Question

我正在处理任务历史记录并尝试查找附加到同一记录的两个日期：1）最近一次任务被批准（最大批准）; 2）批准后的第一个提交日期。

这是我到目前为止所做的：

Select  
a.assn_uid,
max(b.ASSN_TRANS_DATE_ENTERED) as LastApprove, 
e.LastSubmitted 

FROM [PRJDEV_ProjectWebApp].[pub].[MSP_ASSIGNMENT_TRANSACTIONS] a
inner join [PRJDEV_ProjectWebApp].[pub].[MSP_ASSIGNMENT_TRANSACTIONS_COMMENTS] b
on a.ASSN_TRANS_UID = b.ASSN_TRANS_UID



join (select c.assn_uid,
min(d.ASSN_TRANS_DATE_ENTERED) as LastSubmitted

FROM [PRJDEV_ProjectWebApp].[pub].[MSP_ASSIGNMENT_TRANSACTIONS] c
inner join [PRJDEV_ProjectWebApp].[pub].[MSP_ASSIGNMENT_TRANSACTIONS_COMMENTS] d
on c.ASSN_TRANS_UID = d.ASSN_TRANS_UID

where c.ASSN_UID = '499879BC-28B2-E411-8B0A-00059A3C7A00'
and d.[ASSN_TRANS_COMMENT_TYPE_ENUM] = 0

group by c.assn_uid ) e
on e.ASSN_UID = a.ASSN_UID


where a.ASSN_UID = '499879BC-28B2-E411-8B0A-00059A3C7A00'
and b.[ASSN_TRANS_COMMENT_TYPE_ENUM] = 1

group by a.assn_uid, e.LastSubmitted

这很接近，但是，它第一次给我提交任务。我确信我需要使用另一个子查询，我只是不知道如何在同一结果中引用一个列。

这是任务历史。突出显示的是我试图展示的两个日期： enter image description here

Answer 1

我不知道我可以在任何合理的时间内浏览您的查询，但要在特定行之后获取行，您需要执行以下操作：

create table #submissions (
        ID int,
        DateAdded datetime,
        SubmissionType nvarchar(100)
    )
insert #submissions values
    (1, '2010-01-01', 'first ever'),
    (1, '2010-01-02', 'second'),
    (1, '2010-01-03', 'third'),
    (1, '2010-01-04', 'approve'),
    (1, '2010-01-05', 'first after approve'),
    (1, '2010-01-06', 'second after approve'),
    (1, '2010-01-07', 'third after approve')

declare @lastApprovalDate datetime

select @lastApprovalDate = MAX(DateAdded)
    from #submissions
    where
        SubmissionType = 'approve'

declare @firstAfterApprovalDate datetime
select @firstAfterApprovalDate = MIN(DateAdded)
    from #submissions
    where
        DateAdded > @lastApprovalDate

select *
    from #submissions
    where
        DateAdded = @firstAfterApprovalDate

drop table #submissions

一般情况下，使用MAX（）获取最后一个批准日期，然后使用MIN（）获取该日期之后的第一个日期，其中DateAdded＆gt;最大，然后选择该日期的行。我添加了Top 1，以防万一在那时碰巧有多行。不确定您的数据是否可行，但只是为了安全。

Answer 2

在一些帮助下，我们发现我们需要额外的一分钟来处理查询。

SELECT
       final.assn_uid,
       final.LastApprove,
       min(final.SubmissionDate) FirstSubmitted
FROM
       (Select  
       a.assn_uid,
       max(b.ASSN_TRANS_DATE_ENTERED) as LastApprove, 
       e.SubmittedDates SubmissionDate

       FROM [PRJDEV_ProjectWebApp].[pub].[MSP_ASSIGNMENT_TRANSACTIONS] a
       inner join [PRJDEV_ProjectWebApp].[pub].[MSP_ASSIGNMENT_TRANSACTIONS_COMMENTS] b
       on a.ASSN_TRANS_UID = b.ASSN_TRANS_UID



       join (select c.assn_uid,
       (d.ASSN_TRANS_DATE_ENTERED) as SubmittedDates

       FROM [PRJDEV_ProjectWebApp].[pub].[MSP_ASSIGNMENT_TRANSACTIONS] c
       inner join [PRJDEV_ProjectWebApp].[pub].[MSP_ASSIGNMENT_TRANSACTIONS_COMMENTS] d
       on c.ASSN_TRANS_UID = d.ASSN_TRANS_UID

       where c.ASSN_UID = '499879BC-28B2-E411-8B0A-00059A3C7A00'
       and d.[ASSN_TRANS_COMMENT_TYPE_ENUM] = 0

       ) e
       on e.ASSN_UID = a.ASSN_UID


       where a.ASSN_UID = '499879BC-28B2-E411-8B0A-00059A3C7A00'
       and b.[ASSN_TRANS_COMMENT_TYPE_ENUM] = 1
       and e.SubmittedDates > b.ASSN_TRANS_DATE_ENTERED

       group by a.assn_uid, e.SubmittedDates) Final

GROUP BY
       final.assn_uid,
       final.LastApprove

SQL：查找MIN但大于MAX

2 个答案: