lianghsun/bird-text2sql-bench huggingface dataset:
### Database Schema
CREATE TABLE employee
(
    employee_id INTEGER
            primary key,
    first_name  TEXT,
    last_name   TEXT,
    address     TEXT,
    city        TEXT,
    state       TEXT,
    zip         INTEGER,
    phone       TEXT,
    title       TEXT,
    salary      INTEGER,
    supervisor  INTEGER,
    foreign key (supervisor) references employee(employee_id)
)

CREATE TABLE establishment
(
    license_no    INTEGER
            primary key,
    dba_name      TEXT,
    aka_name      TEXT,
    facility_type TEXT,
    risk_level    INTEGER,
    address       TEXT,
    city          TEXT,
    state         TEXT,
    zip           INTEGER,
    latitude      REAL,
    longitude     REAL,
    ward          INTEGER
)

CREATE TABLE inspection
(
    inspection_id   INTEGER
            primary key,
    inspection_date DATE,
    inspection_type TEXT,
    results         TEXT,
    employee_id     INTEGER,
    license_no      INTEGER,
    followup_to     INTEGER,
    foreign key (employee_id) references employee(employee_id),
    foreign key (license_no) references establishment(license_no),
    foreign key (followup_to) references inspection(inspection_id)
)

CREATE TABLE inspection_point
(
    point_id    INTEGER
            primary key,
    Description TEXT,
    category    TEXT,
    code        TEXT,
    fine        INTEGER,
    point_level TEXT
)

CREATE TABLE violation
(
    inspection_id     INTEGER,
    point_id          INTEGER,
    fine              INTEGER,
    inspector_comment TEXT,
    primary key (inspection_id, point_id),
    foreign key (inspection_id) references inspection(inspection_id),
    foreign key (point_id) references inspection_point(point_id)
)

### Question
What is the risk level of the establishment that Bob Benson inspected?

### SQL
SELECT DISTINCT T3.risk_level FROM employee AS T1 INNER JOIN inspection AS T2 ON T1.employee_id = T2.employee_id INNER JOIN establishment AS T3 ON T2.license_no = T3.license_no WHERE T1.first_name = 'Bob' AND T1.last_name = 'Benson'
### Database Schema
CREATE TABLE employee
(
    employee_id INTEGER
            primary key,
    first_name  TEXT,
    last_name   TEXT,
    address     TEXT,
    city        TEXT,
    state       TEXT,
    zip         INTEGER,
    phone       TEXT,
    title       TEXT,
    salary      INTEGER,
    supervisor  INTEGER,
    foreign key (supervisor) references employee(employee_id)
)

CREATE TABLE establishment
(
    license_no    INTEGER
            primary key,
    dba_name      TEXT,
    aka_name      TEXT,
    facility_type TEXT,
    risk_level    INTEGER,
    address       TEXT,
    city          TEXT,
    state         TEXT,
    zip           INTEGER,
    latitude      REAL,
    longitude     REAL,
    ward          INTEGER
)

CREATE TABLE inspection
(
    inspection_id   INTEGER
            primary key,
    inspection_date DATE,
    inspection_type TEXT,
    results         TEXT,
    employee_id     INTEGER,
    license_no      INTEGER,
    followup_to     INTEGER,
    foreign key (employee_id) references employee(employee_id),
    foreign key (license_no) references establishment(license_no),
    foreign key (followup_to) references inspection(inspection_id)
)

CREATE TABLE inspection_point
(
    point_id    INTEGER
            primary key,
    Description TEXT,
    category    TEXT,
    code        TEXT,
    fine        INTEGER,
    point_level TEXT
)

CREATE TABLE violation
(
    inspection_id     INTEGER,
    point_id          INTEGER,
    fine              INTEGER,
    inspector_comment TEXT,
    primary key (inspection_id, point_id),
    foreign key (inspection_id) references inspection(inspection_id),
    foreign key (point_id) references inspection_point(point_id)
)

