jtjt520j/cspider_for_chinese_llama2_1.3b huggingface dataset:
Schema: CREATE TABLE dorm (student_capacity VARCHAR, gender VARCHAR, dorm_name VARCHAR)
user: 名字中包含“慧”字段的宿舍中学生的能力和性别是什么？

assistant: SELECT student_capacity ,  gender FROM dorm WHERE dorm_name LIKE '慧'
Schema: CREATE TABLE dorm (gender VARCHAR, dorm_name VARCHAR, student_capacity INTEGER)
user: 找出容量大于300或小于100的宿舍的名称和性别类型。

assistant: SELECT dorm_name ,  gender FROM dorm WHERE student_capacity  >  300 OR student_capacity  <  100
Schema: CREATE TABLE dorm (gender VARCHAR, dorm_name VARCHAR, student_capacity INTEGER)
user: 容量大于300或小于100的宿舍的名称和类型是什么？

assistant: SELECT dorm_name ,  gender FROM dorm WHERE student_capacity  >  300 OR student_capacity  <  100
Schema: CREATE TABLE student (major VARCHAR, city_code VARCHAR)
user: 找到不同专业和城市的数量。

assistant: SELECT count(DISTINCT major) ,  count(DISTINCT city_code) FROM student
Schema: CREATE TABLE student (major VARCHAR, city_code VARCHAR)
user: 有多少不同的专业，每个学生有多少不同的城市编号？

assistant: SELECT count(DISTINCT major) ,  count(DISTINCT city_code) FROM student
Schema: CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 找到既有“客厅”又有“自习室”作为设施的宿舍的名称。

assistant: SELECT T1.dorm_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T3.amenity_name  =  '客厅' INTERSECT SELECT T1.dorm_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T3.amenity_name  =  '自习室'
Schema: CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 既有“客厅”又有“自习室”的宿舍叫什么名字？

assistant: SELECT T1.dorm_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T3.amenity_name  =  '客厅' INTERSECT SELECT T1.dorm_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T3.amenity_name  =  '自习室'
Schema: CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 找出有“客厅”但没有“自习室”作为设施的宿舍的名字。

assistant: SELECT T1.dorm_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T3.amenity_name  =  '客厅' EXCEPT SELECT T1.dorm_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T3.amenity_name  =  '自习室'
Schema: CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 有“客厅”但没有“自习室”的每间宿舍叫什么名字？

assistant: SELECT T1.dorm_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T3.amenity_name  =  '客厅' EXCEPT SELECT T1.dorm_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T3.amenity_name  =  '自习室'
Schema: CREATE TABLE student (sex VARCHAR, lname VARCHAR, age INTEGER, city_code VARCHAR)
user: 找出那些生活在代号为“武汉”的女性（性别是“女性”）或者年龄在20岁以下的男性（性别是“男”）学生的姓氏。

assistant: SELECT lname FROM student WHERE sex  =  'F' AND city_code  =  '武汉' UNION SELECT lname FROM student WHERE sex  =  '男' AND age  <  20
Schema: CREATE TABLE student (sex VARCHAR, lname VARCHAR, age INTEGER, city_code VARCHAR)
user: 那些生活在代号为“武汉”的女性（性别是“女性”）或者年龄在20岁以下的男性（性别是“男”）学生的姓氏是什么？

assistant: SELECT lname FROM student WHERE sex  =  'F' AND city_code  =  '武汉' UNION SELECT lname FROM student WHERE sex  =  '男' AND age  <  20
Schema: CREATE TABLE dorm (student_capacity VARCHAR, dorm_name VARCHAR)
user: 找到容量最大的宿舍的名字。

assistant: SELECT dorm_name FROM dorm ORDER BY student_capacity DESC LIMIT 1
Schema: CREATE TABLE dorm (student_capacity VARCHAR, dorm_name VARCHAR)
user: 容量最大的宿舍是什么名字？

