jtjt520j/cspider_for_chinese_llama2_1.3b huggingface dataset:
Schema: CREATE TABLE paperkeyphrase (keyphraseid VARCHAR, paperid VARCHAR), CREATE TABLE keyphrase (keyphraseid VARCHAR, keyphrasename VARCHAR), CREATE TABLE writes (authorid VARCHAR, paperid VARCHAR), CREATE TABLE paper (title VARCHAR, paperid VARCHAR), CREATE TABLE author (authorname VARCHAR, authorid VARCHAR)
user: “兰吉特·贾拉”的发表的关于“LH”的论文叫什么名字？

assistant: SELECT DISTINCT t3.title FROM paperkeyphrase AS t2 JOIN keyphrase AS t5 ON t2.keyphraseid  =  t5.keyphraseid JOIN writes AS t4 ON t4.paperid  =  t2.paperid JOIN paper AS t3 ON t4.paperid  =  t3.paperid JOIN author AS t1 ON t4.authorid  =  t1.authorid WHERE t1.authorname LIKE "%兰吉特·贾拉%" AND t5.keyphrasename  =  "LH";
Schema: CREATE TABLE business (name VARCHAR, rating INTEGER)
user: 列出超过4.5颗星的所有餐厅

assistant: SELECT name FROM business WHERE rating  >  4.5;
Schema: CREATE TABLE business (rating VARCHAR, name VARCHAR)
user: 列出所有评级为3.5的餐厅

assistant: SELECT name FROM business WHERE rating  =  3.5;
Schema: CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 列出名字是“米雪”的所有用户ID

assistant: SELECT user_id FROM USER WHERE name  =  "米雪";
Schema: CREATE TABLE business (name VARCHAR, state VARCHAR)
user: 查找所有有“汉堡王”的省份

assistant: SELECT state FROM business WHERE name  =  "汉堡王";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (name VARCHAR, business_id VARCHAR, city VARCHAR)
user: 找出有一个叫“海盗船自助餐”的“餐厅”的所有的城市。

assistant: SELECT t1.city FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.name  =  "海盗船自助餐" AND t2.category_name  =  "餐厅";
Schema: CREATE TABLE business (city VARCHAR, rating INTEGER)
user: 找出低于1.5的餐厅

assistant: SELECT city FROM business WHERE rating  <  1.5;
Schema: CREATE TABLE business (name VARCHAR, city VARCHAR)
user: 找出所有有“南京大牌档”的城市。

assistant: SELECT city FROM business WHERE name  =  "南京大牌档";
Schema: CREATE TABLE review (text VARCHAR, rating INTEGER)
user: 列出所有相关餐厅少于1个的评论

assistant: SELECT text FROM review WHERE rating  <  1;
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR, rating INTEGER)
user: 列出所有超过3.5的“餐馆”

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.rating  >  3.5 AND t2.category_name  =  "餐馆";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (name VARCHAR, business_id VARCHAR, city VARCHAR)
user: 找出所有拥有“南京大牌档”“餐厅”的城市

assistant: SELECT t1.city FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.name  =  "南京大牌档" AND t2.category_name  =  "餐厅";
Schema: CREATE TABLE USER (name VARCHAR, user_id VARCHAR), CREATE TABLE review (text VARCHAR, user_id VARCHAR)
user: 列出“王岩”的所有评论

assistant: SELECT t1.text FROM USER AS t2 JOIN review AS t1 ON t2.user_id  =  t1.user_id WHERE t2.name  =  "王岩";
Schema: CREATE TABLE review (business_id VARCHAR, user_id VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 列出所有由“王岩”评论的餐厅

assistant: SELECT t1.name FROM review AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN USER AS t3 ON t3.user_id  =  t2.user_id WHERE t3.name  =  "王岩";
Schema: CREATE TABLE review (rating VARCHAR, business_id VARCHAR, user_id VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 列出“王岩”评级5的所有业务

assistant: SELECT t1.name FROM review AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN USER AS t3 ON t3.user_id  =  t2.user_id WHERE t2.rating  =  5 AND t3.name  =  "王岩";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR), CREATE TABLE review (text VARCHAR, business_id VARCHAR, user_id VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 列出“米雪”为“意大利”“餐馆”所做的所有评论

