SQL练习题

• sql exercise-1
• like/not like
• in/not in
• exists/not exists
• any/all
  举例：

t_book

t_booktype

t_price

SELECT * FROM t_book WHERE bookTypeId IN (SELECT id FROM t_booktype);

SELECT * FROM t_book WHERE price >=(SELECT price FROM t_price WHERE priceLevel=1);

#只有当后面有查询结果，也就是返回true的时候才会执行前面的查询
SELECT * FROM t_book WHERE EXISTS (SELECT * FROM t_booktype);

#大于等于任意一个即可，即大于等于最小值就可以
SELECT * FROM t_book WHERE price >= ANY(SELECT price FROM t_price);

#大于等于所有的，即大于等于最大值才可以
SELECT * FROM t_book WHERE price >= ALL(SELECT price FROM t_price);

别名

• 字段别名
  select id [as] '编号' from t_item（as 可省略）
• 表别名
  select it.id, it.name from t_item [as] it（as 可省略）
• 结果集别名
  select it.id, it.name from (select * from t_item where price>100) it
  面试的时候做的一道笔试题，有一张项目进度表project_progress，内容如下：

project_name(项目名称)	progress_name(进度)	date(日期)
项目1	申请	2019-1-1
项目2	申请	2019-1-5
项目1	预审	2019-1-10

使用SQL语句将其变成：

项目名称	申请	预审
项目1	2019-1-1	2019-1-10
项目2	2019-1-5

select s1.project_name as 项目名称, s1.date as 申请, s2.date as 预审 from
(
(select project_name, date from project_progress where progress_name = '申请') s1 
left join 
(select project_name, date from project_progress where progress_name = '预审') s2
on s1.project_name = s2.project_name
);

or

select s1.项目名称, 申请,  预审 from
(
(select project_name as 项目名称, date as 申请 from project_progress where progress_name = '申请') s1 
left join 
(select project_name as 项目名称, date as 预审 from project_progress where progress_name = '预审') s2
on s1.项目名称 = s2.项目名称
);