mysql 中如何实现这样的查询 - V2EX
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lusheldon
V2EX    MySQL

mysql 中如何实现这样的查询

  •  
  •   lusheldon 2017-06-21 21:56:10 +08:00 3869 次点击
    这是一个创建于 3038 天前的主题,其中的信息可能已经有所发展或是发生改变。
    mysql>desc salaries
    +-----------+---------+------+-----+---------+-------+
    | Field | Type | Null | Key | Default | Extra |
    +-----------+---------+------+-----+---------+-------+
    | emp_no | int(11) | NO | PRI | NULL | |
    | salary | int(11) | NO | | NULL | |
    | from_date | date | NO | PRI | NULL | |
    | to_date | date | NO | | NULL | |
    +-----------+---------+------+-----+---------+-------+

    某员工的信息
    +--------+--------+------------+------------+
    | emp_no | salary | from_date | to_date |
    +--------+--------+------------+------------+
    | 111400 | 72446 | 1985-01-01 | 1986-01-01 |
    | 111400 | 100416 | 1999-12-29 | 2000-12-28 |
    | 111400 | 99971 | 2000-12-28 | 2001-12-28 |
    | 111400 | 103244 | 2001-12-28 | 9999-01-01 |
    +--------+--------+------------+------------+


    查询所有曾经降过薪的员工 emp_no 以及降薪幅度
    10 条回复    2017-06-22 09:20:19 +08:00
    hustlike
        1
    hustlike  
       2017-06-21 21:59:16 +08:00
    降过多次怎么算
    lusheldon
        2
    lusheldon  
    OP
       2017-06-21 22:02:19 +08:00 via Android
    @hustlike 降多次的情况下,应该也是可以查的吧。咱们先假设每个人最多可能降薪一次吧。
    reus
        3
    reus  
       2017-06-21 22:06:59 +08:00   1
    select
    distinct emp_no
    from salaries a
    join salaries b
    on a.emp_no = b.emp_no and a.to_date > b.to_date and a.salary < b.salary
    lusheldon     4
    lusheldon  
    OP
       2017-06-21 22:09:57 +08:00 via Android
    @reus 这个可能就是我要的答案,不熟悉数据库,没想过可以自己 join 自己。感谢!
    reus
        5
    reus  
       2017-06-21 22:14:43 +08:00
    @lusheldon 但这个降薪幅度要计算的话,还得判断一下 a 和 b 的时间段是不是相邻的
    lusheldon
        6
    lusheldon  
    OP
       2017-06-21 22:19:54 +08:00 via Android
    @reus 加上条件 a.from_date=b.to_date 就能判断了吧。
    reus
        7
    reus  
       2017-06-21 22:23:31 +08:00   1
    @lusheldon 如果没有断的话,是的
    lusheldon
        8
    lusheldon  
    OP
       2017-06-21 22:27:04 +08:00
    select a.emp_no,b.salary-a.salary, a.to_date
    from salaries a join salaries b
    on a.emp_no=b.emp_no and a.to_date>b.to_date and a.salary<b.salary and a.from_date=b.to_date;
    奉上完整答案,再次感谢 @reus
    wwww961h
        9
    wwww961h  
       2017-06-21 22:29:23 +08:00
    用 join,把一个表当两个表用,特别简单
    sohucw
        10
    sohucw  
       2017-06-22 09:20:19 +08:00
    select a.emp_no,b.salary-a.salary, a.to_date
    from salaries a join salaries b
    on a.emp_no=b.emp_no and a.to_date>b.to_date and a.salary<b.salary and a.from_date=b.to_date; 正解啊
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