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  • 求記錄中的最新數據的方法!

    發表于:2007-05-25來源:作者:點擊數: 標簽:方法已知記錄中的最新
    已知數據: Column1 Column2 Column3 Column4 A 10 am 1999-01-01 00:00:00.000 A 11 am 1999-01-02 00:00:00.000 B 12 bm 1999-01-03 00:00:00.000 B 13 bm 1999-01-04 00:00:00.000 C 14 cm 1999-01-05 00:00:00.000 C 15 cm 1999-01-06 00:00:00.000 要求

    已知數據:

    Column1    Column2     Column3    Column4
    A                      10                     am         1999-01-01 00:00:00.000
    A                      11                     am         1999-01-02 00:00:00.000
    B                     12                      bm         1999-01-03 00:00:00.000
    B                     13                      bm         1999-01-04 00:00:00.000
    C                    14                      cm         1999-01-05 00:00:00.000
    C                     15                     cm         1999-01-06 00:00:00.000

    要求得到數據:

    Column1    Column2     Column3    Column4
    A                      11                     am         1999-01-02 00:00:00.000
    B                     13                      bm         1999-01-04 00:00:00.000
    C                     15                     cm         1999-01-06 00:00:00.000

    --數據裝載

    Create Table #T(Column1 varchar(10),Column2 int,Column3 varchar(10),Column4 datetime)
    insert #T select 'A',10,'am','1999-1-1'
    union all select 'A',11,'am','1999-1-2'
    union all select 'B',12,'bm','1999-1-3'
    union all select 'B',13,'bm','1999-1-4'
    union all select 'C',14,'cm','1999-1-5'
    union all select 'C',15,'cm','1999-1-6'

    --測試語句 方法1:

    select a.* from #T a
    where (a.Column4) =
       (select top 1 (Column4) from #T where Column1 = a.Column1 order by Column4 desc)

    --測試結果:

    Column1    Column2     Column3    Column4                                               
    ---------- ----------- ---------- ------------------------------------------------------
    A          11          am         1999-01-02 00:00:00.000
    B          13          bm         1999-01-04 00:00:00.000
    C          15          cm         1999-01-06 00:00:00.000

    --測試語句 方法2:(最佳效率)

    select a.* from #T a join(
     select Column1,Column4=max(Column4) from #T group by Column1
    )b on a.Column1=b.Column1 and a.Column4=b.Column4
    order by a.Column1

    --測試結果 方法2:

    Column1    Column2     Column3    Column4                                               
    ---------- ----------- ---------- ------------------------------------------------------
    A          11          am         1999-01-02 00:00:00.000
    B          13          bm         1999-01-04 00:00:00.000
    C          15          cm         1999-01-06 00:00:00.000


    原文轉自:http://www.kjueaiud.com

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