### Question
Which establishments did Bob Benson inspect in 2010 and what was the results?

### Hint
establishment name refers to dba_name; in 2010 refers to inspection_date like '2010%'

### SQL
SELECT DISTINCT T3.dba_name, T2.results FROM employee AS T1 INNER JOIN inspection AS T2 ON T1.employee_id = T2.employee_id INNER JOIN establishment AS T3 ON T2.license_no = T3.license_no WHERE T1.first_name = 'Bob' AND T1.last_name = 'Benson' AND strftime('%Y', T2.inspection_date) = '2010'
### Database Schema
CREATE TABLE employee
(
    employee_id INTEGER
            primary key,
    first_name  TEXT,
    last_name   TEXT,
    address     TEXT,
    city        TEXT,
    state       TEXT,
    zip         INTEGER,
    phone       TEXT,
    title       TEXT,
    salary      INTEGER,
    supervisor  INTEGER,
    foreign key (supervisor) references employee(employee_id)
)

CREATE TABLE establishment
(
    license_no    INTEGER
            primary key,
    dba_name      TEXT,
    aka_name      TEXT,
    facility_type TEXT,
    risk_level    INTEGER,
    address       TEXT,
    city          TEXT,
    state         TEXT,
    zip           INTEGER,
    latitude      REAL,
    longitude     REAL,
    ward          INTEGER
)

CREATE TABLE inspection
(
    inspection_id   INTEGER
            primary key,
    inspection_date DATE,
    inspection_type TEXT,
    results         TEXT,
    employee_id     INTEGER,
    license_no      INTEGER,
    followup_to     INTEGER,
    foreign key (employee_id) references employee(employee_id),
    foreign key (license_no) references establishment(license_no),
    foreign key (followup_to) references inspection(inspection_id)
)

CREATE TABLE inspection_point
(
    point_id    INTEGER
            primary key,
    Description TEXT,
    category    TEXT,
    code        TEXT,
    fine        INTEGER,
    point_level TEXT
)

CREATE TABLE violation
(
    inspection_id     INTEGER,
    point_id          INTEGER,
    fine              INTEGER,
    inspector_comment TEXT,
    primary key (inspection_id, point_id),
    foreign key (inspection_id) references inspection(inspection_id),
    foreign key (point_id) references inspection_point(point_id)
)

### Question
What is the title of the employee that inspected the establishment with license number 1576687?

### Hint
license number 1576687 refers to license_no = 1576687

### SQL
SELECT DISTINCT T1.title FROM employee AS T1 INNER JOIN inspection AS T2 ON T1.employee_id = T2.employee_id WHERE T2.license_no = 1576687
### Database Schema
CREATE TABLE employee
(
    employee_id INTEGER
            primary key,
    first_name  TEXT,
    last_name   TEXT,
    address     TEXT,
    city        TEXT,
    state       TEXT,
    zip         INTEGER,
    phone       TEXT,
    title       TEXT,
    salary      INTEGER,
    supervisor  INTEGER,
    foreign key (supervisor) references employee(employee_id)
)

CREATE TABLE establishment
(
    license_no    INTEGER
            primary key,
    dba_name      TEXT,
    aka_name      TEXT,
    facility_type TEXT,
    risk_level    INTEGER,
    address       TEXT,
    city          TEXT,
    state         TEXT,
    zip           INTEGER,
    latitude      REAL,
    longitude     REAL,
    ward          INTEGER
)

CREATE TABLE inspection
(
    inspection_id   INTEGER
            primary key,
    inspection_date DATE,
    inspection_type TEXT,
    results         TEXT,
    employee_id     INTEGER,
    license_no      INTEGER,
    followup_to     INTEGER,
    foreign key (employee_id) references employee(employee_id),
    foreign key (license_no) references establishment(license_no),
    foreign key (followup_to) references inspection(inspection_id)
)

CREATE TABLE inspection_point
(
    point_id    INTEGER
            primary key,
    Description TEXT,
    category    TEXT,
    code        TEXT,
    fine        INTEGER,
    point_level TEXT
)

CREATE TABLE violation
(
    inspection_id     INTEGER,
    point_id          INTEGER,
    fine              INTEGER,
    inspector_comment TEXT,
    primary key (inspection_id, point_id),
    foreign key (inspection_id) references inspection(inspection_id),
    foreign key (point_id) references inspection_point(point_id)
)

### Question
How many inspection points with serious point level that have no fine?

### Hint
serious point level refers to point_level = 'Serious '; have no fine refers to fine = 0

### SQL
SELECT COUNT(DISTINCT T2.point_id) FROM inspection_point AS T1 INNER JOIN violation AS T2 ON T1.point_id = T2.point_id WHERE T1.point_level = 'Serious ' AND T2.fine = 0
### Database Schema
CREATE TABLE employee
(
    employee_id INTEGER
            primary key,
    first_name  TEXT,
    last_name   TEXT,
    address     TEXT,
    city        TEXT,
    state       TEXT,
    zip         INTEGER,
    phone       TEXT,
    title       TEXT,
    salary      INTEGER,
    supervisor  INTEGER,
    foreign key (supervisor) references employee(employee_id)
)

CREATE TABLE establishment
(
    license_no    INTEGER
            primary key,
    dba_name      TEXT,
    aka_name      TEXT,
    facility_type TEXT,
    risk_level    INTEGER,
    address       TEXT,
    city          TEXT,
    state         TEXT,
    zip           INTEGER,
    latitude      REAL,
    longitude     REAL,
    ward          INTEGER
)

CREATE TABLE inspection
(
    inspection_id   INTEGER
            primary key,
    inspection_date DATE,
    inspection_type TEXT,
    results         TEXT,
    employee_id     INTEGER,
    license_no      INTEGER,
    followup_to     INTEGER,
    foreign key (employee_id) references employee(employee_id),
    foreign key (license_no) references establishment(license_no),
    foreign key (followup_to) references inspection(inspection_id)
)

CREATE TABLE inspection_point
(
    point_id    INTEGER
            primary key,
    Description TEXT,
    category    TEXT,
    code        TEXT,
    fine        INTEGER,
    point_level TEXT
)

CREATE TABLE violation
(
    inspection_id     INTEGER,
    point_id          INTEGER,
    fine              INTEGER,
    inspector_comment TEXT,
    primary key (inspection_id, point_id),
    foreign key (inspection_id) references inspection(inspection_id),
    foreign key (point_id) references inspection_point(point_id)
)

### Question
What is the percentage of restaurants that paid a fine of 250 among all establishments?

### Hint
a fine of 250 refers to fine = 250; percentage = divide(sum(license_no where fine = 250), count(license_no)) * 100%

### SQL
SELECT CAST(COUNT(CASE WHEN T3.fine = 250 THEN T1.license_no END) AS REAL) * 100 / COUNT(T1.license_no) FROM establishment AS T1 INNER JOIN inspection AS T2 ON T1.license_no = T2.license_no INNER JOIN violation AS T3 ON T2.inspection_id = T3.inspection_id WHERE T1.facility_type = 'Restaurant'
### Database Schema
CREATE TABLE employee
(
    employee_id INTEGER
            primary key,
    first_name  TEXT,
    last_name   TEXT,
    address     TEXT,
    city        TEXT,
    state       TEXT,
    zip         INTEGER,
    phone       TEXT,
    title       TEXT,
    salary      INTEGER,
    supervisor  INTEGER,
    foreign key (supervisor) references employee(employee_id)
)

CREATE TABLE establishment
(
    license_no    INTEGER
            primary key,
    dba_name      TEXT,
    aka_name      TEXT,
    facility_type TEXT,
    risk_level    INTEGER,
    address       TEXT,
    city          TEXT,
    state         TEXT,
    zip           INTEGER,
    latitude      REAL,
    longitude     REAL,
    ward          INTEGER
)

CREATE TABLE inspection
(
    inspection_id   INTEGER
            primary key,
    inspection_date DATE,
    inspection_type TEXT,
    results         TEXT,
    employee_id     INTEGER,
    license_no      INTEGER,
    followup_to     INTEGER,
    foreign key (employee_id) references employee(employee_id),
    foreign key (license_no) references establishment(license_no),
    foreign key (followup_to) references inspection(inspection_id)
)

CREATE TABLE inspection_point
(
    point_id    INTEGER
            primary key,
    Description TEXT,
    category    TEXT,
    code        TEXT,
    fine        INTEGER,
    point_level TEXT
)

CREATE TABLE violation
(
    inspection_id     INTEGER,
    point_id          INTEGER,
    fine              INTEGER,
    inspector_comment TEXT,
    primary key (inspection_id, point_id),
    foreign key (inspection_id) references inspection(inspection_id),
    foreign key (point_id) references inspection_point(point_id)
)

### Question
What is the percentage of establishments with a risk level of 1 among all of the establishments that passed the inspection?

### Hint
a risk level of 1 refers to risk_level = 1; pass the inspection refers to results = 'Pass'; percentage = divide(sum(license_no where risk_level = 1), count(license_no)) * 100% where results = 'Pass'

### SQL
SELECT CAST(COUNT(CASE WHEN T1.risk_level = 1 THEN T1.license_no END) AS REAL) * 100 / COUNT(T1.license_no) FROM establishment AS T1 INNER JOIN inspection AS T2 ON T1.license_no = T2.license_no WHERE T2.results = 'Pass'
### Database Schema
CREATE TABLE employee
(
    employee_id INTEGER
            primary key,
    first_name  TEXT,
    last_name   TEXT,
    address     TEXT,
    city        TEXT,
    state       TEXT,
    zip         INTEGER,
    phone       TEXT,
    title       TEXT,
    salary      INTEGER,
    supervisor  INTEGER,
    foreign key (supervisor) references employee(employee_id)
)

CREATE TABLE establishment
(
    license_no    INTEGER
            primary key,
    dba_name      TEXT,
    aka_name      TEXT,
    facility_type TEXT,
    risk_level    INTEGER,
    address       TEXT,
    city          TEXT,
    state         TEXT,
    zip           INTEGER,
    latitude      REAL,
    longitude     REAL,
    ward          INTEGER
)

CREATE TABLE inspection
(
    inspection_id   INTEGER
            primary key,
    inspection_date DATE,
    inspection_type TEXT,
    results         TEXT,
    employee_id     INTEGER,
    license_no      INTEGER,
    followup_to     INTEGER,
    foreign key (employee_id) references employee(employee_id),
    foreign key (license_no) references establishment(license_no),
    foreign key (followup_to) references inspection(inspection_id)
)

CREATE TABLE inspection_point
(
    point_id    INTEGER
            primary key,
    Description TEXT,
    category    TEXT,
    code        TEXT,
    fine        INTEGER,
    point_level TEXT
)

CREATE TABLE violation
(
    inspection_id     INTEGER,
    point_id          INTEGER,
    fine              INTEGER,
    inspector_comment TEXT,
    primary key (inspection_id, point_id),
    foreign key (inspection_id) references inspection(inspection_id),
    foreign key (point_id) references inspection_point(point_id)
)

### Question
Where does the employee named "Standard Murray" live?

### Hint
address refers to address, city, state

### SQL
SELECT address, city, state FROM employee WHERE first_name = 'Standard' AND last_name = 'Murray'
### Database Schema
CREATE TABLE employee
(
    employee_id INTEGER
            primary key,
    first_name  TEXT,
    last_name   TEXT,
    address     TEXT,
    city        TEXT,
    state       TEXT,
    zip         INTEGER,
    phone       TEXT,
    title       TEXT,
    salary      INTEGER,
    supervisor  INTEGER,
    foreign key (supervisor) references employee(employee_id)
)

