Question

表名：INCLog

UID Incident Number Modified Date   Modified By
1   INC000011193511 2/18/2018 12:04 zuck
1   INC000011193511 2/19/2018 1:42  shei
1   INC000011193511 3/14/2018 5:08  byrr
1   INC000011193511 3/17/2018 5:08  byrr
2   INC000011193513 1/23/2018 2:58  r070
2   INC000011193513 1/27/2018 9:27  r070
2   INC000011193513 2/21/2018 5:42  gont
2   INC000011193513 3/16/2018 6:06  r070
3   INC000011193514 2/1/2018 6:07   shei
3   INC000011193514 2/13/2018 6:07  r070
4   INC000011193515 2/20/2018 21:05 moha
4   INC000011193515 2/21/2018 8:05  moha
4   INC000011193515 3/15/2018 18:34 doss

如何使用Select语句返回过去5天内修改的事件编号？例如; INC000011193511最后一次更新于03/17/2018，该查询应该返回相同票证的整个日志

1   INC000011193511 2/18/2018 12:04 zuck
1   INC000011193511 2/19/2018 1:42  shei
1   INC000011193511 3/14/2018 5:08  byrr
1   INC000011193511 3/17/2018 5:08  byrr

提前感谢您的帮助！

Answer 1

使用最大窗口函数查找每个事件的上次修改日期。然后与当前日期进行比较。您的查询应该是：

select
    *
from (
    select
        *, mDate = max([modified date]) over (partition by [Incident Number])
    from
        myTable
) t
where
    mDate >= dateadd(dd, -5, getdate())

Answer 2

如果您只想要特定交易的最近5天交易，那么您可以使用以下查询。

SELECT * FROM INCLog where IncedentNumber ='INC000011193511' AND ModifiedDate = DATEADD(DAY, - 5,GETDATE () )

Answer 3

我建议这样做：

SELECT Incident_Number, MAX(Modified_Date) maxDate
FROM TempTabel
GROUP BY Incident_Number
HAVING MAX(Modified_Date)) > = dateadd(dd, -5, getdate())

谢谢！

Answer 4

如果我理解正确，您可以使用group by子句查找每个事件编号的最大日期，并将其与过去5天进行比较

SELECT t.* FROM table t
CROSS APPLY (
   SELECT Incident, MAX([Modified Date]) Mdate FROM table
   WHERE Incident = t.Incident
   GROUP BY Incident) c
WHERE c.Mdate >= DATEADD(DD, -5 , getdate())

根据前5天选择标准

4 个答案: