sql-server-2008 - 为每个 GROUP BY 值指定一个序号

Question

我有一个包含很多行的临时表（#TümDATA）。我是带有子句INSERT的另一个临时表（#GrupTOT）中的行。GROUP BY但是我被困在这里，我需要在它们被分组后给它们一个序号。

这是我的 SQL：

INSERT INTO #GrupTOT(AY, BLK, DRE, TOT) 
SELECT J.AY, J.BLK, J.DRE, SUM(J.BORÇ) 
FROM #TümDATA J 
GROUP BY J.AY, J.BLK, J.DRE

Answer 1

您可以使用ROW_NUMBER来获取基于ORDER BY. IDENTITY或者，如果这是您想要的，您可以添加一列以在插入时自动增加一个数字。

ROW_NUMBER方法：

WITH CTE AS
(
    SELECT Col1, Col2, Col3, Count(*) as [COUNT]
    FROM dbo.Table1
    GROUP BY Col1, Col2, Col3
)
INSERT INTO dbo.Table2
    SELECT RowNum = ROW_NUMBER() OVER ( ORDER BY Col1, Col2, Col3, [COUNT] DESC ),
           Col1, Col2, Col3, [COUNT]
    FROM CTE

Answer 2

试试这个：（在 Oracle 中工作）

WITH ORDERS
    AS (SELECT
             TO_DATE ( '2013-09-18 00:00:01',
                     'YYYY-MM-DD HH24:MI:SS' )
                 AS THE_DATE,
             'John' AS NAME
        FROM
             DUAL
        UNION ALL
        SELECT
             TO_DATE ( '2013-09-19 00:00:01',
                     'YYYY-MM-DD HH24:MI:SS' )
                 AS THE_DATE,
             'James' AS NAME
        FROM
             DUAL
        UNION ALL
        SELECT
             TO_DATE ( '2013-09-20 00:00:01',
                     'YYYY-MM-DD HH24:MI:SS' )
                 AS THE_DATE,
             'John' AS NAME
        FROM
             DUAL
        UNION ALL
        SELECT
             TO_DATE ( '2013-09-20 00:00:02',
                     'YYYY-MM-DD HH24:MI:SS' )
                 AS THE_DATE,
             'John' AS NAME
        FROM
             DUAL
        UNION ALL
        SELECT
             TO_DATE ( '2013-09-20 00:00:03',
                     'YYYY-MM-DD HH24:MI:SS' )
                 AS THE_DATE,
             'John' AS NAME
        FROM
             DUAL
        UNION ALL
        SELECT
             TO_DATE ( '2013-09-20 00:00:04',
                     'YYYY-MM-DD HH24:MI:SS' )
                 AS THE_DATE,
             'John' AS NAME
        FROM
             DUAL
        UNION ALL
        SELECT
             TO_DATE ( '2013-09-21 16:00:01',
                     'YYYY-MM-DD HH24:MI:SS' )
                 AS THE_DATE,
             'Jennifer' AS NAME
        FROM
             DUAL)
SELECT
      THE_DATE,
      NAME,
      ROWNUM
FROM
      (SELECT
            TRUNC ( THE_DATE ) THE_DATE,
            NAME,
            COUNT ( 1 )
       FROM
            ORDERS
       GROUP BY
            TRUNC ( THE_DATE ),
            NAME);

原始数据：

9/18/2013 12:00:01 AM   John
9/19/2013 12:00:01 AM   James
9/20/2013 12:00:01 AM   John
9/20/2013 12:00:02 AM   John
9/20/2013 12:00:03 AM   John
9/20/2013 12:00:04 AM   John
9/21/2013 4:00:01 PM    Jennifer

结果：

9/21/2013   Jennifer    1
9/19/2013   James           2
9/20/2013   John            3
9/18/2013   John            4

Answer 3

您可以使用该ROW_NUMBER()功能。

例如

WITH q AS (
    SELECT field_a
    FROM SomeTable
    GROUP BY field_a
)
SELECT ROW_NUMBER() OVER(ORDER BY field_a) AS row_num,
       field_a
FROM q

Answer 4

您可以尝试SELECT INTO使用IDENTITY列。这将创建新的临时表#GrupTOT。

这是一个小提琴示例。

SELECT SeqNo = identity(int,1,1), --Identity column
       AY = J.AY
       BLK = J.BLK, 
       DRE = J.DRE,
       TOT = SUM(J.BORÇ) 
INTO #GrupTOT
FROM #TümDATA J 
GROUP BY J.AY, J.BLK, J.DRE;

--SELECT * FROM #GrupTOT