Question

我有一张记录雇员工作班次的表格。它非常简单，包括（简化为简化）：自动增量primary key，employee_id，job_id，date

为每个轮班工作创建一行，因此如果某个特定员工在某一天没有工作，该员工在该日期将不会有任何行。

在某一天，我需要能够返回每个员工在当天工作的连续工作天数。

我不知道从哪里开始，并希望得到任何帮助。感谢

Answer 1

如果你有一系列数字，那么这些数字和一系列日期之间的差异将是不变的。我们可以使用此事实将工作日分组为具有最小和最大日期的组：

select employee_id, min(date) as mindate, max(date) as maxdate
from (select s.*, date_sub(date, interval seqnum days) as grp
      from (select s.*, (@rn := @rn + 1) as seqnum
            from shifts s cross join
                 (select @rn := 0) const
           ) s
     ) s
group by employee_id, grp;

一旦我们拥有了这个，那么将特定员工的特定日期延长到非常简单：

select employee_id,
       coalesce(datediff(XXX, 1+min(case when XXX between mindate and maxdate then mindate end)),
                0) as SequentialWorkingDays
from (select employee_id, min(date) as mindate, max(date) as maxdate
      from (select s.*, date_sub(date, interval seqnum days) as grp
            from (select s.*, (@rn := @rn + 1) as seqnum
                  from shifts s cross join
                       (select @rn := 0) const
                 ) s
           ) s
      group by employee_id, grp
     ) s
group by employee_id;

Answer 2

假设您的表名为shift，则以下查询将为您提供所提供日期和当前日期之间的行数。

SELECT COUNT(`date`) AS working_days 
FROM `shift` 
WHERE `date` BETWEEN '2014-01-02' AND CURDATE()

Answer 3

如果按排序的员工ID和日期的行号进行调整，您可以尝试计算实例，我认为它看起来像：

SELECT employee_id, job_id, date, days_worked = ROW_NUMBER() OVER (PARTITION BY employee_id, CAST(date) ORDER BY date)
FROM shifts

MySQL：计算具有导致给定日期的连续日期的行

3 个答案: