jtjt520j/cspider_for_chinese_llama2_1.3b huggingface dataset:
Schema: CREATE TABLE Reviewer (title VARCHAR, name VARCHAR), CREATE TABLE Movie (title VARCHAR, name VARCHAR)
user: 所有评论者和电影的名字是什么？

assistant: SELECT name FROM Reviewer UNION SELECT title FROM Movie
Schema: CREATE TABLE Movie (title VARCHAR, mID VARCHAR), CREATE TABLE Rating (mID VARCHAR, rID VARCHAR), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: 找到所有未被“谢晨”评论的电影的标题。

assistant: SELECT DISTINCT title FROM Movie EXCEPT SELECT T2.title FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID JOIN Reviewer AS T3 ON T1.rID  =  T3.rID WHERE T3.name  =  '谢晨'
Schema: CREATE TABLE Movie (title VARCHAR, mID VARCHAR), CREATE TABLE Rating (mID VARCHAR, rID VARCHAR), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: “谢晨”没有评论的所有电影的标题是什么？

assistant: SELECT DISTINCT title FROM Movie EXCEPT SELECT T2.title FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID JOIN Reviewer AS T3 ON T1.rID  =  T3.rID WHERE T3.name  =  '谢晨'
Schema: CREATE TABLE Movie (title VARCHAR, director VARCHAR)
user: 按导演姓名和电影标题排序，返回导演不止一部电影的导演所拍摄的所有电影名称和导演的名字。

assistant: SELECT T1.title ,  T1.director FROM Movie AS T1 JOIN Movie AS T2 ON T1.director  =  T2.director WHERE T1.title != T2.title ORDER BY T1.director ,  T1.title
Schema: CREATE TABLE Movie (title VARCHAR, director VARCHAR)
user: 对于所有导演不止一部电影的导演，他们导演的不同电影名称以及他们的名字都是什么？

assistant: SELECT T1.title ,  T1.director FROM Movie AS T1 JOIN Movie AS T2 ON T1.director  =  T2.director WHERE T1.title != T2.title ORDER BY T1.director ,  T1.title
Schema: CREATE TABLE Movie (title VARCHAR, director VARCHAR)
user: 对于有不止一部电影的导演，返回所有由他们执导的电影的片名和制作年份。

assistant: SELECT T1.title ,  T1.year FROM Movie AS T1 JOIN Movie AS T2 ON T1.director  =  T2.director WHERE T1.title != T2.title
Schema: CREATE TABLE Movie (title VARCHAR, director VARCHAR)
user: 对于每个导演不止一部电影的导演，他们导演的电影的片名和发行日期是什么？

assistant: SELECT T1.title ,  T1.year FROM Movie AS T1 JOIN Movie AS T2 ON T1.director  =  T2.director WHERE T1.title != T2.title
Schema: CREATE TABLE Movie (director VARCHAR)
user: 哪位导演只导演了一部电影？

assistant: SELECT director FROM Movie GROUP BY director HAVING count(*)  =  1
Schema: CREATE TABLE Movie (director VARCHAR)
user: 只制作了一部电影的所有导演的名字是什么？

assistant: SELECT director FROM Movie GROUP BY director HAVING count(*)  =  1
Schema: CREATE TABLE Movie (director VARCHAR)
user: 除了导演为“空”之外，那些只拍了一部电影的导演叫什么名字？

assistant: SELECT director FROM Movie WHERE director != "null" GROUP BY director HAVING count(*)  =  1
Schema: CREATE TABLE Movie (director VARCHAR)
user: 除导演为“空”之外的所有拍过一部电影的导演的名字是什么？

assistant: SELECT director FROM Movie WHERE director != "null" GROUP BY director HAVING count(*)  =  1
Schema: CREATE TABLE Movie (mID VARCHAR, director VARCHAR), CREATE TABLE Rating (mID VARCHAR)
user: 每个导演有多少个电影评论？

assistant: SELECT count(*) ,  T1.director FROM Movie AS T1 JOIN Rating AS T2 ON T1.mID  =  T2.mID GROUP BY T1.director
Schema: CREATE TABLE Movie (mID VARCHAR, director VARCHAR), CREATE TABLE Rating (mID VARCHAR)
user: 对于每个导演，他们收到了多少评论？

assistant: SELECT count(*) ,  T1.director FROM Movie AS T1 JOIN Rating AS T2 ON T1.mID  =  T2.mID GROUP BY T1.director
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 找出平均评级最高的电影。返回电影片名和平均评级。

assistant: SELECT T2.title ,  avg(T1.stars) FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID GROUP BY T1.mID ORDER BY avg(T1.stars) DESC LIMIT 1
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 平均评级最高的电影片名是什么及它们的评级是多少？

assistant: SELECT T2.title ,  avg(T1.stars) FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID GROUP BY T1.mID ORDER BY avg(T1.stars) DESC LIMIT 1
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 平均评级最低的电影的片名和平均评级是多少？

assistant: SELECT T2.title ,  avg(T1.stars) FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID GROUP BY T1.mID ORDER BY avg(T1.stars) LIMIT 1
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 平均评级最低的所有电影的片名是什么？和平均的评级是多少？

assistant: SELECT T2.title ,  avg(T1.stars) FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID GROUP BY T1.mID ORDER BY avg(T1.stars) LIMIT 1
Schema: CREATE TABLE Rating (mID VARCHAR, stars VARCHAR), CREATE TABLE Movie (title VARCHAR, mID VARCHAR, year VARCHAR)
user: 评价排名前三位的电影的片名和发行年份是什么？

assistant: SELECT T2.title ,  T2.year FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID ORDER BY T1.stars DESC LIMIT 3
Schema: CREATE TABLE Rating (mID VARCHAR, stars VARCHAR), CREATE TABLE Movie (title VARCHAR, mID VARCHAR, year VARCHAR)
user: 评价最高的3部电影的片名和放映年限是什么？

assistant: SELECT T2.title ,  T2.year FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID ORDER BY T1.stars DESC LIMIT 3
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER), CREATE TABLE Movie (title VARCHAR, mID VARCHAR, director VARCHAR)
user: 对于每个导演，返回导演的姓名以及他们导演所有电影中评价最高的片名以及该评价的值。忽略导演为“空”的电影。

assistant: SELECT T2.title ,  T1.stars ,  T2.director ,  max(T1.stars) FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID WHERE director != "null" GROUP BY director
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER), CREATE TABLE Movie (title VARCHAR, mID VARCHAR, director VARCHAR)
user: 对于每个导演，他们导演过的所有电影的片名和评级是多少？

assistant: SELECT T2.title ,  T1.stars ,  T2.director ,  max(T1.stars) FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID WHERE director != "null" GROUP BY director
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER, rID VARCHAR), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 找出每位评价者评价最低的电影的片名和星级。

assistant: SELECT T2.title ,  T1.rID ,  T1.stars ,  min(T1.stars) FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID GROUP BY T1.rID
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER, rID VARCHAR), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 对于每个评论者ID，评级最低的电影的名称和评级是什么？

assistant: SELECT T2.title ,  T1.rID ,  T1.stars ,  min(T1.stars) FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID GROUP BY T1.rID
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER), CREATE TABLE Movie (title VARCHAR, mID VARCHAR, director VARCHAR)
user: 找出每个导演执导的所有电影中评价最低的电影的片名和得分。

assistant: SELECT T2.title ,  T1.stars ,  T2.director ,  min(T1.stars) FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID GROUP BY T2.director
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER), CREATE TABLE Movie (title VARCHAR, mID VARCHAR, director VARCHAR)
user: 对于每个导演来说，他们评分最差的电影的片名和得分是什么？

assistant: SELECT T2.title ,  T1.stars ,  T2.director ,  min(T1.stars) FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID GROUP BY T2.director
Schema: CREATE TABLE Rating (mID VARCHAR), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 评价次数最多的电影的名字叫什么？

assistant: SELECT T2.title ,  T1.mID FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID GROUP BY T1.mID ORDER BY count(*) DESC LIMIT 1
Schema: CREATE TABLE Rating (mID VARCHAR), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 最被评论最多次的电影的名字是什么？

assistant: SELECT T2.title ,  T1.mID FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID GROUP BY T1.mID ORDER BY count(*) DESC LIMIT 1
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 星级是在3和5之间的所有电影的标题是什么？

assistant: SELECT T2.title FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID WHERE T1.stars BETWEEN 3 AND 5
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 所有3到5颗星的电影的标题是什么？

assistant: SELECT T2.title FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID WHERE T1.stars BETWEEN 3 AND 5
Schema: CREATE TABLE Rating (stars INTEGER, rID VARCHAR), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: 找出评价都高于3星级的评论人的名字。

assistant: SELECT T2.name FROM Rating AS T1 JOIN Reviewer AS T2 ON T1.rID  =  T2.rID WHERE T1.stars  >  3
Schema: CREATE TABLE Rating (stars INTEGER, rID VARCHAR), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: 给一部电影打过3星以上的评论人叫什么名字？

assistant: SELECT T2.name FROM Rating AS T1 JOIN Reviewer AS T2 ON T1.rID  =  T2.rID WHERE T1.stars  >  3
Schema: CREATE TABLE Rating (mID VARCHAR, rID VARCHAR), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: 找出没有“韦浩”评论每部电影的平均评价星级。

assistant: SELECT mID ,  avg(stars) FROM Rating WHERE mID NOT IN (SELECT T1.mID FROM Rating AS T1 JOIN Reviewer AS T2 ON T1.rID  =  T2.rID WHERE T2.name  =  "韦浩") GROUP BY mID
Schema: CREATE TABLE Rating (mID VARCHAR, rID VARCHAR), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: “韦浩”从未评价的每部电影的平均评价星级是多少？

assistant: SELECT mID ,  avg(stars) FROM Rating WHERE mID NOT IN (SELECT T1.mID FROM Rating AS T1 JOIN Reviewer AS T2 ON T1.rID  =  T2.rID WHERE T2.name  =  "韦浩") GROUP BY mID
Schema: CREATE TABLE Rating (mID VARCHAR, rID VARCHAR), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: “韦浩”没有评价的电影ID是什么？

assistant: SELECT mID FROM Rating EXCEPT SELECT T1.mID FROM Rating AS T1 JOIN Reviewer AS T2 ON T1.rID  =  T2.rID WHERE T2.name  =  "韦浩"
Schema: CREATE TABLE Rating (mID VARCHAR, rID VARCHAR), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: “韦浩”还没有评价过的所有的电影的ID是什么？

assistant: SELECT mID FROM Rating EXCEPT SELECT T1.mID FROM Rating AS T1 JOIN Reviewer AS T2 ON T1.rID  =  T2.rID WHERE T2.name  =  "韦浩"
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER)
user: 找出收到至少2次评论的每部电影的平均评级是多少。

assistant: SELECT mID ,  avg(stars) FROM Rating GROUP BY mID HAVING count(*)  >=  2
Schema: CREATE TABLE Rating (mID VARCHAR, stars INTEGER)
user: 对于每部超过3次评论的电影的平均评级是多少？

assistant: SELECT mID ,  avg(stars) FROM Rating GROUP BY mID HAVING count(*)  >=  2
Schema: CREATE TABLE Rating (rID VARCHAR, stars VARCHAR)
user: 找出那些没有给4星级评论的评价者的ID。

assistant: SELECT rID FROM Rating EXCEPT SELECT rID FROM Rating WHERE stars  =  4
Schema: CREATE TABLE Rating (rID VARCHAR, stars VARCHAR)
user: 没有给4星级评论的所有评价者的ID是什么？

