Question

Sql Query（Oracle DB）;

 select DT.MODULENO,DT.SLOTNO,DT.DID,MAX(DT.DEVICEKEY)as TIME 
    from DEVICETRACE DT
    INNER JOIN PCBTRACE PT on DT.DEVICEKEY=PT.DEVICEKEY 
    where SUBSTR(PT.PCBID,7,6)='019479' 
    group by DT.MODULENO,DT.SLOTNO,DT.DID



 MDN    SLNO    DID                TIME
    1   1   3017230000001705    20171003051515
    1   9   3017260000001437    20171002225453
    1   9   3017260000001438    20171003051515
    1   16  3017230000001681    20171002235012
    1   16  3017230000001680    20171003051515
    1   17  3017270000003284    20171003074307
    1   17  3017270000003249    20171003145129
    1   17  3017270000003253    20171003153600
    1   18  3017240000000274    20171003153600
    1   19  3017200000000148    20171003083715
    1   19  3017200000000147    20171003120846
    1   19  3017200000000146    20171003153420
    1   19  3017200000000145    20171003153600
    1   20  3017230000000876    20171003051515
    1   21  3017250000001389    20171003072136
    1   21  3017250000001388    20171003082830
    1   21  3017250000001395    20171003093001
    1   21  3017250000001408    20171003103917

如何执行以下结果。

MD  SLNO    BEFORE_DID           NOW_DID            TIME
1   1              -            3017230000001705    20171003051515
1   1       3017260000001437    3017260000001438    20171003051515
1   16      3017230000001681    3017230000001680    20171003051515
1   17      3017270000003249    3017270000003253    20171003145129
1   19      3017200000000147    3017200000000146    20171003153420
1   18             -            3017240000000274    20171003153600

我在上面的示例中想要的是将最后两个添加的MDNO和SLNO组的DID并排打印为两列。查询如何获取上面的代码，如上例所示？致谢

Answer 1

类似的东西：

WITH your_query ( MDN, SLNO, DID, time ) AS (
  -- your query here
)
SELECT *
FROM   ( SELECT MDN,
                SLNO,
                DID,
                MAX( time ) OVER ( PARTITION BY MDN, SLNO ) AS time,
                ROW_NUMBER() OVER ( PARTITION BY MDN, SLNO ORDER BY time DESC ) AS rn
         FROM   your_query )
PIVOT  ( MAX( DID ) FOR rn IN ( 2 AS Before_DID, 1 AS Now_DID ) );

sql group按订单时间

1 个答案: