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我有一种情况，我想对两个表之间的差异求和。问题是第二个表中可以存在一行，然后我想将它作为新行插入。

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SELECT T1.seller, T1.code, T1.amount - T2.amount

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4 回答 4

2 
 DECLARE @T1 TABLE(
    seller VARCHAR(10),
    code VARCHAR(3) NULL,
    amount MONEY
)

 DECLARE @T2 TABLE(
    seller VARCHAR(10),
    code VARCHAR(3) NULL,
    amount MONEY
)

INSERT INTO @T1 VALUES
('VL',NULL,1),
('VL','317',70005.6)

INSERT INTO @T2 VALUES
('VL',NULL,0.5),
('VL','500',4450)

SELECT seller,code,SUM(amount) [amount] FROM 
(
SELECT * FROM @T1
UNION ALL
SELECT seller,code,-amount as amount FROM @T2
) T
GROUP BY seller,code
于 2012-12-11T09:20:51.357 回答
1

您需要进行完全外部连接 - 然后总结一下。如果您只运行内部查询，您将获得每个可能的行组合（在 t1 中不包含，在 t1 和 t2 中都存在，在 t2 中不包含） - 然后将其分组并求和。

SELECT  Seller ,
        Code ,
        SUM(Tab1_amt - Tab2_amt) AS Amount
FROM    ( SELECT    COALESCE(tab1.Seller, tab2.Seller) AS Seller ,
                    COALESCE(tab1.code, tab2.code) AS Code ,
                    COALESCE(tab1.amount, 0) AS tab1_amt ,
                    COALESCE(tab2.amount, 0) AS tab2_amt
          FROM      tab1
                    FULL OUTER JOIN tab2 ON tab1.seller = tab2.seller
                                            AND tab1.code = tab2.code
        ) AS Tbl
GROUP BY Seller ,
        Code

请参阅SQLFiddle 演示

于 2012-12-11T09:13:13.250 回答
1

我认为您需要一个FULL JOIN（除非您实际上希望将行插入到第一个表中）

SELECT  COALESCE(t1.Seller, t2.Seller) AS Seller,
        COALESCE(t1.Code, t2.Code) AS Code,
        COALESCE(t1.Amount, 0) - COALESCE(t2.Amount, 0) AS Amount
FROM    Table1 t1
        FULL JOIN Table2 t2
            ON t1.Seller = t2.Seller
            AND COALESCE(t1.Code, 0) = COALESCE(t2.Code, 0);

如果确实需要将行插入表 1，则需要执行 2 次操作，首先插入，然后选择：

INSERT Table1 (Seller, Code, Amount)
SELECT  t2.Seller, t2.Code, 0 AS Amount
FROM    Table2 t2
WHERE   NOT EXISTS
        (   SELECT  1
            FROM    Table1 t1
            WHERE   t1.Seller = t2.Seller
            AND     COALESCE(t1.Code, 0) = COALESCE(t2.Code, 0)
        );

SELECT  t1.Seller, 
        t1.Code, 
        t1.Amount - COALESCE(t2.Amount, 0) AS Amount
FROM    Table1 t1
        LEFT JOIN Table2 t2
            ON t1.Seller = t2.Seller
            AND COALESCE(t1.Code, 0) = COALESCE(t2.Code, 0);

编辑

如果每个表中的行不是唯一的并且您需要对它们求和，那么您将需要在子查询中进行求和，因为 JOIN 将引入交叉连接：

考虑这些数据

Table1
Seller  Code    Amount
VL      500     10
VL      500     20

Table2
Seller  Code    Amount
VL      500     30
VL      500     5

加入此活动您将获得：

t1.Seller   t1.Code t1.Amount   t2.Seller   t2.Code t2.Amount
VL          500     10          VL          500     30
VL          500     10          VL          500     5
VL          500     20          VL          500     30
VL          500     20          VL          500     5

差的总和是 -10 而不是 -5。

SELECT  COALESCE(t1.Seller, t2.Seller) AS Seller,
        COALESCE(t1.Code, t2.Code) AS Code,
        COALESCE(t1.Amount, 0) - COALESCE(t2.Amount, 0) AS Amount
FROM    (   SELECT  Seller, Code, SUM(Amount) AS Amount
            FROM    Table1 
            GROUP BY Seller, Code
        ) t1
        FULL JOIN 
        (   SELECT  Seller, Code, SUM(Amount) AS Amount
            FROM    Table2 
            GROUP BY Seller, Code
        ) t2
            ON t1.Seller = t2.Seller
            AND COALESCE(t1.Code, 0) = COALESCE(t2.Code, 0);

编辑 2

丹尼尔答案中的UNION方法将比 FULL JOIN 执行得更好：

SELECT  Seller, Code, Amount = SUM(Amount)
FROM    (   SELECT  Seller, Code, Amount
            FROM    Table1
            UNION
            SELECT  Seller, Code, -Amount
            FROM    Table2
        ) t
GROUP BY Seller, Code
于 2012-12-11T09:13:48.883 回答
1

像这样的东西？你不知道要插入什么表到“因此tableX”

insert into tableX (seller,code,amount) values (T1.seller, T1.code, T1.amount -T2.amount)
select count(*) from table2 having count(*) < 1

请注意，您在插入以应用过滤器之后将过滤器作为选择语句放置您没有指定您需要什么类型的 sql，所以我不能说这是否有效

于 2012-12-11T09:16:36.077 回答