assistant: SELECT dorm_name FROM dorm ORDER BY student_capacity DESC LIMIT 1
Schema: CREATE TABLE dorm_amenity (amenity_name VARCHAR)
user: 按字母顺序列出所有不同的宿舍设施。

assistant: SELECT amenity_name FROM dorm_amenity ORDER BY amenity_name
Schema: CREATE TABLE dorm_amenity (amenity_name VARCHAR)
user: 不同的宿舍设施名称按字母顺序排列是什么？

assistant: SELECT amenity_name FROM dorm_amenity ORDER BY amenity_name
Schema: CREATE TABLE student (city_code VARCHAR)
user: 找出大多数学生居住的城市代码。

assistant: SELECT city_code FROM student GROUP BY city_code ORDER BY count(*) DESC LIMIT 1
Schema: CREATE TABLE student (city_code VARCHAR)
user: 最多数量的学生居住的城市代码是什么？

assistant: SELECT city_code FROM student GROUP BY city_code ORDER BY count(*) DESC LIMIT 1
Schema: CREATE TABLE student (fname VARCHAR, lname VARCHAR, age INTEGER)
user: 找出年龄小于平均年龄的学生的名字和姓氏。

assistant: SELECT fname ,  lname FROM student WHERE age  <  (SELECT avg(age) FROM student)
Schema: CREATE TABLE student (fname VARCHAR, lname VARCHAR, age INTEGER)
user: 所有年龄小于平均水平的学生的名字和姓氏是什么？

assistant: SELECT fname ,  lname FROM student WHERE age  <  (SELECT avg(age) FROM student)
Schema: CREATE TABLE student (fname VARCHAR, lname VARCHAR, age VARCHAR, city_code VARCHAR)
user: 根据编码“香港”列出不在该城市居住的学生的名字和姓氏，并按年龄对结果进行排序。

assistant: SELECT fname ,  lname FROM student WHERE city_code != '香港' ORDER BY age
Schema: CREATE TABLE student (fname VARCHAR, lname VARCHAR, age VARCHAR, city_code VARCHAR)
user: 按年龄排序的所有未在“香港”生活的学生的姓氏和名字是什么？

assistant: SELECT fname ,  lname FROM student WHERE city_code != '香港' ORDER BY age
Schema: CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR)
user: 按字母顺序对结果进行排序列出“礼堂”的所有设施的名称。

assistant: SELECT T1.amenity_name FROM dorm_amenity AS T1 JOIN has_amenity AS T2 ON T2.amenid  =  T1.amenid JOIN dorm AS T3 ON T2.dormid  =  T3.dormid WHERE T3.dorm_name  =  '礼堂' ORDER BY T1.amenity_name
Schema: CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR)
user: 按字母顺序列出的“礼堂”的所有设施有哪些？

assistant: SELECT T1.amenity_name FROM dorm_amenity AS T1 JOIN has_amenity AS T2 ON T2.amenid  =  T1.amenid JOIN dorm AS T3 ON T2.dormid  =  T3.dormid WHERE T3.dorm_name  =  '礼堂' ORDER BY T1.amenity_name
Schema: CREATE TABLE dorm (gender VARCHAR, student_capacity INTEGER)
user: 找出宿舍的数量和每个性别的总容量。

assistant: SELECT count(*) ,  sum(student_capacity) ,  gender FROM dorm GROUP BY gender
Schema: CREATE TABLE dorm (gender VARCHAR, student_capacity INTEGER)
user: 有多少个宿舍且每个性别的总容量是多少？

assistant: SELECT count(*) ,  sum(student_capacity) ,  gender FROM dorm GROUP BY gender
Schema: CREATE TABLE student (sex VARCHAR, age INTEGER)
user: 找出不同性别学生的平均年龄和最高年龄。

assistant: SELECT avg(age) ,  max(age) ,  sex FROM student GROUP BY sex
Schema: CREATE TABLE student (sex VARCHAR, age INTEGER)
user: 每个性别的平均年龄和最高龄是多少？

assistant: SELECT avg(age) ,  max(age) ,  sex FROM student GROUP BY sex
Schema: CREATE TABLE student (major VARCHAR)
user: 找出每个专业学生的人数。

assistant: SELECT count(*) ,  major FROM student GROUP BY major
Schema: CREATE TABLE student (major VARCHAR)
user: 每个专业有多少学生？

assistant: SELECT count(*) ,  major FROM student GROUP BY major
Schema: CREATE TABLE student (age INTEGER, city_code VARCHAR)
user: 找到居住在每个城市的学生的数量和平均年龄。

assistant: SELECT count(*) ,  avg(age) ,  city_code FROM student GROUP BY city_code
Schema: CREATE TABLE student (age INTEGER, city_code VARCHAR)
user: 每个城市有多少学生且对应的平均年龄是多少？

assistant: SELECT count(*) ,  avg(age) ,  city_code FROM student GROUP BY city_code
Schema: CREATE TABLE student (age INTEGER, sex VARCHAR, city_code VARCHAR)
user: 找出每个城市男生（性别是“男”）的平均年龄和数量。

assistant: SELECT count(*) ,  avg(age) ,  city_code FROM student WHERE sex  =  '男' GROUP BY city_code
Schema: CREATE TABLE student (age INTEGER, sex VARCHAR, city_code VARCHAR)
user: 每个城市的平均年龄和男学生有多少？

assistant: SELECT count(*) ,  avg(age) ,  city_code FROM student WHERE sex  =  '男' GROUP BY city_code
Schema: CREATE TABLE student (city_code VARCHAR)
user: 找到拥有一个以上学生的城市其对应的学生人数。

assistant: SELECT count(*) ,  city_code FROM student GROUP BY city_code HAVING count(*)  >  1
Schema: CREATE TABLE student (city_code VARCHAR)
user: 每个城市有多少学生，哪个城市有不止一个城市？

assistant: SELECT count(*) ,  city_code FROM student GROUP BY city_code HAVING count(*)  >  1
Schema: CREATE TABLE student (fname VARCHAR, lname VARCHAR, major VARCHAR)
user: 查找不属于最大专业的学生的姓氏和姓氏。

assistant: SELECT fname ,  lname FROM student WHERE major != (SELECT major FROM student GROUP BY major ORDER BY count(*) DESC LIMIT 1)
Schema: CREATE TABLE student (fname VARCHAR, lname VARCHAR, major VARCHAR)
user: 不是最大的专业的学生的名字和姓氏是什么？

assistant: SELECT fname ,  lname FROM student WHERE major != (SELECT major FROM student GROUP BY major ORDER BY count(*) DESC LIMIT 1)
Schema: CREATE TABLE student (sex VARCHAR, age INTEGER)
user: 找出每个性别中年龄大于平均年龄的学生人数。

assistant: SELECT count(*) ,  sex FROM student WHERE age  >  (SELECT avg(age) FROM student) GROUP BY sex
Schema: CREATE TABLE student (sex VARCHAR, age INTEGER)
user: 每个性别中大于平均年龄的学生人数有多少？

assistant: SELECT count(*) ,  sex FROM student WHERE age  >  (SELECT avg(age) FROM student) GROUP BY sex
Schema: CREATE TABLE student (age INTEGER, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR)
user: 找到每个宿舍学生的平均年龄和宿舍名称。

assistant: SELECT avg(T1.age) ,  T3.dorm_name FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid GROUP BY T3.dorm_name
Schema: CREATE TABLE student (age INTEGER, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR)
user: 每个宿舍的平均年龄是多少，每个宿舍的名字是什么？

assistant: SELECT avg(T1.age) ,  T3.dorm_name FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid GROUP BY T3.dorm_name
Schema: CREATE TABLE dorm (dormid VARCHAR, student_capacity INTEGER), CREATE TABLE has_amenity (dormid VARCHAR)
user: 找出每个可以容纳100多名学生的宿舍的设施数量。

assistant: SELECT count(*) ,  T1.dormid FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid WHERE T1.student_capacity  >  100 GROUP BY T1.dormid
Schema: CREATE TABLE dorm (dormid VARCHAR, student_capacity INTEGER), CREATE TABLE has_amenity (dormid VARCHAR)
user: 每个宿舍都有多少设施？

assistant: SELECT count(*) ,  T1.dormid FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid WHERE T1.student_capacity  >  100 GROUP BY T1.dormid
Schema: CREATE TABLE student (stuid VARCHAR, age INTEGER), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR)
user: 找到每个宿舍里年龄大于20岁的学生。

assistant: SELECT count(*) ,  T3.dorm_name FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid WHERE T1.age  >  20 GROUP BY T3.dorm_name
Schema: CREATE TABLE student (stuid VARCHAR, age INTEGER), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR)
user: 每个宿舍有多少学生年龄大于20岁？

assistant: SELECT count(*) ,  T3.dorm_name FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid WHERE T1.age  >  20 GROUP BY T3.dorm_name
Schema: CREATE TABLE student (fname VARCHAR, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR)
user: 找到住在“大礼堂”的学生的名字。

assistant: SELECT T1.fname FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid WHERE T3.dorm_name  =  '大礼堂'
Schema: CREATE TABLE student (fname VARCHAR, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR)
user: “大礼堂”所有学生的名字是什么？

assistant: SELECT T1.fname FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid WHERE T3.dorm_name  =  '大礼堂'
Schema: CREATE TABLE student (age INTEGER, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (student_capacity INTEGER)
user: 找出住容量最大的宿舍的学生的平均年龄。

assistant: SELECT avg(T1.age) FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid WHERE T3.student_capacity  =  (SELECT max(student_capacity) FROM dorm)
Schema: CREATE TABLE student (age INTEGER, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (student_capacity INTEGER)
user: 住在容量最大的宿舍的学生的平均年龄是多少？

assistant: SELECT avg(T1.age) FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid WHERE T3.student_capacity  =  (SELECT max(student_capacity) FROM dorm)
Schema: CREATE TABLE student (stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, gender VARCHAR)
user: 找出住在男生宿舍的学生总数（性别是“男”）。

assistant: SELECT count(*) FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid WHERE T3.gender  =  '男'
Schema: CREATE TABLE student (stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, gender VARCHAR)
user: 住在“男”宿舍里的学生总数是多少？

assistant: SELECT count(*) FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid WHERE T3.gender  =  '男'
Schema: CREATE TABLE student (stuid VARCHAR, sex VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR)
user: 找出居住在“大礼堂”的女学生（性别是“F”）

assistant: SELECT count(*) FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid WHERE T3.dorm_name  =  '大礼堂' AND T1.sex  =  'F'
Schema: CREATE TABLE student (stuid VARCHAR, sex VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR)
user: “大礼堂”有多少“女”学生居住？

assistant: SELECT count(*) FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid JOIN dorm AS T3 ON T3.dormid  =  T2.dormid WHERE T3.dorm_name  =  '大礼堂' AND T1.sex  =  'F'
Schema: CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 找出“大礼堂”宿舍设施的名称。

assistant: SELECT T3.amenity_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T1.dorm_name  =  '大礼堂'
Schema: CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: “大礼堂”的设施名称是什么？

assistant: SELECT T3.amenity_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T1.dorm_name  =  '大礼堂'
Schema: CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 找出“大礼堂”宿舍的名称。根据设施名称对结果进行排序。

assistant: SELECT T3.amenity_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T1.dorm_name  =  '大礼堂' ORDER BY T3.amenity_name
Schema: CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: “大礼堂”按字母顺序有哪些设施？

assistant: SELECT T3.amenity_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T1.dorm_name  =  '大礼堂' ORDER BY T3.amenity_name
Schema: CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR)
user: 找出所有宿舍中最常见的设施的名字。

assistant: SELECT T1.amenity_name FROM dorm_amenity AS T1 JOIN has_amenity AS T2 ON T1.amenid  =  T2.amenid GROUP BY T2.amenid ORDER BY count(*) DESC LIMIT 1
Schema: CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR)
user: 宿舍里最常见的设施是什么？

assistant: SELECT T1.amenity_name FROM dorm_amenity AS T1 JOIN has_amenity AS T2 ON T1.amenid  =  T2.amenid GROUP BY T2.amenid ORDER BY count(*) DESC LIMIT 1
Schema: CREATE TABLE student (fname VARCHAR, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR)
user: 找到住在设施最多的宿舍的学生的名字。

assistant: SELECT T1.fname FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid WHERE T2.dormid IN (SELECT T2.dormid FROM dorm AS T3 JOIN has_amenity AS T4 ON T3.dormid  =  T4.dormid JOIN dorm_amenity AS T5 ON T4.amenid  =  T5.amenid GROUP BY T3.dormid ORDER BY count(*) DESC LIMIT 1)
Schema: CREATE TABLE student (fname VARCHAR, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE dorm (dormid VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR)
user: 住在设施最多的宿舍的所有学生的名字是什么？

assistant: SELECT T1.fname FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid WHERE T2.dormid IN (SELECT T2.dormid FROM dorm AS T3 JOIN has_amenity AS T4 ON T3.dormid  =  T4.dormid JOIN dorm_amenity AS T5 ON T4.amenid  =  T5.amenid GROUP BY T3.dormid ORDER BY count(*) DESC LIMIT 1)
Schema: CREATE TABLE dorm (dormid VARCHAR, student_capacity VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR)
user: 找出设施最少的宿舍的名称和容量。

assistant: SELECT T1.dorm_name ,  T1.student_capacity FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid GROUP BY T2.dormid ORDER BY count(*) LIMIT 1
Schema: CREATE TABLE dorm (dormid VARCHAR, student_capacity VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR)
user: 有最少设施的宿舍的名称和容量是什么？

assistant: SELECT T1.dorm_name ,  T1.student_capacity FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid GROUP BY T2.dormid ORDER BY count(*) LIMIT 1
Schema: CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 找到没有“客厅”设施的宿舍的名字。

assistant: SELECT dorm_name FROM dorm EXCEPT SELECT T1.dorm_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T3.amenity_name  =  '客厅'
Schema: CREATE TABLE dorm (dormid VARCHAR, dorm_name VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 没有“客厅”设施的宿舍的名字是什么？

assistant: SELECT dorm_name FROM dorm EXCEPT SELECT T1.dorm_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid WHERE T3.amenity_name  =  '客厅'
Schema: CREATE TABLE student (fname VARCHAR, lname VARCHAR, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 找出住在有“客厅”设施的宿舍的学生的名字和姓氏。

assistant: SELECT T1.fname ,  T1.lname FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid WHERE T2.dormid IN (SELECT T3.dormid FROM has_amenity AS T3 JOIN dorm_amenity AS T4 ON T3.amenid  =  T4.amenid WHERE T4.amenity_name  =  '客厅')
Schema: CREATE TABLE student (fname VARCHAR, lname VARCHAR, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 住在拥有“客厅”设施的宿舍里的所有学生的名字和姓氏是什么？

assistant: SELECT T1.fname ,  T1.lname FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid WHERE T2.dormid IN (SELECT T3.dormid FROM has_amenity AS T3 JOIN dorm_amenity AS T4 ON T3.amenid  =  T4.amenid WHERE T4.amenity_name  =  '客厅')
Schema: CREATE TABLE student (fname VARCHAR, age VARCHAR, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 找到住在没有“客厅”设施的宿舍的学生的名字和年龄。

assistant: SELECT T1.fname ,  T1.age FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid WHERE T2.dormid NOT IN (SELECT T3.dormid FROM has_amenity AS T3 JOIN dorm_amenity AS T4 ON T3.amenid  =  T4.amenid WHERE T4.amenity_name  =  '客厅')
Schema: CREATE TABLE student (fname VARCHAR, age VARCHAR, stuid VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR)
user: 每个住在“客厅”设施的宿舍的学生的名字和年龄是多少？

assistant: SELECT T1.fname ,  T1.age FROM student AS T1 JOIN lives_in AS T2 ON T1.stuid  =  T2.stuid WHERE T2.dormid NOT IN (SELECT T3.dormid FROM has_amenity AS T3 JOIN dorm_amenity AS T4 ON T3.amenid  =  T4.amenid WHERE T4.amenity_name  =  '客厅')
Schema: CREATE TABLE dorm (dormid VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE student (stuid VARCHAR, lname VARCHAR)
user: 找到姓“帅”的学生住的宿舍的设施的名称。

assistant: SELECT T3.amenity_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid JOIN lives_in AS T4 ON T4.dormid  =  T1.dormid JOIN student AS T5 ON T5.stuid  =  T4.stuid WHERE T5.lname  =  '帅'
Schema: CREATE TABLE dorm (dormid VARCHAR), CREATE TABLE has_amenity (amenid VARCHAR, dormid VARCHAR), CREATE TABLE dorm_amenity (amenid VARCHAR, amenity_name VARCHAR), CREATE TABLE lives_in (stuid VARCHAR, dormid VARCHAR), CREATE TABLE student (stuid VARCHAR, lname VARCHAR)
user: 姓氏为“帅”的学生住在宿舍里有什么设施？

assistant: SELECT T3.amenity_name FROM dorm AS T1 JOIN has_amenity AS T2 ON T1.dormid  =  T2.dormid JOIN dorm_amenity AS T3 ON T2.amenid  =  T3.amenid JOIN lives_in AS T4 ON T4.dormid  =  T1.dormid JOIN student AS T5 ON T5.stuid  =  T4.stuid WHERE T5.lname  =  '帅'
Schema: CREATE TABLE customers (Id VARCHAR)
user: 那里有多少顾客？

assistant: SELECT count(*) FROM customers
Schema: CREATE TABLE customers (Id VARCHAR)
user: 统计客户的数量。

assistant: SELECT count(*) FROM customers
Schema: CREATE TABLE customers (phone_number VARCHAR, email_address VARCHAR)
user: 根据电子邮件地址和电话号码排序查找所有客户的电子邮件和电话号码。

assistant: SELECT email_address ,  phone_number FROM customers ORDER BY email_address ,  phone_number
Schema: CREATE TABLE customers (phone_number VARCHAR, email_address VARCHAR)
user: 按电子邮件地址和电话号码排序的所有客户的电子邮件和电话号码是什么？

assistant: SELECT email_address ,  phone_number FROM customers ORDER BY email_address ,  phone_number
Schema: CREATE TABLE customers (customer_type_code VARCHAR, town_city VARCHAR)
user: 哪个城市的类型代码是“良好信用评级”的客户数量最少？

assistant: SELECT town_city FROM customers WHERE customer_type_code  =  "良好信用评级" GROUP BY town_city ORDER BY count(*) LIMIT 1
Schema: CREATE TABLE customers (customer_type_code VARCHAR, town_city VARCHAR)
user: 返回有最少客户的客户类型代码为“良好信用评级”的城市。

assistant: SELECT town_city FROM customers WHERE customer_type_code  =  "良好信用评级" GROUP BY town_city ORDER BY count(*) LIMIT 1
Schema: CREATE TABLE products (product_id VARCHAR, product_name VARCHAR), CREATE TABLE complaints (product_id VARCHAR)
user: 列出所有产品的名称以及其收到的投诉数量。

assistant: SELECT t1.product_name ,  count(*) FROM products AS t1 JOIN complaints AS t2 ON t1.product_id  =  t2.product_id GROUP BY t1.product_name