Question

以下是数据：

empID Date     Type
----- -------- ----
    1 1/1/2012  u
    1 1/2/2012  u
    1 1/3/2012  u
    1 2/2/2012  u
    4 1/1/2012  u
    4 1/3/2012  u
    4 1/4/2012  u
    4 1/6/2012  u

会回来：

empID count
----- -----
    1     2
    4     3

当两个日期“在一起”时，它们计为一次出现，如果日期被分开，则它们计为两次出现。这是为了跟踪员工的出勤情况......如何将SQL语句看起来按“合并”日期进行分组，并将它们计为1 ...我真的很难理解逻辑。

Answer 1

SELECT 
    empID
  , COUNT(*) AS cnt
FROM 
    tableX AS x
WHERE
    NOT EXISTS
        ( SELECT *
          FROM tableX AS y
          WHERE y.empID = x.empID
            AND DATEADD ("d", -1, x.[Date]) = y.[Date]
        )
GROUP BY
    empID ;

Answer 2

试试这个：

;WITH CTE as
(select *,ROW_NUMBER() over (partition by empID order by date) as rn from test2 t1)

select empID,COUNT(*) as count
from CTE c1
where isnull((DATEDIFF(day,(select date from CTE where c1.rn=rn+1 and empID=c1.empID ),c1.date)),0) <> 1
group by empID

SQL计数非连续日期

2 个答案: