Question

我想获取最近两个月所有员工的薪水差额。

预期输出：

我尝试过的查询：

SELECT tab1.name, tab1.emp_id 
FROM a_test tab1 join a_test tab2
on tab1.id = tab2.id
group by tab1.emp_id

Answer 1

通过使用join，正如您在过去两个月所说的那样，为什么我使用now()和month函数，以及月份列是否包含月份号

select t1.emp_id,t1.name,(t1.salary-t2.salary) as sal_diff from
    (
    select * 
    from a_test
    where month=Month(now())
    ) as t1
    left join 
    (
    select * from a_test
    where month=month(now())-1
      ) as t2
      on t1.emp_id=t2.emp_id

但是，如果您使用月份名称，则在月份列中使用MONTHNAME，查询将为

select t1.emp_id,t1.name,(t1.salary-t2.salary) as sal_diff from
        (
        select * 
        from a_test
        where month=MONTHNAME(STR_TO_DATE(month(now()), '%m'))
        ) as t1
        left join 
        (
        select * from a_test
        where month=MONTHNAME(STR_TO_DATE(month(now())-1, '%m'))
          ) as t2
          on t1.emp_id=t2.emp_id

Answer 2

正如我在评论中指出的那样，加入似乎是对此的最佳解决方案：

SELECT a.emp_id,
       a.salary-b.salary AS diff, 
       CONCAT(a.salary,' - ',b.salary) as diff_str
FROM a_test AS a
JOIN a_test AS b 
ON a.emp_id = b.emp_id AND a.month_id-1 = b.month_id
WHERE a.month_id = (SELECT MAX(month_id) FROM a_test)

我添加了最后一行，仅返回最近两个月的差额。我本可以做WHERE a.month_id = MONTH(NOW())，但我选择从数据表中选择上个月。请注意，这里的子查询并不是低效的，因为它只运行了一次。

结果是：

emp_id    diff    diff_str
1       -26000    27000 - 53000
2         9000    32000 - 23000

http://sqlfiddle.com/#!9/02bfb/11/0

https://dev.mysql.com/doc/refman/8.0/en/join.html

Answer 3

您可以尝试以下方法：

select (parent.salary - (select ch.salary from a_test as ch where ch.emp_id = parent.emp_id order by id desc limit 1,1)) as diff, emp_id, name, salary, month from a_test as parent where month = 'feb';

Sql Fiddle：http://sqlfiddle.com/#!9/52a019/2

您可以添加/更新主查询和子查询中的where条件，以动态获取月份。如下所示：

where month = MONTHNAME(CURRENT_DATE())

MySQL-过去两个月的员工薪金差异

3 个答案: