使用CTE更新上一行值的值

时间：2012-09-25 14:59:25

标签： sql sql-server-2008 tsql

这个问题是我的旧问题here的变体。我希望用一个例子来解释这个问题。所以

示例数据
以下是要使用的示例数据：

DECLARE @Test TABLE (GID      int,             Seq     int, 
                     IsLive   bit,             Eff     date, 
                     Name     varchar(50),     Salary  decimal) 

INSERT INTO @Test VALUES (1, 1, 1, '01-08-2012', 'RTS', NULL)
INSERT INTO @Test VALUES (1, 2, 0, '01-09-2012', 'RTA', NULL)
INSERT INTO @Test VALUES (1, 3, 1, '01-10-2012', 'FSA', NULL)
INSERT INTO @Test VALUES (1, 4, 0, '01-11-2012',  NULL, NULL)
INSERT INTO @Test VALUES (1, 5, 1, '01-12-2012', 'FSA', NULL)
INSERT INTO @Test VALUES (2, 1, 1, '01-08-2012', 'RTS', NULL)
INSERT INTO @Test VALUES (2, 2, 0, '01-09-2012', 'RTA', NULL)
INSERT INTO @Test VALUES (2, 3, 1, '01-10-2012', 'FSA', NULL)
INSERT INTO @Test VALUES (2, 4, 0, '01-11-2012', 'GSM', NULL)
INSERT INTO @Test VALUES (2, 5, 1, '01-12-2012', 'FSA', NULL)
INSERT INTO @Test VALUES (3, 1, 1, '01-01-2012', 'FSA', NULL)
INSERT INTO @Test VALUES (3, 2, 0, '01-02-2012', NULL, NULL)
INSERT INTO @Test VALUES (4, 1, 1, '01-01-2012', NULL, NULL)
INSERT INTO @Test VALUES (4, 2, 0, '01-02-2012', 'FSA', NULL)
INSERT INTO @Test VALUES (4, 3, 0, '01-03-2012', NULL, NULL)
INSERT INTO @Test VALUES (5, 1, 0, '01-01-2012', NULL, NULL)
INSERT INTO @Test VALUES (5, 2, 1, '01-02-2012', 'LSI', NULL)
INSERT INTO @Test VALUES (5, 3, 0, '01-03-2012', NULL, NULL)
INSERT INTO @Test VALUES (6, 1, 1, '01-01-2012', NULL, NULL)
INSERT INTO @Test VALUES (6, 2, 0, '01-02-2012', 'LSI', NULL)
INSERT INTO @Test VALUES (6, 3, 1, '01-03-2012', NULL, NULL)

SELECT * FROM @Test

以下是两个示例结果集。虽然代码段显示插入，但要点是显示可接受的输出集的样子：

示例输出＃1
在下面的数据集中，当行有IsLive=0时，其列中的值必须超过其下面IsLive=1跳过NULL值的行上的相同列的值。忽略第一行IsLive=1行之前的所有IsLive=0行。

    INSERT INTO @Test VALUES (1, 1, 1, '01-08-2012', 'RTS', NULL)
    INSERT INTO @Test VALUES (1, 2, 0, '01-09-2012', 'RTA', NULL)
    INSERT INTO @Test VALUES (1, 3, 1, '01-10-2012', 'RTA', NULL)
    INSERT INTO @Test VALUES (1, 4, 0, '01-11-2012',  NULL, NULL)
    INSERT INTO @Test VALUES (1, 5, 1, '01-12-2012', 'RTA', NULL)
    INSERT INTO @Test VALUES (2, 1, 1, '01-08-2012', 'RTS', NULL)
    INSERT INTO @Test VALUES (2, 2, 0, '01-09-2012', 'RTA', NULL)
    INSERT INTO @Test VALUES (2, 3, 1, '01-10-2012', 'RTA', NULL)
    INSERT INTO @Test VALUES (2, 4, 0, '01-11-2012', 'GSM', NULL)
    INSERT INTO @Test VALUES (2, 5, 1, '01-12-2012', 'GSM', NULL)
    INSERT INTO @Test VALUES (3, 1, 1, '01-01-2012', 'FSA', NULL)
    INSERT INTO @Test VALUES (3, 2, 0, '01-02-2012', NULL, NULL)
    INSERT INTO @Test VALUES (4, 1, 1, '01-01-2012', NULL, NULL)
    INSERT INTO @Test VALUES (4, 2, 0, '01-02-2012', 'FSA', NULL)
    INSERT INTO @Test VALUES (4, 3, 0, '01-03-2012', NULL, NULL)
    INSERT INTO @Test VALUES (5, 1, 0, '01-01-2012', NULL, NULL)
    INSERT INTO @Test VALUES (5, 2, 1, '01-02-2012', 'LSI', NULL)
    INSERT INTO @Test VALUES (5, 3, 0, '01-03-2012', NULL, NULL)
    INSERT INTO @Test VALUES (6, 1, 1, '01-01-2012', NULL, NULL)
    INSERT INTO @Test VALUES (6, 2, 0, '01-02-2012', 'LSI', NULL)
    INSERT INTO @Test VALUES (6, 3, 1, '01-03-2012', 'LSI', NULL)

    SELECT * FROM @Test AS FakedOutput_1

示例输出＃2
在下面的数据集中，当行有IsLive=0时，其列中的值必须超过其下方IsLive=1的行上相同列的值。具有NULL值的列采用上一行的值。

    INSERT INTO @Test VALUES (1, 1, 1, '01-08-2012', 'RTS', NULL)
    INSERT INTO @Test VALUES (1, 2, 0, '01-09-2012', 'RTA', NULL)
    INSERT INTO @Test VALUES (1, 3, 1, '01-10-2012', 'RTA', NULL)
    -- <- the following row is different from prev
    INSERT INTO @Test VALUES (1, 4, 0, '01-11-2012', 'RTA', NULL) 
    INSERT INTO @Test VALUES (1, 5, 1, '01-12-2012', 'RTA', NULL)
    INSERT INTO @Test VALUES (2, 1, 1, '01-08-2012', 'RTS', NULL)
    INSERT INTO @Test VALUES (2, 2, 0, '01-09-2012', 'RTA', NULL)
    INSERT INTO @Test VALUES (2, 3, 1, '01-10-2012', 'RTA', NULL)
    INSERT INTO @Test VALUES (2, 4, 0, '01-11-2012', 'GSM', NULL)
    INSERT INTO @Test VALUES (2, 5, 1, '01-12-2012', 'GSM', NULL)
    INSERT INTO @Test VALUES (3, 1, 1, '01-01-2012', 'FSA', NULL)
    INSERT INTO @Test VALUES (3, 2, 0, '01-02-2012', 'FSA', NULL)
    INSERT INTO @Test VALUES (4, 1, 1, '01-01-2012', NULL, NULL)
    INSERT INTO @Test VALUES (4, 2, 0, '01-02-2012', 'FSA', NULL)
    INSERT INTO @Test VALUES (4, 3, 0, '01-03-2012', 'FSA', NULL)
    INSERT INTO @Test VALUES (5, 1, 0, '01-01-2012', NULL, NULL)
    INSERT INTO @Test VALUES (5, 2, 1, '01-02-2012', 'LSI', NULL)
    INSERT INTO @Test VALUES (5, 3, 0, '01-03-2012', 'LSI', NULL)
    INSERT INTO @Test VALUES (6, 1, 1, '01-01-2012', NULL, NULL)
    INSERT INTO @Test VALUES (6, 2, 0, '01-02-2012', 'LSI', NULL)
    INSERT INTO @Test VALUES (6, 3, 1, '01-03-2012', 'LSI', NULL)

