Question

ClaimID ClaimStatusCode StatusDate
11,     Closed,         2010-05-10 12:19:00.000
11,     Open,           2010-05-25 09:30:00.000
11,     Closed,         2011-06-01 00:00:00.000
11,     Open,           2011-06-10 00:00:00.000
22,     Closed,         2011-03-14 00:00:00.000
22,     Open,           2011-05-04 00:00:00.000
33,     Closed,         2007-12-19 17:19:00.000
33,     Open,           2007-12-24 12:07:00.000
**33,   Open,           2008-09-08 15:36:00.000**
44,     Closed,         2008-11-19 17:19:00.000
44,     Open,           2008-12-24 12:07:00.000
44,     Closed,         2009-07-28 15:36:00.000

对于任何ClaimID，如果打开声明状态，则不需要第二个打开状态行。这就是每次封闭都是由Open打开的。这里为ClaimID 33的最后一行，StatusDate 2008-09-08 15：36：00.000不是必需的（加星标以供参考）。我尝试使用rownumber函数，但没有运气。

答案应该是：

ClaimID ClaimStatusCode StatusDate 
11,     Closed,         2010-05-10 12:19:00.000
11,     Open,           2010-05-25 09:30:00.000
11,     Closed,         2011-06-01 00:00:00.000
11,     Open,           2011-06-10 00:00:00.000
22,     Closed,         2011-03-14 00:00:00.000
22,     Open,           2011-05-04 00:00:00.000
33,     Closed,         2007-12-19 17:19:00.000
33,     Open,           2007-12-24 12:07:00.000
44,     Closed,         2008-11-19 17:19:00.000
44,     Open,           2008-12-24 12:07:00.000
44,     Closed,         2009-07-28 15:36:00.000

Answer 1

我使用Cross Apply获取行的上一个状态，然后检查prev状态是否与当前状态相同：

delete ct
from
    claimtest ct
    cross apply (
        select top 1 prevstatus = ClaimStatusCode 
        from claimtest sub 
        where ct.claimid = sub.claimid and sub.statusdate < ct.statusdate 
        order by statusdate desc
    ) p
where 
    ct.ClaimStatusCode = p.prevstatus

Answer 2

    SELECT
       ClaimID, ClaimStatusCode, MIN(StatusDate)
    FROM
       Table
   GROUP BY
      ClaimID, ClaimStatusCode
   HAVING
      SUM(CASE claimID WHEN 'Open' THEN 1 ELSE -1 END) <= 1

编辑每个yoru评论 - 添加了HAVING条款

PS - 此选择为您获取“好”记录。如果你想改变它只获得“坏”记录（候选删除，我假设），然后使用MAX而不是MIN，并使用＆gt;有条款中的1。

删除连续的行

2 个答案: