lianghsun/bird-text2sql-bench huggingface dataset:
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
Between 2011 to 2016, in which countries can you find the universities where at least 50% of its students are international students?

### Hint
Between 2011 to 2016 refers to year BETWEEN 2011 AND 2016; at least 50% of its students are international students refers to pct_international_students > 50; which country refers to country_name

### SQL
SELECT DISTINCT T3.country_name FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id INNER JOIN country AS T3 ON T3.id = T1.country_id WHERE T2.pct_international_students > 50 AND T2.year BETWEEN 2011 AND 2016
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
How many universities have no less than 20,000 female students in 2016? Identify how many of the said universities are located in the United States of America.

### Hint
have no less than 20,000 female students refers to DIVIDE(MULTIPLY(pct_female_students, num_students), 100) > 20000; in 2016 refers to year = 2016; located in the United States of America refers to country_name = 'United States of America'

### SQL
SELECT COUNT(*) , SUM(CASE WHEN T3.country_name = 'United States of America' THEN 1 ELSE 0 END) AS nums_in_usa FROM university AS T1 INNER JOIN university_year AS T2 ON T1.id = T2.university_id INNER JOIN country AS T3 ON T3.id = T1.country_id WHERE T2.year = 2016 AND T2.num_students * T2.pct_female_students / 100 > 20000
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
What are the names of the top 5 universities with the highest number of international students?

### Hint
highest number of international students refers to MAX(DIVIDE(MULTIPLY(num_students, pct_international_students), 100)); name of university refers to university_name;

### SQL
SELECT DISTINCT T2.university_name FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id ORDER BY (CAST(T1.num_students * T1.pct_international_students AS REAL) / 100) DESC LIMIT 5
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
What is the university ID of the university with the largest student staff ratio?

### Hint
the largest student staff ratio refers to max(student_staff_ratio)

### SQL
SELECT university_id FROM university_year ORDER BY student_staff_ratio DESC LIMIT 1
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
Give the year where a university had the lowest number of students.

### Hint
had the lowest number of students refers to MIN(num_students)

### SQL
SELECT year FROM university_year ORDER BY num_students ASC LIMIT 1
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
Compute the average percentage of female students.

### Hint
average percentage of female students refers to avg(pct_female_students)

### SQL
SELECT AVG(pct_female_students) FROM university_year
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
Provide the number of international students and number of students in 2013 in university ID 20.

### Hint
number of international students refers to DIVIDE(MULTIPLY(pct_international_students, num_students), 100); in 2013 refers to year = 2013

### SQL
SELECT pct_international_students * num_students, num_students FROM university_year WHERE year = 2013 AND university_id = 20
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
What is the university ID of Harvard University?

### Hint
of Harvard University refers to university_name = 'Harvard University';

### SQL
SELECT id FROM university WHERE university_name = 'Harvard University'
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
List the university ID of the university that scored 100 in 2011.

### Hint
in 2011 refers to year = 2011; score = 100

### SQL
SELECT university_id FROM university_ranking_year WHERE score = 100 AND year = 2011
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
Provide the ranking system of the ranking criteria named Quality of Education Rank.

### Hint
criteria named Quality of Education Rank refers to criteria_name = 'Quality of Education Rank'; ranking system refers to system_name;

### SQL
SELECT T1.system_name FROM ranking_system AS T1 INNER JOIN ranking_criteria AS T2 ON T1.id = T2.ranking_system_id WHERE T2.criteria_name = 'Quality of Education Rank'
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
What is the student staff ratio of Harvard University in 2012?

### Hint
Harvard University refers to university_name = 'Harvard University'; in 2012 refers to year = 2012

### SQL
SELECT T1.student_staff_ratio FROM university_year AS T1 INNER JOIN university AS T2 ON T1.university_id = T2.id WHERE T2.university_name = 'Harvard University' AND T1.year = 2012
### Database Schema
CREATE TABLE country
(
    id           INTEGER not null
            primary key,
    country_name TEXT default NULL
)

CREATE TABLE ranking_system
(
    id          INTEGER not null
            primary key,
    system_name TEXT default NULL
)

CREATE TABLE ranking_criteria
(
    id                INTEGER not null
            primary key,
    ranking_system_id INTEGER          default NULL,
    criteria_name     TEXT default NULL,
    foreign key (ranking_system_id) references ranking_system(id)
)

CREATE TABLE university
(
    id              INTEGER not null
            primary key,
    country_id      INTEGER          default NULL,
    university_name TEXT default NULL,
    foreign key (country_id) references country(id)
)

CREATE TABLE university_ranking_year
(
    university_id       INTEGER default NULL,
    ranking_criteria_id INTEGER default NULL,
    year                INTEGER default NULL,
    score               INTEGER default NULL,
    foreign key (ranking_criteria_id) references ranking_criteria(id),
    foreign key (university_id) references university(id)
)

CREATE TABLE university_year
(
    university_id              INTEGER           default NULL,
    year                       INTEGER           default NULL,
    num_students               INTEGER           default NULL,
    student_staff_ratio        REAL default NULL,
    pct_international_students INTEGER           default NULL,
    pct_female_students        INTEGER           default NULL,
    foreign key (university_id) references university(id)
)

### Question
Give the location of the university ID 112.

### Hint
location refers to country_name

### SQL
SELECT T2.country_name FROM university AS T1 INNER JOIN country AS T2 ON T1.country_id = T2.id WHERE T1.id = 112