assistant: SELECT rID FROM Rating EXCEPT SELECT rID FROM Rating WHERE stars  =  4
Schema: CREATE TABLE Rating (rID VARCHAR, stars VARCHAR)
user: 找到至少一次没有给过4颗星评论的评价者的ID。

assistant: SELECT rID FROM Rating WHERE stars != 4
Schema: CREATE TABLE Rating (rID VARCHAR, stars VARCHAR)
user: 至少一次没有给过4颗星评论的所有评价者的ID是什么？

assistant: SELECT rID FROM Rating WHERE stars != 4
Schema: CREATE TABLE Rating (mID VARCHAR, rID VARCHAR), CREATE TABLE Movie (title VARCHAR, mID VARCHAR, year INTEGER), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: 2000年以后制作或是由“韦浩”评论的电影的名字是什么？

assistant: SELECT DISTINCT T2.title FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID JOIN Reviewer AS T3 ON T1.rID  =  T3.rID WHERE T3.name  =  '韦浩' OR T2.year  >  2000
Schema: CREATE TABLE Rating (mID VARCHAR, rID VARCHAR), CREATE TABLE Movie (title VARCHAR, mID VARCHAR, year INTEGER), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: 2000年以后制作或“韦浩”评论过的所有电影的名字是什么？

assistant: SELECT DISTINCT T2.title FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID JOIN Reviewer AS T3 ON T1.rID  =  T3.rID WHERE T3.name  =  '韦浩' OR T2.year  >  2000
Schema: CREATE TABLE Movie (title VARCHAR, YEAR INTEGER, director VARCHAR)
user: 在1980年之前制作的或者由“姜文”导演的电影的名字是什么？

assistant: SELECT title FROM Movie WHERE director  =  "姜文" OR YEAR   <  1980
Schema: CREATE TABLE Movie (title VARCHAR, YEAR INTEGER, director VARCHAR)
user: 1980年以前拍的电影或由“姜文”担任导演的电影都有哪些名字？

assistant: SELECT title FROM Movie WHERE director  =  "姜文" OR YEAR   <  1980
Schema: CREATE TABLE Rating (rID VARCHAR, stars VARCHAR), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: 曾经给出3星级和4星级评价的评价者的名字是什么？

assistant: SELECT T2.name FROM Rating AS T1 JOIN Reviewer AS T2 ON T1.rID  =  T2.rID WHERE T1.stars  =  3 INTERSECT SELECT T2.name FROM Rating AS T1 JOIN Reviewer AS T2 ON T1.rID  =  T2.rID WHERE T1.stars  =  4
Schema: CREATE TABLE Rating (rID VARCHAR, stars VARCHAR), CREATE TABLE Reviewer (name VARCHAR, rID VARCHAR)
user: 给出3颗星或4颗星评价的评论者的叫什么名字？

assistant: SELECT T2.name FROM Rating AS T1 JOIN Reviewer AS T2 ON T1.rID  =  T2.rID WHERE T1.stars  =  3 INTERSECT SELECT T2.name FROM Rating AS T1 JOIN Reviewer AS T2 ON T1.rID  =  T2.rID WHERE T1.stars  =  4
Schema: CREATE TABLE Rating (mID VARCHAR, stars VARCHAR), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 有3颗星和4颗星评价的电影叫什么名字？

assistant: SELECT T2.title FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID WHERE T1.stars  =  3 INTERSECT SELECT T2.title FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID WHERE T1.stars  =  4
Schema: CREATE TABLE Rating (mID VARCHAR, stars VARCHAR), CREATE TABLE Movie (title VARCHAR, mID VARCHAR)
user: 所有收到3和4颗星评价的电影的名字是什么？

assistant: SELECT T2.title FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID WHERE T1.stars  =  3 INTERSECT SELECT T2.title FROM Rating AS T1 JOIN Movie AS T2 ON T1.mID  =  T2.mID WHERE T1.stars  =  4
Schema: CREATE TABLE county_public_safety (Id VARCHAR)
user: 有多少个省？

