROM Aircraft
Show ids for all aircrafts with more than 1000 distance.
CREATE TABLE Aircraft (aid VARCHAR, distance INTEGER)
SELECT aid FROM Aircraft WHERE distance > 1000
How many aircrafts have distance between 1000 and 5000?
CREATE TABLE Aircraft (distance INTEGER)
SELECT COUNT(*) FROM Aircraft WHERE distance BETWEEN 1000 AND 5000
What is the name and distance for aircraft with id 12?
CREATE TABLE Aircraft (name VARCHAR, distance VARCHAR, aid VARCHAR)
SELECT name, distance FROM Aircraft WHERE aid = 12
What is the minimum, average, and maximum distance of all aircrafts.
CREATE TABLE Aircraft (distance INTEGER)
SELECT MIN(distance), AVG(distance), MAX(distance) FROM Aircraft
Show the id and name of the aircraft with the maximum distance.
CREATE TABLE Aircraft (aid VARCHAR, name VARCHAR, distance VARCHAR)
SELECT aid, name FROM Aircraft ORDER BY distance DESC LIMIT 1
Show the name of aircrafts with top three lowest distances.
CREATE TABLE Aircraft (name VARCHAR, distance VARCHAR)
SELECT name FROM Aircraft ORDER BY distance LIMIT 3
Show names for all aircrafts with distances more than the average.
CREATE TABLE Aircraft (name VARCHAR, distance INTEGER)
SELECT name FROM Aircraft WHERE distance > (SELECT AVG(distance) FROM Aircraft)
How many employees do we have?
CREATE TABLE Employee (Id VARCHAR)
SELECT COUNT(*) FROM Employee
Show name and salary for all employees sorted by salary.
CREATE TABLE Employee (name VARCHAR, salary VARCHAR)
SELECT name, salary FROM Employee ORDER BY salary
Show ids for all employees with at least 100000 salary.
CREATE TABLE Employee (eid VARCHAR, salary INTEGER)
SELECT eid FROM Employee WHERE salary > 100000
How many employees have salary between 100000 and 200000?
CREATE TABLE Employee (salary INTEGER)
SELECT COUNT(*) FROM Employee WHERE salary BETWEEN 100000 AND 200000
What is the name and salary for employee with id 242518965?
CREATE TABLE Employee (name VARCHAR, salary VARCHAR, eid VARCHAR)
SELECT name, salary FROM Employee WHERE eid = 242518965
What is average and maximum salary of all employees.
CREATE TABLE Employee (salary INTEGER)
SELECT AVG(salary), MAX(salary) FROM Employee
Show the id and name of the employee with maximum salary.
CREATE TABLE Employee (eid VARCHAR, name VARCHAR, salary VARCHAR)
SELECT eid, name FROM Employee ORDER BY salary DESC LIMIT 1
Show the name of employees with three lowest salaries.
CREATE TABLE Employee (name VARCHAR, salary VARCHAR)
SELECT name FROM Employee ORDER BY salary LIMIT 3
Show names for all employees with salary more than the average.
CREATE TABLE Employee (name VARCHAR, salary INTEGER)
SELECT name FROM Employee WHERE salary > (SELECT AVG(salary) FROM Employee)
Show the id and salary of Mark Young.
CREATE TABLE Employee (eid VARCHAR, salary VARCHAR, name VARCHAR)
SELECT eid, salary FROM Employee WHERE name = 'Mark Young'
How many flights do we have?
CREATE TABLE Flight (Id VARCHAR)
SELECT COUNT(*) FROM Flight
Show flight number, origin, destination of all flights in the alphabetical order of the departure cities.
CREATE TABLE Flight (flno VARCHAR, origin VARCHAR, destination VARCHAR)
SELECT flno, origin, destination FROM Flight ORDER BY origin
Show all flight number from Los Angeles.
CREATE TABLE Flight (flno VARCHAR, origin VARCHAR)
SELECT flno FROM Flight WHERE origin = "Los Angeles"
Show origins of all flights with destination Honolulu.
CREATE TABLE Flight (origin VARCHAR, destination VARCHAR)
SELECT origin FROM Flight WHERE destination = "Honolulu"
Show me the departure date and arrival date for all flights from Los Angeles to Honolulu.
CREATE TABLE Flight (departure_date VARCHAR, arrival_date VARCHAR, origin VARCHAR, destination VARCHAR)
SELECT departure_date, arrival_date FROM Flight WHERE origin = "Los Angeles" AND destination = "Honolulu"
Show flight number for all flights with more than 2000 distance.
CREATE TABLE Flight (flno VARCHAR, distance INTEGER)
SELECT flno FROM Flight WHERE distance > 2000
What is the average price for flights from Los Angeles to Honolulu.
CREATE TABLE Flight (price INTEGER, origin VARCHAR, destination VARCHAR)
SELECT AVG(price) FROM Flight WHERE origin = "Los Angeles" AND destination = "Honolulu"
Show origin and destination for flights with price higher than 300.
CREATE TABLE Flight (origin VARCHAR, destination VARCHAR, price INTEGER)
SELECT origin, destination FROM Flight WHERE price > 300
Show the flight number and distance of the flight with maximum price.
CREATE TABLE Flight (flno VARCHAR, distance VARCHAR, price VARCHAR)
SELECT flno, distance FROM Flight ORDER BY price DESC LIMIT 1
Show the flight number of flights with three lowest distances.
CREATE TABLE Flight (flno VARCHAR, distance VARCHAR)
SELECT flno FROM Flight ORDER BY distance LIMIT 3
What is the average distance and average price for flights from Los Angeles.
CREATE TABLE Flight (distance INTEGER, price INTEGER, origin VARCHAR)
SELECT AVG(distance), AVG(price) FROM Flight WHERE origin = "Los Angeles"
Show all origins and the number of flights from each origin.
CREATE TABLE Flight (origin VARCHAR)
SELECT origin, COUNT(*) FROM Flight GROUP BY origin
Show all destinations and the number of flights to each destination.
CREATE TABLE Flight (destination VARCHAR)
SELECT destination, COUNT(*) FROM Flight GROUP BY destination
Which origin has most number of flights?
CREATE TABLE Flight (origin VARCHAR)
SELECT origin FROM Flight GROUP BY origin ORDER BY COUNT(*) DESC LIMIT 1
Which destination has least number of flights?
CREATE TABLE Flight (destination VARCHAR)
SELECT destination FROM Flight GROUP BY destination ORDER BY COUNT(*) LIMIT 1
What is the aircraft name for the flight with number 99
CREATE TABLE Flight (aid VARCHAR, flno VARCHAR); CREATE TABLE Aircraft (name VARCHAR, aid VARCHAR)
SELECT T2.name FROM Flight AS T1 JOIN Aircraft AS T2 ON T1.aid = T2.aid WHERE T1.flno = 99
Show all flight numbers with aircraft Airbus A340-300.
CREATE TABLE Flight (flno VARCHAR, aid VARCHAR); CREATE TABLE Aircraft (aid VARCHAR, name VARCHAR)
SELECT T1.flno FROM Flight AS T1 JOIN Aircraft AS T2 ON T1.aid = T2.aid WHERE T2.name = "Airbus A340-300"
Show aircraft names and number of flights for each aircraft.
CREATE TABLE Aircraft (name VARCHAR, aid VARCHAR); CREATE TABLE Flight (aid VARCHAR)
SELECT T2.name, COUNT(*) FROM Flight AS T1 JOIN Aircraft AS T2 ON T1.aid = T2.aid GROUP BY T1.aid
Show names for all aircraft with at least two flights.
CREATE TABLE Aircraft (name VARCHAR, aid VARCHAR); CREATE TABLE Flight (aid VARCHAR)
SELECT T2.name FROM Flight AS T1 JOIN Aircraft AS T2 ON T1.aid = T2.aid GROUP BY T1.aid HAVING COUNT(*) >= 2
How many employees have certificate.
CREATE TABLE Certificate (eid VARCHAR)
SELECT COUNT(DISTINCT eid) FROM Certificate
Show ids for all employees who don't have a certificate.
CREATE TABLE Employee (eid VARCHAR); CREATE TABLE Certificate (eid VARCHAR)
SELECT eid FROM Employee EXCEPT SELECT eid FROM Certificate
Show names for all aircrafts of which John Williams has certificates.
CREATE TABLE Aircraft (name VARCHAR, aid VARCHAR); CREATE TABLE Employee (eid VARCHAR, name VARCHAR); CREATE TABLE Certificate (eid VARCHAR, aid VARCHAR)
SELECT T3.name FROM Employee AS T1 JOIN Certificate AS T2 ON T1.eid = T2.eid JOIN Aircraft AS T3 ON T3.aid = T2.aid WHERE T1.name = "John Williams"
Show names for all employees who have certificate of Boeing 737-800.
CREATE TABLE Employee (name VARCHAR, eid VARCHAR); CREATE TABLE Certificate (eid VARCHAR, aid VARCHAR); CREATE TABLE Aircraft (aid VARCHAR, name VARCHAR)
SELECT T1.name FROM Employee AS T1 JOIN Certificate AS T2 ON T1.eid = T2.eid JOIN Aircraft AS T3 ON T3.aid = T2.aid WHERE T3.name = "Boeing 737-800"
Show names for all employees who have certificates on both Boeing 737-800 and Airbus A340-300.
CREATE TABLE Employee (name VARCHAR, eid VARCHAR); CREATE TABLE Certificate (eid VARCHAR, aid VARCHAR); CREATE TABLE Aircraft (aid VARCHAR, name VARCHAR)
SELECT T1.name FROM Employee AS T1 JOIN Certificate AS T2 ON T1.eid = T2.eid JOIN Aircraft AS T3 ON T3.aid = T2.aid WHERE T3.name = "Boeing 737-800" INTERSECT SELECT T1.name FROM Employee AS T1 JOIN Certificate AS T2 ON T1.eid = T2.eid JOIN Aircraft AS T3 ON T3.aid = T2.aid WHERE T3.name = "Airbus A340-300"
Show names for all employees who do not have certificate of Boeing 737-800.
CREATE TABLE Certificate (eid VARCHAR, aid VARCHAR); CREATE TABLE Employee (name VARCHAR, eid VARCHAR); CREATE TABLE Employee (name VARCHAR); CREATE TABLE Aircraft (aid VARCHAR, name VARCHAR)
SELECT name FROM Employee EXCEPT SELECT T1.name FROM Employee AS T1 JOIN Certificate AS T2 ON T1.eid = T2.eid JOIN Aircraft AS T3 ON T3.aid = T2.aid WHERE T3.name = "Boeing 737-800"
Show the name of aircraft which fewest people have its certificate.
CREATE TABLE Certificate (aid VARCHAR); CREATE TABLE Aircraft (name VARCHAR, aid VARCHAR)
SELECT T2.name FROM Certificate AS T1 JOIN Aircraft AS T2 ON T2.aid = T1.aid GROUP BY T1.aid ORDER BY COUNT(*) DESC LIMIT 1
Show the name and distance of the aircrafts with more than 5000 distance and which at least 5 people have its certificate.
CREATE TABLE Certificate (aid VARCHAR); CREATE TABLE Aircraft (name VARCHAR, aid VARCHAR, distance INTEGER)
SELECT T2.name FROM Certificate AS T1 JOIN Aircraft AS T2 ON T2.aid = T1.aid WHERE T2.distance > 5000 GROUP BY T1.aid ORDER BY COUNT(*) >= 5
what is the salary and name of the employee who has the most number of aircraft certificates?
CREATE TABLE Certificate (eid VARCHAR); CREATE TABLE Employee (name VARCHAR, salary VARCHAR, eid VARCHAR)
SELECT T1.name, T1.salary FROM Employee AS T1 JOIN Certificate AS T2 ON T1.eid = T2.eid GROUP BY T1.eid ORDER BY COUNT(*) DESC LIMIT 1
What is the salary and name of the employee who has the most number of certificates on aircrafts with distance more than 5000?
CREATE TABLE Aircraft (aid VARCHAR, distance INTEGER); CREATE TABLE Employee (name VARCHAR, eid VARCHAR); CREATE TABLE Certificate (eid VARCHAR, aid VARCHAR)
SELECT T1.name FROM Employee AS T1 JOIN Certificate AS T2 ON T1.eid = T2.eid JOIN Aircraft AS T3 ON T3.aid = T2.aid WHERE T3.distance > 5000 GROUP BY T1.eid ORDER BY COUNT(*) DESC LIMIT 1
How many allergies are there?
CREATE TABLE Allergy_type (allergy VARCHAR)
SELECT COUNT(DISTINCT allergy) FROM Allergy_type
How many different allergy types exist?
CREATE TABLE Allergy_type (allergytype VARCHAR)
SELECT COUNT(DISTINCT allergytype) FROM Allergy_type
Show all allergy types.
CREATE TABLE Allergy_type (allergytype VARCHAR)
SELECT DISTINCT allergytype FROM Allergy_type
Show all allergies and their types.
CREATE TABLE Allergy_type (allergy VARCHAR, allergytype VARCHAR)
SELECT allergy, allergytype FROM Allergy_type
Show all allergies with type food.
CREATE TABLE Allergy_type (allergy VARCHAR, allergytype VARCHAR)
SELECT DISTINCT allergy FROM Allergy_type WHERE allergytype = "food"
What is the type of allergy Cat?
CREATE TABLE Allergy_type (allergytype VARCHAR, allergy VARCHAR)
SELECT allergytype FROM Allergy_type WHERE allergy = "Cat"
How many allergies have type animal?
CREATE TABLE Allergy_type (allergytype VARCHAR)
SELECT COUNT(*) FROM Allergy_type WHERE allergytype = "animal"
Show all allergy types and the number of allergies in each type.
CREATE TABLE Allergy_type (allergytype VARCHAR)
SELECT allergytype, COUNT(*) FROM Allergy_type GROUP BY allergytype
Which allergy type has most number of allergies?
CREATE TABLE Allergy_type (allergytype VARCHAR)
SELECT allergytype FROM Allergy_type GROUP BY allergytype ORDER BY COUNT(*) DESC LIMIT 1
Which allergy type has least number of allergies?
CREATE TABLE Allergy_type (allergytype VARCHAR)
SELECT allergytype FROM Allergy_type GROUP BY allergytype ORDER BY COUNT(*) LIMIT 1
How many students are there?
CREATE TABLE Student (Id VARCHAR)
SELECT COUNT(*) FROM Student
Show first name and last name for all students.
CREATE TABLE Student (Fname VARCHAR, Lname VARCHAR)
SELECT Fname, Lname FROM Student
How many different advisors are listed?
CREATE TABLE Student (advisor VARCHAR)
SELECT COUNT(DISTINCT advisor) FROM Student
Show all majors.
CREATE TABLE Student (Major VARCHAR)
SELECT DISTINCT Major FROM Student
Show all cities where students live.
CREATE TABLE Student (city_code VARCHAR)
SELECT DISTINCT city_code FROM Student
Show first name, last name, age for all female students. Their sex is F.
CREATE TABLE Student (Fname VARCHAR, Lname VARCHAR, Age VARCHAR, Sex VARCHAR)
SELECT Fname, Lname, Age FROM Student WHERE Sex = 'F'
Show student ids for all male students.
CREATE TABLE Student (StuID VARCHAR, Sex VARCHAR)
SELECT StuID FROM Student WHERE Sex = 'M'
How many students are age 18?
CREATE TABLE Student (age VARCHAR)
SELECT COUNT(*) FROM Student WHERE age = 18
Show all student ids who are older than 20.
CREATE TABLE Student (StuID VARCHAR, age INTEGER)
SELECT StuID FROM Student WHERE age > 20
Which city does the student whose last name is "Kim" live in?
CREATE TABLE Student (city_code VARCHAR, LName VARCHAR)
SELECT city_code FROM Student WHERE LName = "Kim"
Who is the advisor of student with ID 1004?
CREATE TABLE Student (Advisor VARCHAR, StuID VARCHAR)
SELECT Advisor FROM Student WHERE StuID = 1004
How many students live in HKG or CHI?
CREATE TABLE Student (city_code VARCHAR)
SELECT COUNT(*) FROM Student WHERE city_code = "HKG" OR city_code = "CHI"
Show the minimum, average, and maximum age of all students.
CREATE TABLE Student (age INTEGER)
SELECT MIN(age), AVG(age), MAX(age) FROM Student
What is the last name of the youngest student?
CREATE TABLE Student (LName VARCHAR, age INTEGER)
SELECT LName FROM Student WHERE age = (SELECT MIN(age) FROM Student)
Show the student id of the oldest student.
CREATE TABLE Student (StuID VARCHAR, age INTEGER)
SELECT StuID FROM Student WHERE age = (SELECT MAX(age) FROM Student)
Show all majors and corresponding number of students.
CREATE TABLE Student (major VARCHAR)
SELECT major, COUNT(*) FROM Student GROUP BY major
Which major has most number of students?
CREATE TABLE Student (major VARCHAR)
SELECT major FROM Student GROUP BY major ORDER BY COUNT(*) DESC LIMIT 1
Show all ages and corresponding number of students.
CREATE TABLE Student (age VARCHAR)
SELECT age, COUNT(*) FROM Student GROUP BY age
Show the average age for male and female students.
CREATE TABLE Student (sex VARCHAR, age INTEGER)
SELECT AVG(age), sex FROM Student GROUP BY sex
Show all cities and corresponding number of students.
CREATE TABLE Student (city_code VARCHAR)
SELECT city_code, COUNT(*) FROM Student GROUP BY city_code
Show all advisors and corresponding number of students.
CREATE TABLE Student (advisor VARCHAR)
SELECT advisor, COUNT(*) FROM Student GROUP BY advisor
Which advisor has most number of students?
CREATE TABLE Student (advisor VARCHAR)
SELECT advisor FROM Student GROUP BY advisor ORDER BY COUNT(*) DESC LIMIT 1
How many students have cat allergies?
CREATE TABLE Has_allergy (Allergy VARCHAR)
SELECT COUNT(*) FROM Has_allergy WHERE Allergy = "Cat"
Show all student IDs who have at least two allergies.
CREATE TABLE Has_allergy (StuID VARCHAR)
SELECT StuID FROM Has_allergy GROUP BY StuID HAVING COUNT(*) >= 2
What are the student ids of students who don't have any allergies?
CREATE TABLE Has_allergy (StuID VARCHAR); CREATE TABLE Student (StuID VARCHAR)
SELECT StuID FROM Student EXCEPT SELECT StuID FROM Has_allergy
How many female students have milk or egg allergies?
CREATE TABLE Student (StuID VARCHAR, sex VARCHAR); CREATE TABLE has_allergy (StuID VARCHAR, allergy VARCHAR)
SELECT COUNT(*) FROM has_allergy AS T1 JOIN Student AS T2 ON T1.StuID = T2.StuID WHERE T2.sex = "F" AND T1.allergy = "Milk" OR T1.allergy = "Eggs"
How many students have a food allergy?
CREATE TABLE Has_allergy (allergy VARCHAR); CREATE TABLE Allergy_type (allergy VARCHAR, allergytype VARCHAR)
SELECT COUNT(*) FROM Has_allergy AS T1 JOIN Allergy_type AS T2 ON T1.allergy = T2.allergy WHERE T2.allergytype = "food"
Which allergy has most number of students affected?
CREATE TABLE Has_allergy (Allergy VARCHAR)
SELECT Allergy FROM Has_allergy GROUP BY Allergy ORDER BY COUNT(*) DESC LIMIT 1
Show all allergies with number of students affected.
CREATE TABLE Has_allergy (Allergy VARCHAR)
SELECT Allergy, COUNT(*) FROM Has_allergy GROUP BY Allergy
Show all allergy type with number of students affected.
CREATE TABLE Has_allergy (allergy VARCHAR); CREATE TABLE Allergy_type (allergytype VARCHAR, allergy VARCHAR)
SELECT T2.allergytype, COUNT(*) FROM Has_allergy AS T1 JOIN Allergy_type AS T2 ON T1.allergy = T2.allergy GROUP BY T2.allergytype
Find the last name and age of the student who has allergy to both milk and cat.
CREATE TABLE Has_allergy (lname VARCHAR, age VARCHAR, StuID VARCHAR, Allergy VARCHAR); CREATE TABLE Student (lname VARCHAR, age VARCHAR, StuID VARCHAR, Allergy VARCHAR)
SELECT lname, age FROM Student WHERE StuID IN (SELECT StuID FROM Has_allergy WHERE Allergy = "Milk" INTERSECT SELECT StuID FROM Has_allergy WHERE Allergy = "Cat")
What are the allergies and their types that the student with first name Lisa has? And order the result by name of allergies.
CREATE TABLE Has_allergy (Allergy VARCHAR, StuID VARCHAR); CREATE TABLE Student (StuID VARCHAR, Fname VARCHAR); CREATE TABLE Allergy_type (Allergy VARCHAR, AllergyType VARCHAR)
SELECT T1.Allergy, T1.AllergyType FROM Allergy_type AS T1 JOIN Has_allergy AS T2 ON T1.Allergy = T2.Allergy JOIN Student AS T3 ON T3.StuID = T2.StuID WHERE T3.Fname = "Lisa" ORDER BY T1.Allergy
Find the first name and gender of the student who has allergy to milk but not cat.
CREATE TABLE Student (fname VARCHAR, sex VARCHAR, StuID VARCHAR, Allergy VARCHAR); CREATE TABLE Has_allergy (fname VARCHAR, sex VARCHAR, StuID VARCHAR, Allergy VARCHAR)
SELECT fname, sex FROM Student WHERE StuID IN (SELECT StuID FROM Has_allergy WHERE Allergy = "Milk" EXCEPT SELECT StuID FROM Has_allergy WHERE Allergy = "Cat")
Find the average age of the students who have allergies with food and animal types.
CREATE TABLE Student (age INTEGER, StuID VARCHAR); CREATE TABLE Allergy_Type (Allergy VARCHAR, allergytype VARCHAR); CREATE TABLE Has_allergy (StuID VARCHAR, Allergy VARCHAR)
SELECT AVG(age) FROM Student WHERE StuID IN (SELECT T1.StuID FROM Has_allergy AS T1 JOIN Allergy_Type AS T2 ON T1.Allergy = T2.Allergy WHERE T2.allergytype = "food" INTERSECT SELECT T1.StuID FROM Has_allergy AS T1 JOIN Allergy_Type AS T2 ON T1.Allergy = T2.Allergy WHERE T2.allergytype = "animal")
List the first and last name of the students who do not have any food type allergy.
CREATE TABLE Student (fname VARCHAR, lname VARCHAR, StuID VARCHAR); CREATE TABLE Allergy_Type (Allergy VARCHAR, allergytype VARCHAR); CREATE TABLE Has_allergy (StuID VARCHAR, Allergy VARCHAR)
SELECT fname, lname FROM Student WHERE NOT StuID IN (SELECT T1.StuID FROM Has_allergy AS T1 JOIN Allergy_Type AS T2 ON T1.Allergy = T2.Allergy WHERE T2.allergytype = "food")
Find the number of male (sex is 'M') students who have some food type allery.
CREATE TABLE Has_allergy (Allergy VARCHAR); CREATE TABLE Allergy_Type (Allergy VARCHAR, allergytype VARCHAR); CREATE TABLE Student (sex VARCHAR, StuID VARCHAR)
SELECT COUNT(*) FROM Student WHERE sex = "M" AND StuID IN (SELECT StuID FROM Has_allergy AS T1 JOIN Allergy_Type AS T2 ON T1.Allergy = T2.Allergy WHERE T2.allergytype = "food")
Find the different first names and cities of the students who have allergy to milk or cat.
CREATE TABLE Has_Allergy (stuid VARCHAR, Allergy VARCHAR); CREATE TABLE Student (fname VARCHAR, city_code VARCHAR, stuid VARCHAR)
SELECT DISTINCT T1.fname, T1.city_code FROM Student AS T1 JOIN Has_Allergy AS T2 ON T1.stuid = T2.stuid WHERE T2.Allergy = "Milk" OR T2.Allergy = "Cat"
Find the number of students who are older than 18 and do not have allergy to either food or animal.
CREATE TABLE Allergy_Type (Allergy VARCHAR, allergytype VARCHAR); CREATE TABLE Has_allergy (Allergy VARCHAR); CREATE TABLE Student (age VARCHAR, StuID VARCHAR)
SELECT COUNT(*) FROM Student WHERE age > 18 AND NOT StuID IN (SELECT StuID FROM Has_allergy AS T1 JOIN Allergy_Type AS T2 ON T1.Allergy = T2.Allergy WHERE T2.allergytype = "food" OR T2.allergytype = "animal")
Find the first name and major of the students who are not allegry to soy.
CREATE TABLE Has_allergy (fname VARCHAR, major VARCHAR, StuID VARCHAR, Allergy VARCHAR); CREATE TABLE Student (fname VARCHAR, major VARCHAR, StuID VARCHAR, Allergy VARCHAR)
SELECT fname, major FROM Student WHERE NOT StuID IN (SELECT StuID FROM Has_allergy WHERE Allergy = "Soy")