    SELECT * FROM @Test AS FakedOutput_2

尝试解决方案
这是我到目前为止提出的问题，但是我的第一个测试用例（GID=1）

失败了 
;WITH CTE AS ( 
    -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
    SELECT  T.GID, T.SEQ, T.IsLive, NULL cGuid, NULL cSEQ, 
            cast(0 as bit) cIsLive, T.Name, T.Salary 
    FROM    @Test T
    JOIN    @Test S ON T.GID = S.GID AND T.Seq = S.Seq AND S.IsLive = 0
    -- - - - - - - 
      UNION ALL 
    -- - - - - - - 
    SELECT  t.GID, t.SEQ, T.IsLive, c.GID cGID, c.Seq cSEQ, 
             c.IsLive cIsLive, ISNULL(C.Name, T.Name), 
             ISNULL(t.Salary, c.Salary) 
    FROM    CTE c 
    JOIN    @Test t ON    t.GID     = c.GID   AND 
                          t.Seq     > c.Seq   AND 
                          t.IsLive  = 1       AND 
                          c.IsLive  = 0
    -- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 
) 
--SELECT * FROM CTE ORDER BY CTE.GID, CTE.Seq
UPDATE  t 
SET     Name = c.Name, Salary = c.Salary
FROM    @Test t 
JOIN    CTE c     ON c.GID = t.GID AND c.Seq = t.SEQ
WHERE   C.cIsLive IS NOT NULL

2 个答案:

答案 0 :(得分：1)

使用APPLY适用于您的测试用例。以下与您的解决方案2相同

SELECT  t1.GID,
        t1.Seq,
        t1.IsLive,
        t1.Eff,
        CASE WHEN t1.IsLive = 0 THEN COALESCE(t1.Name, t3.Name) ELSE COALESCE(t3.Name, t1.Name) END AS Name,
        Salary
FROM    @Test T1
        OUTER APPLY
        (   SELECT  TOP 1 Name
            FROM    @Test T2
            WHERE   T2.GID = T1.GID
            AND     T2.Seq < T1.Seq
            AND     t2.IsLive = 0
            AND     t2.Name IS NOT NULL
            ORDER BY Seq DESC
        ) t3

修改

注意到需要UPDATE：

UPDATE  @Test
SET     Name = CASE WHEN t1.IsLive = 0 THEN COALESCE(t1.Name, t3.Name) ELSE COALESCE(t3.Name, t1.Name) END
FROM    @Test T1
        OUTER APPLY
        (   SELECT  TOP 1 Name
            FROM    @Test T2
            WHERE   T2.GID = T1.GID
            AND     T2.Seq < T1.Seq
            AND     t2.IsLive = 0
            AND     t2.Name IS NOT NULL
            ORDER BY Seq DESC
        ) t3

编辑2

我稍微更改了apply中的查询，现在它会尝试找到最接近的行，其中live = 0且名称不为null，如果没有live = 1的行（如GID = 4）将采用名称不为null的最近一行：

UPDATE  @Test
SET     Name = CASE WHEN t1.IsLive = 0 THEN COALESCE(t1.Name, t3.Name) ELSE COALESCE(t3.Name, t1.Name) END
FROM    @Test T1
        OUTER APPLY
        (   SELECT  TOP 1 Name
            FROM    @Test T2
            WHERE   T2.GID = T1.GID
            AND     T2.Seq < T1.Seq
            AND     t2.Name IS NOT NULL
            ORDER BY t2.IsLive, Seq DESC
        ) t3

答案 1 :(得分：1)

我花了一段时间才意识到这个问题只是我上一个问题的一个小变化。离键盘只有一段时间帮助我看到了答案！ @GarethDs的回答也为此做出了贡献。

;WITH CTE AS ( 
    SELECT  T.GID, T.SEQ, T.IsLive, Name, Salary 
    FROM    @Test T
    JOIN    ( SELECT   GID, MIN(Seq) Seq 
              FROM     @Test 
              GROUP BY GID     
            ) S ON T.GID = S.GID AND T.Seq = S.Seq

    UNION ALL 

    SELECT t.GID, t.SEQ, T.IsLive,
           CASE WHEN T.IsLive = 0 THEN COALESCE(T.Name, C.Name) 
                ELSE COALESCE(C.Name, T.Name) END,
           CASE WHEN T.IsLive = 0 THEN COALESCE(T.Salary, C.Salary) 
                ELSE COALESCE(C.Salary, T.Salary) END 
    FROM   CTE C
    JOIN   @Test T ON T.GID = C.GID AND T.SEQ = C.SEQ+1 
) 
--SELECT * FROM CTE ORDER BY CTE.GID, CTE.Seq
UPDATE T 
SET    Name   = C.Name,  
       Salary =  C.Salary
FROM   @Test T
JOIN   CTE C ON C.GID = T.GID AND C.Seq = T.SEQ
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