CREATE TABLE establishment
(
    license_no    INTEGER
            primary key,
    dba_name      TEXT,
    aka_name      TEXT,
    facility_type TEXT,
    risk_level    INTEGER,
    address       TEXT,
    city          TEXT,
    state         TEXT,
    zip           INTEGER,
    latitude      REAL,
    longitude     REAL,
    ward          INTEGER
)

CREATE TABLE inspection
(
    inspection_id   INTEGER
            primary key,
    inspection_date DATE,
    inspection_type TEXT,
    results         TEXT,
    employee_id     INTEGER,
    license_no      INTEGER,
    followup_to     INTEGER,
    foreign key (employee_id) references employee(employee_id),
    foreign key (license_no) references establishment(license_no),
    foreign key (followup_to) references inspection(inspection_id)
)

CREATE TABLE inspection_point
(
    point_id    INTEGER
            primary key,
    Description TEXT,
    category    TEXT,
    code        TEXT,
    fine        INTEGER,
    point_level TEXT
)

CREATE TABLE violation
(
    inspection_id     INTEGER,
    point_id          INTEGER,
    fine              INTEGER,
    inspector_comment TEXT,
    primary key (inspection_id, point_id),
    foreign key (inspection_id) references inspection(inspection_id),
    foreign key (point_id) references inspection_point(point_id)
)

### Question
What is the facility type of the establishment named "Kinetic Playground"?

### Hint
the establishment named "Kinetic Playground" refers to dba_name = 'Kinetic Playground'

### SQL
SELECT facility_type FROM establishment WHERE dba_name = 'Kinetic Playground'
### Database Schema
CREATE TABLE employee
(
    employee_id INTEGER
            primary key,
    first_name  TEXT,
    last_name   TEXT,
    address     TEXT,
    city        TEXT,
    state       TEXT,
    zip         INTEGER,
    phone       TEXT,
    title       TEXT,
    salary      INTEGER,
    supervisor  INTEGER,
    foreign key (supervisor) references employee(employee_id)
)

CREATE TABLE establishment
(
    license_no    INTEGER
            primary key,
    dba_name      TEXT,
    aka_name      TEXT,
    facility_type TEXT,
    risk_level    INTEGER,
    address       TEXT,
    city          TEXT,
    state         TEXT,
    zip           INTEGER,
    latitude      REAL,
    longitude     REAL,
    ward          INTEGER
)

CREATE TABLE inspection
(
    inspection_id   INTEGER
            primary key,
    inspection_date DATE,
    inspection_type TEXT,
    results         TEXT,
    employee_id     INTEGER,
    license_no      INTEGER,
    followup_to     INTEGER,
    foreign key (employee_id) references employee(employee_id),
    foreign key (license_no) references establishment(license_no),
    foreign key (followup_to) references inspection(inspection_id)
)

CREATE TABLE inspection_point
(
    point_id    INTEGER
            primary key,
    Description TEXT,
    category    TEXT,
    code        TEXT,
    fine        INTEGER,
    point_level TEXT
)

CREATE TABLE violation
(
    inspection_id     INTEGER,
    point_id          INTEGER,
    fine              INTEGER,
    inspector_comment TEXT,
    primary key (inspection_id, point_id),
    foreign key (inspection_id) references inspection(inspection_id),
    foreign key (point_id) references inspection_point(point_id)
)

### Question
How much salary does Jessica Anthony receive?

### SQL
SELECT salary FROM employee WHERE first_name = 'Jessica' AND last_name = 'Anthony'
### Database Schema
CREATE TABLE employee
(
    employee_id INTEGER
            primary key,
    first_name  TEXT,
    last_name   TEXT,
    address     TEXT,
    city        TEXT,
    state       TEXT,
    zip         INTEGER,
    phone       TEXT,
    title       TEXT,
    salary      INTEGER,
    supervisor  INTEGER,
    foreign key (supervisor) references employee(employee_id)
)