assistant: SELECT count(*) FROM county_public_safety
Schema: CREATE TABLE county_public_safety (Id VARCHAR)
user: 统计省的数量。

assistant: SELECT count(*) FROM county_public_safety
Schema: CREATE TABLE county_public_safety (Name VARCHAR, Population VARCHAR)
user: 按人口降序列出各个省的名称。

assistant: SELECT Name FROM county_public_safety ORDER BY Population DESC
Schema: CREATE TABLE county_public_safety (Name VARCHAR, Population VARCHAR)
user: 按人口下降顺序排列的省的名称是什么？

assistant: SELECT Name FROM county_public_safety ORDER BY Population DESC
Schema: CREATE TABLE county_public_safety (LOCATION VARCHAR, Police_force VARCHAR)
user: 列出不在东边的省的不同警察机关。

assistant: SELECT DISTINCT Police_force FROM county_public_safety WHERE LOCATION != "冬"
Schema: CREATE TABLE county_public_safety (LOCATION VARCHAR, Police_force VARCHAR)
user: 不位于东部的省有哪些不同的警察机关？

assistant: SELECT DISTINCT Police_force FROM county_public_safety WHERE LOCATION != "冬"
Schema: CREATE TABLE county_public_safety (Crime_rate INTEGER)
user: 所有省的最低犯罪率和最高犯罪率是多少？

assistant: SELECT min(Crime_rate) ,  max(Crime_rate) FROM county_public_safety
Schema: CREATE TABLE county_public_safety (Crime_rate INTEGER)
user: 返回所有省的最低和最高犯罪率。

assistant: SELECT min(Crime_rate) ,  max(Crime_rate) FROM county_public_safety
Schema: CREATE TABLE county_public_safety (Crime_rate VARCHAR, Police_officers VARCHAR)
user: 按警官数量的升序显示各省的犯罪率。

assistant: SELECT Crime_rate FROM county_public_safety ORDER BY Police_officers ASC
Schema: CREATE TABLE county_public_safety (Crime_rate VARCHAR, Police_officers VARCHAR)
user: 按警官数目升序排列的每个省的犯罪率是多少？

assistant: SELECT Crime_rate FROM county_public_safety ORDER BY Police_officers ASC
Schema: CREATE TABLE city (Name VARCHAR)
user: 按城市字母顺序排列的城市的名字是什么？

assistant: SELECT Name FROM city ORDER BY Name ASC
Schema: CREATE TABLE city (Name VARCHAR)
user: 返回按字母顺序排列的城市名称。

assistant: SELECT Name FROM city ORDER BY Name ASC
Schema: CREATE TABLE city (Hispanic VARCHAR, Black INTEGER)
user: 在黑人比例高于10的城市，西班牙裔的比例是多少？

assistant: SELECT Hispanic FROM city WHERE Black  >  10
Schema: CREATE TABLE city (Hispanic VARCHAR, Black INTEGER)
user: 返回黑人百分比大于10的城市的西班牙裔百分比。

assistant: SELECT Hispanic FROM city WHERE Black  >  10
Schema: CREATE TABLE county_public_safety (Name VARCHAR, Population VARCHAR)
user: 列出人口最多的省的名字。

assistant: SELECT Name FROM county_public_safety ORDER BY Population DESC LIMIT 1
Schema: CREATE TABLE county_public_safety (Name VARCHAR, Population VARCHAR)
user: 人口最多的省叫什么名字？

assistant: SELECT Name FROM county_public_safety ORDER BY Population DESC LIMIT 1
Schema: CREATE TABLE city (White VARCHAR, Name VARCHAR)
user: 列出白人百分比最大的5个的城市名称。

assistant: SELECT Name FROM city ORDER BY White DESC LIMIT 5
Schema: CREATE TABLE city (White VARCHAR, Name VARCHAR)
user: 白人比例最大的五个城市叫什么名字？

assistant: SELECT Name FROM city ORDER BY White DESC LIMIT 5
Schema: CREATE TABLE city (County_ID VARCHAR, Name VARCHAR), CREATE TABLE county_public_safety (County_ID VARCHAR, Name VARCHAR)
user: 显示城市名称和所在省的名称。

assistant: SELECT T1.Name ,  T2.Name FROM city AS T1 JOIN county_public_safety AS T2 ON T1.County_ID  =  T2.County_ID
Schema: CREATE TABLE city (County_ID VARCHAR, Name VARCHAR), CREATE TABLE county_public_safety (County_ID VARCHAR, Name VARCHAR)
user: 城市的名称以及它们所对应的省的名称分别是什么？

assistant: SELECT T1.Name ,  T2.Name FROM city AS T1 JOIN county_public_safety AS T2 ON T1.County_ID  =  T2.County_ID
Schema: CREATE TABLE city (White VARCHAR, County_ID VARCHAR), CREATE TABLE county_public_safety (Crime_rate VARCHAR, County_ID VARCHAR)
user: 显示城市的白人百分比和他们所在的省的犯罪率。

assistant: SELECT T1.White ,  T2.Crime_rate FROM city AS T1 JOIN county_public_safety AS T2 ON T1.County_ID  =  T2.County_ID
Schema: CREATE TABLE city (White VARCHAR, County_ID VARCHAR), CREATE TABLE county_public_safety (Crime_rate VARCHAR, County_ID VARCHAR)
user: 城市的白人比例是多少，它们所对应的省的犯罪率是多少？

assistant: SELECT T1.White ,  T2.Crime_rate FROM city AS T1 JOIN county_public_safety AS T2 ON T1.County_ID  =  T2.County_ID
Schema: CREATE TABLE city (Police_officers VARCHAR, name VARCHAR, county_ID VARCHAR), CREATE TABLE county_public_safety (Police_officers VARCHAR, name VARCHAR, county_ID VARCHAR)
user: 显示在警察数量最多的省的城市名称。

assistant: SELECT name FROM city WHERE county_ID  =  (SELECT county_ID FROM county_public_safety ORDER BY Police_officers DESC LIMIT 1)
Schema: CREATE TABLE city (Police_officers VARCHAR, name VARCHAR, county_ID VARCHAR), CREATE TABLE county_public_safety (Police_officers VARCHAR, name VARCHAR, county_ID VARCHAR)
user: 省内警察最多的城市叫什么名字？

assistant: SELECT name FROM city WHERE county_ID  =  (SELECT county_ID FROM county_public_safety ORDER BY Police_officers DESC LIMIT 1)
Schema: CREATE TABLE city (population INTEGER, county_ID VARCHAR), CREATE TABLE county_public_safety (population INTEGER, county_ID VARCHAR)
user: 显示人口超过20000的省的城市数量。

assistant: SELECT count(*) FROM city WHERE county_ID IN (SELECT county_ID FROM county_public_safety WHERE population  >  20000)
Schema: CREATE TABLE city (population INTEGER, county_ID VARCHAR), CREATE TABLE county_public_safety (population INTEGER, county_ID VARCHAR)
user: 人口超过20000的省有多少城市？

assistant: SELECT count(*) FROM city WHERE county_ID IN (SELECT county_ID FROM county_public_safety WHERE population  >  20000)
Schema: CREATE TABLE city (County_ID VARCHAR, White INTEGER), CREATE TABLE county_public_safety (Crime_rate VARCHAR, County_ID VARCHAR)
user: 显示一个白人比例超过90的城市的犯罪率。

assistant: SELECT T2.Crime_rate FROM city AS T1 JOIN county_public_safety AS T2 ON T1.County_ID  =  T2.County_ID WHERE T1.White  >  90
Schema: CREATE TABLE city (County_ID VARCHAR, White INTEGER), CREATE TABLE county_public_safety (Crime_rate VARCHAR, County_ID VARCHAR)
user: 白人比例超过90的城市的省的犯罪率是多少？

assistant: SELECT T2.Crime_rate FROM city AS T1 JOIN county_public_safety AS T2 ON T1.County_ID  =  T2.County_ID WHERE T1.White  >  90
Schema: CREATE TABLE county_public_safety (Police_force VARCHAR)
user: 请显示警察机关和每个警察机关管辖的省的数目。

assistant: SELECT Police_force ,  COUNT(*) FROM county_public_safety GROUP BY Police_force
Schema: CREATE TABLE county_public_safety (Police_force VARCHAR)
user: 每个警察机关对应多少个省？

assistant: SELECT Police_force ,  COUNT(*) FROM county_public_safety GROUP BY Police_force
Schema: CREATE TABLE county_public_safety (LOCATION VARCHAR)
user: 什么地方有数量最多的省？

assistant: SELECT LOCATION FROM county_public_safety GROUP BY LOCATION ORDER BY COUNT(*) DESC LIMIT 1
Schema: CREATE TABLE county_public_safety (LOCATION VARCHAR)
user: 哪个地方有最多相对应的省？

assistant: SELECT LOCATION FROM county_public_safety GROUP BY LOCATION ORDER BY COUNT(*) DESC LIMIT 1
Schema: CREATE TABLE county_public_safety (County_ID VARCHAR, Name VARCHAR), CREATE TABLE city (County_ID VARCHAR, Name VARCHAR)
user: 列出没有城市的省的名字。

assistant: SELECT Name FROM county_public_safety WHERE County_ID NOT IN (SELECT County_ID FROM city)
Schema: CREATE TABLE county_public_safety (County_ID VARCHAR, Name VARCHAR), CREATE TABLE city (County_ID VARCHAR, Name VARCHAR)
user: 不包含任何城市的省的名称是什么？

assistant: SELECT Name FROM county_public_safety WHERE County_ID NOT IN (SELECT County_ID FROM city)
Schema: CREATE TABLE county_public_safety (LOCATION VARCHAR, Police_force VARCHAR)
user: 显示东部和西部共同管辖的警察机构。

assistant: SELECT Police_force FROM county_public_safety WHERE LOCATION  =  "冬" INTERSECT SELECT Police_force FROM county_public_safety WHERE LOCATION  =  "西"
Schema: CREATE TABLE county_public_safety (LOCATION VARCHAR, Police_force VARCHAR)
user: 在东部和西部的两个省都有哪些警察机构同时管辖？

assistant: SELECT Police_force FROM county_public_safety WHERE LOCATION  =  "冬" INTERSECT SELECT Police_force FROM county_public_safety WHERE LOCATION  =  "西"
Schema: CREATE TABLE city (county_id VARCHAR, name VARCHAR, Crime_rate INTEGER), CREATE TABLE county_public_safety (county_id VARCHAR, name VARCHAR, Crime_rate INTEGER)
user: 显示犯罪率低于100的省的城市名称。

assistant: SELECT name FROM city WHERE county_id IN (SELECT county_id FROM county_public_safety WHERE Crime_rate  <  100)
Schema: CREATE TABLE city (county_id VARCHAR, name VARCHAR, Crime_rate INTEGER), CREATE TABLE county_public_safety (county_id VARCHAR, name VARCHAR, Crime_rate INTEGER)
user: 犯罪率低于100的省的城市名称是什么？

assistant: SELECT name FROM city WHERE county_id IN (SELECT county_id FROM county_public_safety WHERE Crime_rate  <  100)
Schema: CREATE TABLE county_public_safety (Population VARCHAR, Case_burden VARCHAR)
user: 按人口递减顺序显示省的人口负担。

assistant: SELECT Case_burden FROM county_public_safety ORDER BY Population DESC
Schema: CREATE TABLE county_public_safety (Population VARCHAR, Case_burden VARCHAR)
user: 按人口降序，每个省的人口负担是多少？

assistant: SELECT Case_burden FROM county_public_safety ORDER BY Population DESC
Schema: CREATE TABLE Rooms (decor VARCHAR, beds VARCHAR, roomName VARCHAR, basePrice INTEGER)
user: 找到所有底价低于160美元且包含两张床的现代风格的房间的名称。

assistant: SELECT roomName FROM Rooms WHERE basePrice  <  160 AND beds =  2 AND decor  =  '现代化';
Schema: CREATE TABLE Rooms (decor VARCHAR, beds VARCHAR, roomName VARCHAR, basePrice INTEGER)
user: 那些底价低于160美元并包含两张床的现代风格的房间叫什么名字？

assistant: SELECT roomName FROM Rooms WHERE basePrice  <  160 AND beds =  2 AND decor  =  '现代化';
Schema: CREATE TABLE Rooms (maxOccupancy INTEGER, RoomId VARCHAR, roomName VARCHAR, basePrice INTEGER)
user: 找到所有价格高于160且可以容纳2人以上的房间的房间名称和ID。

assistant: SELECT roomName ,  RoomId FROM Rooms WHERE basePrice  >  160 AND maxOccupancy  >  2;
Schema: CREATE TABLE Rooms (maxOccupancy INTEGER, RoomId VARCHAR, roomName VARCHAR, basePrice INTEGER)
user: 请问超过花费超过160且可以住两个人的所有房间的名称和ID是什么？

assistant: SELECT roomName ,  RoomId FROM Rooms WHERE basePrice  >  160 AND maxOccupancy  >  2;
Schema: CREATE TABLE Reservations (Room VARCHAR), CREATE TABLE Rooms (RoomId VARCHAR, roomName VARCHAR)
user: 找到酒店里最受欢迎即预订数量最多的房间。

assistant: SELECT T2.roomName FROM Reservations AS T1 JOIN Rooms AS T2 ON T1.Room  =  T2.RoomId GROUP BY T1.Room ORDER BY count(*) DESC LIMIT 1;
Schema: CREATE TABLE Reservations (Room VARCHAR), CREATE TABLE Rooms (RoomId VARCHAR, roomName VARCHAR)
user: 预订次数最多的房间是哪一间？

assistant: SELECT T2.roomName FROM Reservations AS T1 JOIN Rooms AS T2 ON T1.Room  =  T2.RoomId GROUP BY T1.Room ORDER BY count(*) DESC LIMIT 1;
Schema: CREATE TABLE Reservations (kids VARCHAR, FirstName VARCHAR, LastName VARCHAR)
user: 有多少孩子待在“罗”“锋”预定的房间里？

assistant: SELECT kids FROM Reservations WHERE FirstName = "罗" AND LastName  =  "锋";
Schema: CREATE TABLE Reservations (kids VARCHAR, FirstName VARCHAR, LastName VARCHAR)
user: 找出名字是“罗”“锋”的人预订的房间中孩子的数量。

assistant: SELECT kids FROM Reservations WHERE FirstName = "罗" AND LastName  =  "锋";
Schema: CREATE TABLE Reservations (FirstName VARCHAR, LastName VARCHAR)
user: “罗”“锋”曾预订过多少次房间。

assistant: SELECT count(*) FROM Reservations WHERE FirstName = "罗" AND LastName  =  "锋";
Schema: CREATE TABLE Reservations (FirstName VARCHAR, LastName VARCHAR)
user: 找出“罗”“锋”预订过房间的次数。

assistant: SELECT count(*) FROM Reservations WHERE FirstName = "罗" AND LastName  =  "锋";
Schema: CREATE TABLE Reservations (Rate VARCHAR, Room VARCHAR, CheckOut VARCHAR, CheckIn VARCHAR), CREATE TABLE Rooms (RoomId VARCHAR, roomName VARCHAR)
user: 列出房价最高的房间的全名、房价、入住和退房日期。

assistant: SELECT T2.roomName ,  T1.Rate ,  T1.CheckIn ,  T1.CheckOut FROM Reservations AS T1 JOIN Rooms AS T2 ON T1.Room  =  T2.RoomId GROUP BY T1.Room ORDER BY T1.Rate DESC LIMIT 1;