assistant: SELECT t4.text FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN category AS t3 ON t3.business_id  =  t1.business_id JOIN review AS t4 ON t4.business_id  =  t1.business_id JOIN USER AS t5 ON t5.user_id  =  t4.user_id WHERE t2.category_name  =  "意大利" AND t3.category_name  =  "餐馆" AND t5.name  =  "米雪";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR, state VARCHAR), CREATE TABLE review (text VARCHAR, business_id VARCHAR)
user: 找出在“云南”的“星巴克”“餐厅”的评论数量

assistant: SELECT COUNT ( DISTINCT t3.text ) FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN review AS t3 ON t3.business_id  =  t1.business_id WHERE t1.name  =  "星巴克" AND t1.state  =  "云南" AND t2.category_name  =  "餐厅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (rating VARCHAR, business_id VARCHAR, name VARCHAR)
user: 列出5星级“意大利”“餐厅”

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN category AS t3 ON t3.business_id  =  t1.business_id WHERE t1.rating  =  5 AND t2.category_name  =  "意大利" AND t3.category_name  =  "餐厅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, city VARCHAR), CREATE TABLE neighbourhood (neighbourhood_name VARCHAR, business_id VARCHAR)
user: 列出在“成都”的“意大利”“餐厅”的街区

assistant: SELECT t1.neighbourhood_name FROM category AS t3 JOIN business AS t2 ON t3.business_id  =  t2.business_id JOIN category AS t4 ON t4.business_id  =  t2.business_id JOIN neighbourhood AS t1 ON t1.business_id  =  t2.business_id WHERE t2.city  =  "成都" AND t3.category_name  =  "意大利" AND t4.category_name  =  "餐厅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, city VARCHAR, rating INTEGER), CREATE TABLE neighbourhood (neighbourhood_name VARCHAR, business_id VARCHAR)
user: 列出“成都”所有拥有评级低于2.5的“意大利”“餐厅”的街区

assistant: SELECT t1.neighbourhood_name FROM category AS t3 JOIN business AS t2 ON t3.business_id  =  t2.business_id JOIN category AS t4 ON t4.business_id  =  t2.business_id JOIN neighbourhood AS t1 ON t1.business_id  =  t2.business_id WHERE t2.city  =  "Madison" AND t2.rating  <  2.5 AND t3.category_name  =  "意大利" AND t4.category_name  =  "餐厅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR, state VARCHAR)
user: 找出“黑龙江”所有的“餐馆”

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.state  =  "黑龙江" AND t2.category_name  =  "餐馆";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR, state VARCHAR)
user: 列出所有在“黑龙江”的“餐馆”。

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.state  =  "黑龙江" AND t2.category_name  =  "餐馆";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, review_count INTEGER), CREATE TABLE review (text VARCHAR, business_id VARCHAR)
user: 找出所有超过100个评论的“宠物梳理师”

assistant: SELECT t3.text FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN review AS t3 ON t3.business_id  =  t1.business_id WHERE t1.review_count  >  100 AND t2.category_name  =  "宠物梳理师";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR)
user: 在“东莞”的“啤酒厂”有哪些？

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.city  =  "东莞" AND t2.category_name  =  "啤酒厂";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR)
user: 找出“东莞”的所有“啤酒厂”

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.city  =  "东莞" AND t2.category_name  =  "啤酒厂";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR)
user: 中找处在“厦门”所有的“啤酒厂”

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.city  =  "厦门" AND t2.category_name  =  "啤酒厂";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR), CREATE TABLE review (business_id VARCHAR, user_id VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 找出所有评论“沙县小吃”“餐厅”的用户

assistant: SELECT t4.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN review AS t3 ON t3.business_id  =  t1.business_id JOIN USER AS t4 ON t4.user_id  =  t3.user_id WHERE t1.name  =  "沙县小吃" AND t2.category_name  =  "餐厅";
Schema: CREATE TABLE business (name VARCHAR, city VARCHAR, full_address VARCHAR)
user: 列出在“东莞”所有“沃尔玛”的地址

assistant: SELECT full_address FROM business WHERE city  =  "东莞" AND name  =  "沃尔玛";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR), CREATE TABLE review (business_id VARCHAR, user_id VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 找出“赵毅”评论的所有在“重庆”的“餐厅”

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN review AS t3 ON t3.business_id  =  t1.business_id JOIN USER AS t4 ON t4.user_id  =  t3.user_id WHERE t1.city  =  "重庆" AND t2.category_name  =  "餐厅" AND t4.name  =  "赵毅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR), CREATE TABLE review (business_id VARCHAR, user_id VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: “重庆”的哪些“餐厅”被“赵毅”评论过？

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN review AS t3 ON t3.business_id  =  t1.business_id JOIN USER AS t4 ON t4.user_id  =  t3.user_id WHERE t1.city  =  "重庆" AND t2.category_name  =  "餐厅" AND t4.name  =  "赵毅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR), CREATE TABLE review (business_id VARCHAR, user_id VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 找出“赵毅”评论的所有的“酒吧”

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN review AS t3 ON t3.business_id  =  t1.business_id JOIN USER AS t4 ON t4.user_id  =  t3.user_id WHERE t2.category_name  =  "酒吧" AND t4.name  =  "赵毅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR, rating INTEGER), CREATE TABLE review (business_id VARCHAR, user_id VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 找出所有由“赵毅”评论的至少有3颗星的“酒吧”。

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN review AS t3 ON t3.business_id  =  t1.business_id JOIN USER AS t4 ON t4.user_id  =  t3.user_id WHERE t1.rating  >=  3 AND t2.category_name  =  "酒吧" AND t4.name  =  "赵毅";
Schema: CREATE TABLE tip (year VARCHAR, business_id VARCHAR, user_id VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 找出在2015年所有为“喜茶”写过评论的用户

assistant: SELECT t3.name FROM tip AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN USER AS t3 ON t3.user_id  =  t2.user_id WHERE t1.name  =  "喜茶" AND t2.year  =  2015;
Schema: CREATE TABLE business (state VARCHAR, name VARCHAR, rating INTEGER)
user: 在“云南”找出所有低于2分的企业

assistant: SELECT name FROM business WHERE rating  <  2 AND state  =  "云南";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR)
user: 找出所有在“厦门”的“海鲜”“餐厅”

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN category AS t3 ON t3.business_id  =  t1.business_id WHERE t1.city  =  "厦门" AND t2.category_name  =  "海鲜" AND t3.category_name  =  "餐厅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR)
user: 列出“厦门”所有的“海鲜”“餐厅”

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN category AS t3 ON t3.business_id  =  t1.business_id WHERE t1.city  =  "厦门" AND t2.category_name  =  "海鲜" AND t3.category_name  =  "餐厅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR)
user: 找出“厦门”供应“海鲜”的“餐厅”

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN category AS t3 ON t3.business_id  =  t1.business_id WHERE t1.city  =  "厦门" AND t2.category_name  =  "海鲜" AND t3.category_name  =  "餐厅";
Schema: CREATE TABLE USER (name VARCHAR, user_id VARCHAR), CREATE TABLE review (text VARCHAR, rating INTEGER, user_id VARCHAR)
user: 找出评分高于4分的“赵毅”的所有评论，

assistant: SELECT t1.text FROM USER AS t2 JOIN review AS t1 ON t2.user_id  =  t1.user_id WHERE t1.rating  >  4 AND t2.name  =  "赵毅";
Schema: CREATE TABLE business (name VARCHAR, business_id VARCHAR, city VARCHAR)
user: 在“厦门”中找出所有“苹果店”

assistant: SELECT business_id FROM business WHERE city  =  "厦门" AND name  =  "苹果店";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR, rating INTEGER)
user: 找出评级高于4.5分的“成都”“餐厅”

assistant: SELECT t1.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.city  =  "成都" AND t1.rating  >  4.5 AND t2.category_name  =  "餐厅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR), CREATE TABLE neighbourhood (neighbourhood_name VARCHAR, business_id VARCHAR)
user: “海底捞”附近的“餐馆”是什么？

assistant: SELECT t1.neighbourhood_name FROM category AS t3 JOIN business AS t2 ON t3.business_id  =  t2.business_id JOIN neighbourhood AS t1 ON t1.business_id  =  t2.business_id WHERE t2.name  =  "海底捞" AND t3.category_name  =  "餐馆";
Schema: CREATE TABLE tip (text VARCHAR, business_id VARCHAR, likes INTEGER), CREATE TABLE business (business_id VARCHAR, name VARCHAR)
user: 找出关于收到超过9个赞的“便宜坊”的所有提示

assistant: SELECT t2.text FROM tip AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.name  =  "便宜坊" AND t2.likes  >  9;
Schema: CREATE TABLE review (year VARCHAR, text VARCHAR, business_id VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR)
user: 找出2014年度“探鱼”的所有评论

assistant: SELECT t2.text FROM review AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.name  =  "探鱼" AND t2.year  =  2014;
Schema: CREATE TABLE tip (business_id VARCHAR, user_id VARCHAR), CREATE TABLE business (business_id VARCHAR, city VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 找出所有写在“成都”的餐馆的评论的用户

assistant: SELECT t3.name FROM tip AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN USER AS t3 ON t3.user_id  =  t2.user_id WHERE t1.city  =  "成都";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (name VARCHAR, business_id VARCHAR, city VARCHAR, state VARCHAR)
user: 在“云南”省的所有城市里找出一家名为“海盗船自助餐”的“餐厅”。

assistant: SELECT t1.city FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.name  =  "海盗船自助餐" AND t1.state  =  "云南" AND t2.category_name  =  "餐厅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR), CREATE TABLE tip (business_id VARCHAR, user_id VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 找出“宠物梳理师”评论的用户

assistant: SELECT t4.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN tip AS t3 ON t3.business_id  =  t1.business_id JOIN USER AS t4 ON t4.user_id  =  t3.user_id WHERE t2.category_name  =  "宠物梳理师";
Schema: CREATE TABLE tip (text VARCHAR, business_id VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR, state VARCHAR)
user: 找出“云南”“星巴克”的所有评论。

assistant: SELECT t2.text FROM tip AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.name  =  "星巴克" AND t1.state  =  "云南";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR), CREATE TABLE review (business_id VARCHAR, user_id VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 列出所有审查“餐馆”业务的用户。

assistant: SELECT t4.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN review AS t3 ON t3.business_id  =  t1.business_id JOIN USER AS t4 ON t4.user_id  =  t3.user_id WHERE t2.category_name  =  "餐馆";
Schema: CREATE TABLE tip (year VARCHAR, text VARCHAR, business_id VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR, state VARCHAR)
user: 列出2010在“黑龙江”的“星巴克”的所有提示。

assistant: SELECT t2.text FROM tip AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.name  =  "星巴克" AND t1.state  =  "黑龙江" AND t2.year  =  2010;
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR), CREATE TABLE review (year VARCHAR, business_id VARCHAR, user_id VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: 列出所有在2010的做“餐馆”生意的用户。

assistant: SELECT t4.name FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN review AS t3 ON t3.business_id  =  t1.business_id JOIN USER AS t4 ON t4.user_id  =  t3.user_id WHERE t2.category_name  =  "餐馆" AND t3.year  =  2010;
Schema: CREATE TABLE USER (user_id VARCHAR), CREATE TABLE review (year VARCHAR, user_id VARCHAR), CREATE TABLE tip (text VARCHAR, user_id VARCHAR)
user: 找出在2012编写评论的用户的所有提示

assistant: SELECT t2.text FROM USER AS t3 JOIN review AS t1 ON t3.user_id  =  t1.user_id JOIN tip AS t2 ON t3.user_id  =  t2.user_id WHERE t1.year  =  2012;
Schema: CREATE TABLE review (text VARCHAR, business_id VARCHAR), CREATE TABLE business (rating VARCHAR, business_id VARCHAR)
user: 查找所有评分为2.5的企业的评论

assistant: SELECT t2.text FROM review AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.rating  =  2.5;
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR)
user: 在“成都”找出“逃生游戏”的数量

assistant: SELECT COUNT ( DISTINCT t1.name ) FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.city  =  "成都" AND t2.category_name  =  "逃生游戏";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR)
user: “成都”“逃生游戏”的数量是多少？

assistant: SELECT COUNT ( DISTINCT t1.name ) FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.city  =  "成都" AND t2.category_name  =  "逃生游戏";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR)
user: “成都”有多少“逃生游戏”

assistant: SELECT COUNT ( DISTINCT t1.name ) FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.city  =  "成都" AND t2.category_name  =  "逃生游戏";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR)
user: 在“成都”中“逃生游戏”的数量是多少？

assistant: SELECT COUNT ( DISTINCT t1.name ) FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.city  =  "成都" AND t2.category_name  =  "逃生游戏";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (city VARCHAR, business_id VARCHAR, name VARCHAR)
user: “成都”有多少家“逃生游戏”？

assistant: SELECT COUNT ( DISTINCT t1.name ) FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.city  =  "成都" AND t2.category_name  =  "逃生游戏";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR, rating INTEGER)
user: 找出评分超过3.5的“餐厅”

assistant: SELECT COUNT ( DISTINCT t1.name ) FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.rating  >  3.5 AND t2.category_name  =  "餐厅";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, city VARCHAR), CREATE TABLE checkin (business_id VARCHAR, count INTEGER)
user: 找出在“厦门”的“摩洛哥”“餐厅”总访问数

assistant: SELECT SUM ( t4.count ) FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN category AS t3 ON t3.business_id  =  t1.business_id JOIN checkin AS t4 ON t4.business_id  =  t1.business_id WHERE t1.city  =  "厦门" AND t2.category_name  =  "餐厅" AND t3.category_name  =  "摩洛哥";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, city VARCHAR), CREATE TABLE checkin (business_id VARCHAR, day VARCHAR, count INTEGER)
user: 找出“星期五”在“厦门”“摩洛哥”“餐厅”的总访问数

assistant: SELECT SUM ( t4.count ) FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN category AS t3 ON t3.business_id  =  t1.business_id JOIN checkin AS t4 ON t4.business_id  =  t1.business_id WHERE t1.city  =  "厦门" AND t2.category_name  =  "摩洛哥" AND t3.category_name  =  "餐厅" AND t4.day  =  "星期五";
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, city VARCHAR), CREATE TABLE checkin (business_id VARCHAR, day VARCHAR, count INTEGER)
user: 找出每天在“厦门”“摩洛哥”“餐厅”的总访问数

assistant: SELECT t4.day  ,  SUM ( t4.count ) FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN category AS t3 ON t3.business_id  =  t1.business_id JOIN checkin AS t4 ON t4.business_id  =  t1.business_id WHERE t1.city  =  "厦门" AND t2.category_name  =  "摩洛哥" AND t3.category_name  =  "餐厅" GROUP BY t4.day;
Schema: CREATE TABLE category (business_id VARCHAR, category_name VARCHAR), CREATE TABLE business (business_id VARCHAR, state VARCHAR), CREATE TABLE checkin (business_id VARCHAR, day VARCHAR, count INTEGER)
user: 找出“星期日”在每一个省的“意大利”“餐厅”的总访问数

assistant: SELECT t1.state  ,  SUM ( t4.count ) FROM category AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN category AS t3 ON t3.business_id  =  t1.business_id JOIN checkin AS t4 ON t4.business_id  =  t1.business_id WHERE t2.category_name  =  "意大利" AND t3.category_name  =  "餐馆" AND t4.day  =  "星期日" GROUP BY t1.state;
Schema: CREATE TABLE USER (name VARCHAR, user_id VARCHAR), CREATE TABLE review (year VARCHAR, text VARCHAR, user_id VARCHAR)
user: 在2015年“王岩”写了多少评论

assistant: SELECT COUNT ( DISTINCT t1.text ) FROM USER AS t2 JOIN review AS t1 ON t2.user_id  =  t1.user_id WHERE t1.year  =  2015 AND t2.name  =  "王岩";
Schema: CREATE TABLE USER (name VARCHAR, user_id VARCHAR), CREATE TABLE review (user_id VARCHAR, rating INTEGER)
user: “米雪”的评论中的平均评级是多少？

assistant: SELECT AVG ( t1.rating ) FROM USER AS t2 JOIN review AS t1 ON t2.user_id  =  t1.user_id WHERE t2.name  =  "米雪";
Schema: CREATE TABLE checkin (count VARCHAR, business_id VARCHAR, day VARCHAR), CREATE TABLE business (business_id VARCHAR, name VARCHAR)
user: “星巴克”在“星期五”的访问数数是多少？

assistant: SELECT t2.count FROM checkin AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id WHERE t1.name  =  "星巴克" AND t2.day  =  "星期五";
Schema: CREATE TABLE review (business_id VARCHAR, user_id VARCHAR), CREATE TABLE business (name VARCHAR, business_id VARCHAR, city VARCHAR), CREATE TABLE USER (name VARCHAR, user_id VARCHAR)
user: “杭州”有多少网友评论“外婆家”

assistant: SELECT COUNT ( DISTINCT t3.name ) FROM review AS t2 JOIN business AS t1 ON t2.business_id  =  t1.business_id JOIN USER AS t3 ON t3.user_id  =  t2.user_id WHERE t1.city  =  "杭州" AND t1.name  =  "外婆家";