CREATE TABLE establishment
(
    license_no    INTEGER
            primary key,
    dba_name      TEXT,
    aka_name      TEXT,
    facility_type TEXT,
    risk_level    INTEGER,
    address       TEXT,
    city          TEXT,
    state         TEXT,
    zip           INTEGER,
    latitude      REAL,
    longitude     REAL,
    ward          INTEGER
)

CREATE TABLE inspection
(
    inspection_id   INTEGER
            primary key,
    inspection_date DATE,
    inspection_type TEXT,
    results         TEXT,
    employee_id     INTEGER,
    license_no      INTEGER,
    followup_to     INTEGER,
    foreign key (employee_id) references employee(employee_id),
    foreign key (license_no) references establishment(license_no),
    foreign key (followup_to) references inspection(inspection_id)
)

CREATE TABLE inspection_point
(
    point_id    INTEGER
            primary key,
    Description TEXT,
    category    TEXT,
    code        TEXT,
    fine        INTEGER,
    point_level TEXT
)

CREATE TABLE violation
(
    inspection_id     INTEGER,
    point_id          INTEGER,
    fine              INTEGER,
    inspector_comment TEXT,
    primary key (inspection_id, point_id),
    foreign key (inspection_id) references inspection(inspection_id),
    foreign key (point_id) references inspection_point(point_id)
)

### Question
What is the restaurant's name at "41.9532864854" latitude and "-87.7673790701422" longitude?

### Hint
restaurant refers to facility_type = 'Restaurant'; name refers to dba_name; "41.9532864854" latitude and "-87.7673790701422" longitude refers to latitude = 41.9532864854 and longitude = -87.7673790701422

### SQL
SELECT dba_name FROM establishment WHERE latitude = 41.9532864854 AND longitude = -87.7673790701422 AND facility_type = 'Restaurant'
### Database Schema
CREATE TABLE employee
(
    employee_id INTEGER
            primary key,
    first_name  TEXT,
    last_name   TEXT,
    address     TEXT,
    city        TEXT,
    state       TEXT,
    zip         INTEGER,
    phone       TEXT,
    title       TEXT,
    salary      INTEGER,
    supervisor  INTEGER,
    foreign key (supervisor) references employee(employee_id)
)

CREATE TABLE establishment
(
    license_no    INTEGER
            primary key,
    dba_name      TEXT,
    aka_name      TEXT,
    facility_type TEXT,
    risk_level    INTEGER,
    address       TEXT,
    city          TEXT,
    state         TEXT,
    zip           INTEGER,
    latitude      REAL,
    longitude     REAL,
    ward          INTEGER
)

CREATE TABLE inspection
(
    inspection_id   INTEGER
            primary key,
    inspection_date DATE,
    inspection_type TEXT,
    results         TEXT,
    employee_id     INTEGER,
    license_no      INTEGER,
    followup_to     INTEGER,
    foreign key (employee_id) references employee(employee_id),
    foreign key (license_no) references establishment(license_no),
    foreign key (followup_to) references inspection(inspection_id)
)

CREATE TABLE inspection_point
(
    point_id    INTEGER
            primary key,
    Description TEXT,
    category    TEXT,
    code        TEXT,
    fine        INTEGER,
    point_level TEXT
)

CREATE TABLE violation
(
    inspection_id     INTEGER,
    point_id          INTEGER,
    fine              INTEGER,
    inspector_comment TEXT,
    primary key (inspection_id, point_id),
    foreign key (inspection_id) references inspection(inspection_id),
    foreign key (point_id) references inspection_point(point_id)
)

### Question
Among the list of employees, what is the total number of supervisors?

### Hint
supervisor refers to title = 'Supervisor'

### SQL
SELECT COUNT(employee_id) FROM employee WHERE title = 'Supervisor'