Question

我有一张包含以下数据的表格：

 empid | officeid |     doj
-------+----------+------------
  19   |    112   | 2012-05-06
  26   |    112   | 2012-10-16
  35   |    112   | 2014-05-01
  17   |    112   | 2015-12-14
  19   |    169   | 2016-10-07
  26   |    146   | 2015-08-07
  42   |    102   | 2016-06-07
  35   |    135   | 2016-10-15
  26   |    112   | 2016-10-20

（doj =加入日期）。我需要在特定时期内获得一份员工的报告，其中包括在办公室工作的天数。例如，我想要 Office ID 112 的报告 2016年10月。所以结果应该是这样的：

 empid |  days
-------+-------
  35   |   14  (moved to another office on October 15th)
  17   |   31  (complete month)
  26   |   11  (October 20 to 31)
  42   |   31  (complete month)

员工19不应该来，因为他现在不在办公室112。我尝试了什么：

select sh.empid,datediff(
    greatest(last_day('2016-10-01'),
    (select doj from servicehistory where empid=sh.empid and date<sh.date
        order by date desc limit 1)
    ),date
) as days from servicehistory sh where sh.office='112' and date=(
    if(
        ((select count(date) from servicehistory where empid=sh.empid and
            date between '2016-10-01' and '2016-10-31')>0),
        (select date from servicehistory where empid=sh.empid and
            date between '2016-10-01' and '2016-10-31' limit 1),
        (select max(date) from servicehistory where empid=sh.empid and
            office=sh.office and date<=last_day('2016-10-01'))
    )
)

但它没有给出理想的结果。它提供的日期超过31天，员工目前不在办公室。

Answer 1

select (case when tmp.doj<='2016-10-01' then DATEDIFF(last_day('2016-10-01'),'2016-10-01')
else DATEDIFF(last_day('2016-10-01'),tmp.doj) end )as days,tmp.emp_id from (
select emp_id,max(doj)as doj from emp e group by emp_id)tmp

尝试此操作并根据您的要求进行更改。

MySQL - 日期范围内的工作天数

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