lianghsun/bird-text2sql-bench huggingface dataset:
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
What is the description of the low risk violation of Tiramisu Kitchen on 2014/1/14?

### Hint
Tiramisu Kitchen is the name of the business; 2014/1/14 refers to date = '2014-01-14'; low risk violations refer to risk_category = 'Low Risk';

### SQL
SELECT T1.description FROM violations AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T1.`date` = '2014-01-14' AND T2.name = 'Tiramisu Kitchen' AND T1.risk_category = 'Low Risk'
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
Please list the descriptions of all the high risk violations of Tiramisu Kitchen.

### Hint
Tiramisu Kitchen is the name of the business; high risk violations refer to risk_category = 'High Risk';

### SQL
SELECT DISTINCT T1.description FROM violations AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T1.risk_category = 'High Risk' AND T2.name = 'Tiramisu Kitchen'
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
How many routine inspections did Tiramisu Kitchen have?

### Hint
Tiramisu Kitchen is the name of the business; routine inspections refer to type = 'Routine - Unscheduled';

### SQL
SELECT COUNT(T1.business_id) FROM inspections AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T1.type = 'Routine - Unscheduled' AND T2.name = 'Tiramisu Kitchen'
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
Among the routine inspections of Tiramisu Kitchen, how many of them have a score of over 70?

### Hint
Tiramisu Kitchen is the name of the business; routine inspections refer to type = 'Routine - Unscheduled'; score of over 70 refers to score > 70;

### SQL
SELECT COUNT(T2.business_id) FROM inspections AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T2.name = 'Tiramisu Kitchen' AND T1.type = 'Routine - Unscheduled' AND T1.score > 70
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
Which restaurant had more low risk violation in inspections, Tiramisu Kitchen or OMNI S.F. Hotel - 2nd Floor Pantry?

### Hint
Tiramisu Kitchen and OMNI S.F. Hotel - 2nd Floor Pantry are names of the business; more low risk violations refer to MAX(COUNT(risk_category = 'Low Risk'));

### SQL
SELECT CASE WHEN SUM(CASE WHEN T2.name = 'OMNI S.F. Hotel - 2nd Floor Pantry' THEN 1 ELSE 0 END) > SUM(CASE WHEN T2.name = 'Tiramisu Kitchen' THEN 1 ELSE 0 END) THEN 'OMNI S.F. Hotel - 2nd Floor Pantry' ELSE 'Tiramisu Kitchen' END AS result FROM violations AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T1.risk_category = 'Low Risk'
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
How many high risk violations do the restaurants in San Francisco have in total?

### Hint
restaurants in San Francisco refer to business_id where city in ('San Francisco', 'SF', 'S.F.', 'SAN FRANCISCO'); high risk violations refer to risk_category = 'High Risk';

### SQL
SELECT COUNT(T2.business_id) FROM violations AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T2.city IN ('San Francisco', 'SF', 'S.F.', 'SAN FRANCISCO') AND T1.risk_category = 'High Risk'
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
Which restaurant has the highest total number of high risk violations?

### Hint
the highest total number of high risk violations refer to MAX(COUNT(risk_category = 'High Risk'));

### SQL
SELECT T2.name FROM violations AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T1.risk_category = 'High Risk' GROUP BY T2.name ORDER BY COUNT(T2.name) DESC LIMIT 1
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
What is the average scores of Tiramisu Kitchen in all inspections?

### Hint
avg(score);

### SQL
SELECT AVG(T1.score) FROM inspections AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T2.name = 'Tiramisu Kitchen'
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
Which business had the most number of inspections? Give the Id number for that business.

### Hint
the most number of inspections refers to MAX(COUNT(business_id)); Id number for that business refers to business_id;

### SQL
SELECT business_id FROM inspections GROUP BY business_id ORDER BY COUNT(business_id) DESC LIMIT 1
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
Tell the Id number of the business with the most number of violations.

