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
Where in Chicago does the restaurant named "Old Timers Rest & Lounge" located?

### Hint
in Chicago refers to city = 'CHICAGO'; restaurant refers to facility_type = 'Restaurant'; "Old Timers Rest & Lounge" refers to dba_name = 'OLD TIMERS REST & LOUNGE'; location refers to address

### SQL
SELECT address FROM establishment WHERE city = 'CHICAGO' AND dba_name = 'OLD TIMERS REST & LOUNGE' 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
How many employees are living in Hoffman Estates, IL?

### Hint
in Hoffman Estates refers to city = 'Hoffman Estates'; IL refers to state = 'IL'

### SQL
SELECT COUNT(employee_id) FROM employee WHERE state = 'IL' AND city = 'Hoffman Estates'
### 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 total number of establishments with the highest risk level?

### Hint
total number of establishments with the highest risk level = count(max(risk_level))

### SQL
SELECT COUNT(license_no) FROM establishment WHERE 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
Who is the employee that receives 82700 as their salary?

### Hint
employee name refers to first_name, last_name; receives 82700 as salary refers to salary = 82700

### SQL
SELECT first_name, last_name FROM employee WHERE salary = 82700
### 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
Provide the last name of the employee involved in the inspection ID 52256.

### SQL
SELECT DISTINCT T1.last_name FROM employee AS T1 INNER JOIN inspection AS T2 ON T1.employee_id = T2.employee_id WHERE T2.inspection_id = 52256
### 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 names of taverns that paid a $100 fine upon inspection.

### Hint
name refers to dba_name; tavern refers to facility_type = 'Tavern'; a $100 fine refers to fine = 100

### SQL
SELECT DISTINCT T1.dba_name 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 = 'Tavern' AND T3.fine = 100
### 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
List point level of inspections with no fine.

### Hint
no fine refers to fine = 0

### SQL
SELECT DISTINCT T1.point_level FROM inspection_point AS T1 INNER JOIN violation AS T2 ON T1.point_id = T2.point_id WHERE 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
Provide the facility type and license number of establishments with the lowest risk level but failed the inspection.

### Hint
license number refers to license_no; the lowest risk level refers to min(risk_level); failed the inspection refers to results = 'Fail'

### SQL
SELECT DISTINCT T1.facility_type, T1.license_no FROM establishment AS T1 INNER JOIN inspection AS T2 ON T1.license_no = T2.license_no WHERE T1.risk_level = 1 AND T2.results = 'Fail'
### 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 result of the February 24, 2010 inspection involving the employee named "Arnold Holder"?

### Hint
February 24, 2010 refers to inspection_date = '2010-02-24'

### SQL
SELECT DISTINCT T2.results FROM employee AS T1 INNER JOIN inspection AS T2 ON T1.employee_id = T2.employee_id WHERE T2.inspection_date = '2010-02-24' AND T1.first_name = 'Arnold' AND T1.last_name = 'Holder'
### 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 restaurants failed the inspection in April 2010?

### Hint
restaurant refers to facility_type = 'Restaurant'; failed the inspection refers to results = 'Fail'; in April 2010 refers to inspection_date like '2010-04%'

### SQL
SELECT COUNT(DISTINCT T1.license_no) FROM establishment AS T1 INNER JOIN inspection AS T2 ON T1.license_no = T2.license_no WHERE strftime('%Y-%m', T2.inspection_date) = '2010-04' AND T1.facility_type = 'Restaurant' AND T2.results = 'Fail'
### 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
List all inspection IDs where the employee named "Rosemary Kennedy" was involved.

### SQL
SELECT DISTINCT T2.inspection_id FROM employee AS T1 INNER JOIN inspection AS T2 ON T1.employee_id = T2.employee_id WHERE T1.first_name = 'Rosemary' AND T1.last_name = 'Kennedy'