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
Among the facilities that have undergone at least one inspection in 2010, how many of them are restaurants or cafeterias?

### Hint
in 2010 refers to inspection_date like '2010%'; restaurant or cafeteria refers to facility_type = 'Restaurant'

### SQL
SELECT COUNT(DISTINCT T1.license_no) FROM inspection AS T1 INNER JOIN establishment AS T2 ON T1.license_no = T2.license_no WHERE strftime('%Y', T1.inspection_date) = '2010' AND T2.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
Please list the location coordinates of all the facilities that had an inspection on 2010/5/11.

### Hint
location coordinates refers to latitude, longitude; on 2010/5/11 refers to inspection_date = '2010-05-11'

### SQL
SELECT DISTINCT T2.latitude, T2.longitude FROM inspection AS T1 INNER JOIN establishment AS T2 ON T1.license_no = T2.license_no WHERE T1.inspection_date = '2010-05-11'
### 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 facilities that have undergone at least one inspection in 2010, how many of them are in ward no.42?

### Hint
in 2010 refers to inspection_date like '2010%'; in ward no.42 refers to ward = 42

### SQL
SELECT COUNT(DISTINCT T1.license_no) FROM inspection AS T1 INNER JOIN establishment AS T2 ON T1.license_no = T2.license_no WHERE strftime('%Y', T1.inspection_date) = '2010' AND T2.ward = 42
### 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
Please list the full names of all the sanitarians who have inspected the facility Burbank.

### Hint
full name refers to first_name, last_name; the facility Burbank refers to dba_name = 'Burbank'

### SQL
SELECT DISTINCT T1.first_name, T1.last_name 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 T3.dba_name = 'Burbank' AND T1.title = 'Sanitarian'
### 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
Please list the assumed name of all the facilities that failed an inspection in 2010.

### Hint
assumed name refers to dba_name; failed an inspection refers to results = 'Fail'; in 2010 refers to inspection_date like '2010%'

### SQL
SELECT DISTINCT T2.dba_name FROM inspection AS T1 INNER JOIN establishment AS T2 ON T1.license_no = T2.license_no WHERE T1.results = 'Fail' AND strftime('%Y', T1.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 full name of the sanitarian who inspected Amundsen High School on 2010/5/11?

### Hint
full name refers to first_name, last_name;  Amundsen High School refers to dba_name = 'AMUNDSEN HIGH SCHOOL'; on 2010/5/11 refers to inspection_date = '2010-05-11'

### SQL
SELECT T1.first_name, T1.last_name 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 T2.inspection_date = '2010-05-11' AND T3.dba_name = 'AMUNDSEN HIGH SCHOOL' AND T1.title = 'Sanitarian'
### 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 inspections done by sanitarian Joshua Rosa, how many of them have the result of "pass"?

### Hint
have the result of "pass" refers to results = 'Pass'

### SQL
SELECT COUNT(T2.inspection_id) FROM employee AS T1 INNER JOIN inspection AS T2 ON T1.employee_id = T2.employee_id WHERE T2.results = 'Pass' AND T1.first_name = 'Joshua' AND T1.last_name = 'Rosa'
### 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
After Azha Restaurant Inc. passed the inspection on 2010/1/21, when was the follow-up inspection done?

### Hint
Azha Restaurant Inc. refers to dba_name = 'Azha Restaurant Inc.'; on 2010/1/21 refers to inspection_date = '2010-01-21'; follow-up inspection date refers to followup_to

### SQL
SELECT T1.followup_to FROM inspection AS T1 INNER JOIN establishment AS T2 ON T1.license_no = T2.license_no WHERE T2.dba_name = 'Azha Restaurant Inc.' AND T1.results = 'Pass' AND T1.inspection_date = '2010-01-21'
### 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 facilities that had undergone at least one inspection in 2010, how many of them have the most serious food safety issues?

### Hint
in 2010 refers to inspection_date like '2010%'; the most serious food safety issues refers to risk_level = 3

### SQL
SELECT COUNT(DISTINCT T2.license_no) FROM inspection AS T1 INNER JOIN establishment AS T2 ON T1.license_no = T2.license_no WHERE strftime('%Y', T1.inspection_date) = '2010' AND T2.risk_level = 3
### 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 average number of inspections carried out in the year 2010 by a sanitarian whose salary is over 70000?

### Hint
in the year 2010 refers to inspection_date like '2010%'; salary is over 70000 refers to salary > 70000; average number = divide(sum(inspection where inspection_date like '2010%'), sum(employee_id where salary > 70000))

### SQL
SELECT CAST(SUM(CASE WHEN T2.inspection_date LIKE '2010%' THEN 1 ELSE 0 END) AS REAL) / SUM(CASE WHEN T1.salary > 70000 THEN 1 ELSE 0 END) FROM employee AS T1 INNER JOIN inspection AS T2 ON T1.employee_id = T2.employee_id
### 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 point level of "Refrigeration and metal stem thermometers provided and conspicuous"?

### Hint
"Refrigeration and metal stem thermometers provided and conspicuous" refers to Description = 'Refrigeration and metal stem thermometers provided and conspicuous '

### SQL
SELECT point_level FROM inspection_point WHERE Description = 'Refrigeration and metal stem thermometers provided and conspicuous '