检测重叠日期并更新最新记录SQL Server 2008

Question

我希望能够检测所有记录（一些重复）并将重叠的记录（之后上传的记录）标记为OVER。为此，我有SELECT返回现有的重叠记录，CTE将此列设置为OVER。

我的问题是调整select查询以使用此值标记最新值并将其存储在cte中，因为我不熟悉SQL。

选择：

select t.* 
    from testtable t where exists 
    (select 1 from testtable t2 
    where t.idd = t2.idd
            AND t.id<>t2.id
            AND t2.beg <= t.end
            AND t.beg <= t2.end)

完成CTE的一半：

;with cte
        as (select t.*, Row_number() over (partition by idd order by date_uploaded desc) RN 
            from testtable as t)
    update cte set overlapped = 'OVER'
    where RN > 1
    and (overlapped is null or overlapped <> 'UNIQUE')

示例数据，它应该是什么样子：

overlapped ID   idd    iduser iddate name beg end date_uploaded
UNIQUE  52  -1907372231 666 201802  sol 2018-09-01  2018-09-10  2018-09-12 
OVER    53  -1907372231 666 201802  sol 2018-09-10  2018-09-12  2018-09-13 

注意第53行如何将BEG日期与END重叠

对我的问题有任何帮助，非常感谢。

Answer 1

当您从中选择时，请使用CASE表达式，而不是更新CTE：

;with cte
        as (select t.*, Row_number() over (partition by idd order by date_uploaded desc) RN 
            from testtable as t)
    SELECT CASE 
      WHEN RN > 1 and (overlapped is null or overlapped <> 'UNIQUE') THEN 'OVER'
      ELSE overlapped
    END AS overlapped
FROM cte