### Hint
Id number for that business refers to business_id; the most number of violations refers to MAX(COUNT(business_id));

### SQL
SELECT business_id FROM violations GROUP BY business_id ORDER BY COUNT(business_id) DESC LIMIT 1
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
Give the name of the business which met all the required standards during the unscheduled routine inspection on 2016/9/28.

### Hint
met all the required standards refers to score = 100; unscheduled routine inspection on 2016/9/28 refers to type = 'Routine - Unscheduled' where date = '2016-09-28';

### SQL
SELECT T2.name FROM inspections AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T1.score = 100 AND T1.`date` = '2016-09-28' AND T1.type = 'Routine - Unscheduled'
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
Which business had the most number of high risk violations? Give the name of the business.

### Hint
the most number of high risk violations refers to MAX(COUNT(business_id)) where risk_category = 'High';

### SQL
SELECT T2.name FROM violations AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T1.risk_category = 'High Risk' GROUP BY T2.name ORDER BY COUNT(T2.name) DESC LIMIT 1
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
How many kinds of violations did "Stacks Restaurant" have on 2016/10/4?

### Hint
"Stacks Restaurant" is the name of the business; date = '2016-10-04';

### SQL
SELECT COUNT(DISTINCT T1.violation_type_id) FROM violations AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T2.name = 'Stacks Restaurant' AND T1.`date` = '2016-10-04'
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
Give the description of the moderate risk violation which "Chez Fayala, Inc." had on 2016/7/1.

### Hint
"Chez Fayala, Inc." is the name of the business; moderate risk violation refers to risk_category = 'Moderate Risk'; date = '2016-07-01';

### SQL
SELECT T1.description FROM violations AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE T2.name = 'Chez Fayala, Inc.' AND T1.`date` = '2016-07-01' AND T1.risk_category = 'Moderate Risk'
### Database Schema
CREATE TABLE `businesses` (
  `business_id` INTEGER NOT NULL,
  `name` TEXT NOT NULL,
  `address` TEXT DEFAULT NULL,
  `city` TEXT DEFAULT NULL,
  `postal_code` TEXT DEFAULT NULL,
  `latitude` REAL DEFAULT NULL,
  `longitude` REAL DEFAULT NULL,
  `phone_number` INTEGER DEFAULT NULL,
  `tax_code` TEXT DEFAULT NULL,
  `business_certificate` INTEGER NOT NULL,
  `application_date` DATE DEFAULT NULL,
  `owner_name` TEXT NOT NULL,
  `owner_address` TEXT DEFAULT NULL,
  `owner_city` TEXT DEFAULT NULL,
  `owner_state` TEXT DEFAULT NULL,
  `owner_zip` TEXT DEFAULT NULL,
  PRIMARY KEY (`business_id`)
)

CREATE TABLE `inspections` (
  `business_id` INTEGER NOT NULL,
  `score` INTEGER DEFAULT NULL,
  `date` DATE NOT NULL,
  `type` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

CREATE TABLE `violations` (
  `business_id` INTEGER NOT NULL,
  `date` DATE NOT NULL,
  `violation_type_id` TEXT NOT NULL,
  `risk_category` TEXT NOT NULL,
  `description` TEXT NOT NULL,
  FOREIGN KEY (`business_id`) REFERENCES `businesses` (`business_id`)
)

### Question
Which business had the lowest score for the unscheduled routine inspection on 2016/9/26? Give the name of the business.

### Hint
the lowest score for unscheduled routine inspection refers to type = 'Routine - Unscheduled' where MIN(score); date = '2016-09-26';

### SQL
SELECT T2.name FROM inspections AS T1 INNER JOIN businesses AS T2 ON T1.business_id = T2.business_id WHERE score = ( SELECT MIN(score) FROM inspections WHERE `date` = '2016-09-26' AND type = 'Routine - Unscheduled' ) AND T1.`date` = '2016-09-26' AND T1.type = 'Routine - Unscheduled'