Question

不确定如何显示我想要的结果图表，但请参阅下面我尝试做的事情。我试图创建一个rcnew列来取得rc所在的位置＆＃39; 098000＆＃39;或者＆＃39; 999998＆＃39;并将其从最接近的先前工资日转换为rc，其中rc不在＆＃39; 098000＆＃39;或者＆＃39; 999998＆＃39;。正如您所看到的，我的代码在paydate相同的位置，因为我无法弄清楚如何解决这个问题。如果没有平等的工资，我想采取一个明确的。下面结果中的NULL应为200000。

ern rc      paydate     rcnew
123 098000  20151221    NULL
123 200000  20151214    200000
456 098000  20151221    200000
456 200000  20151221    200000



   SELECT
    ern, rc, paydate, 
    CASE WHEN fss1.rc in ('098000', '999998') 
         THEN (SELECT TOP 1 rc
               FROM [FIN_DataMart].[dbo].[FSSpaydetl] fss
               WHERE fss.ern = fss1.ern
               AND rc not in ('098000', '999998')
               AND fss.paydate = fss1.paydate
               ORDER BY paydate DESC)
         ELSE fss1.rc
    END rcnew
FROM [FIN_DataMart].[dbo].[FSSpaydetl] fss1

GROUP BY ern, rc, paydate
ORDER BY ern, paydate desc

Answer 1

如果一行中有多个特殊值，则

lag()将无效。另一种选择是outer apply：

select fss.ern, rss.rc, fss.paydate,
       (case when fss.rc in ('098000', '999998') then fss2.rc else fss.rc end)
from FIN_DataMart].[dbo].[FSSpaydetl] fss outer apply
     (select top 1 t2.*
      from FIN_DataMart].[dbo].[FSSpaydetl] fss fss2
      where fss2.paydate < fss2.paydate and
            fss2.ern = fss.ern and
            fss2.rc not in ('098000', '999998')
      order by fss2.paydate desc
     ) fss2;

Answer 2

您可以使用LAG功能，如果您的意思是“rc来自最近的paydate”，但正如@GordonLinoff正确设计的那样，您实际上正在寻找“{ {1}}距离最近的rc paydate不是'098000', '999998'，此解决方案不起作用。

DECLARE @Table AS TABLE (enc INT, rc CHAR(6), paydate DATE)
INSERT INTO @Table (enc, rc, paydate)
VALUES (123,'098000','20151221')
,(123,'200000','20151214')
,(456,'098000','20151221')
,(456,'200000','20151221')

SELECT
    *
    ,LagResult = LAG(rc,1,0) OVER (ORDER BY paydate)
    ,rcnew = CASE
       WHEN rc IN ('098000','999998') THEN LAG(rc,1,0) OVER (ORDER BY paydate)
       ELSE rc
    END
FROM
    @Table
ORDER BY
    paydate

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2 个答案: