如果不存在，则插入两个表的总和

Question

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，然后我想将其插入为新行。

伪

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

看图说明

LINK

Answer 1

 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

Answer 2

您需要进行完整的外部连接 - 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

见SQLFiddle Demo

Answer 3

这样的？你没有告诉要插入什么表到“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，您将过滤器作为选择语句放在插入应用过滤器之后 你没有指定你需要什么类型的 sql，所以我不能说这是否有效

Answer 4

我认为您需要一个 FULL JOIN（除非您确实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);

如果您确实需要向表 1 插入行，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

当你加入这个你会得到：

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

Daniel's answer 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