Question

我有一种情况，我希望总结两个表之间的差异。问题是第二个表中可以存在一行，然后我想将其作为新行插入。

伪

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

查看图片以获得解释

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

你需要做一个完整的外部联接 - 然后加以总结。如果你只运行内部查询，你将获得每个可能的行组合（在t1中是独占的，在t1和t2中都存在，在t2中是exlclusive） - 然后你将它组合起来并做总和。

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 Demo

Answer 3

我认为您需要FULL JOIN（除非您确实希望将行插入第一个表格）

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);

如果确实需要插入表1的行，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);

修改

如果每个表中的行都不是unqiue而你需要对它们求和，那么你需要在子查询中进行求和，因为JOIN将引入交叉连接：

考虑这个数据

Table1 Seller Code Amount VL 500 10 VL 500 20 Table2 Seller Code Amount VL 500 30 VL 500 5

当你加入这个时，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

差值的总和是-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方法的效果会比完全加入好得多：

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

Answer 4

这样的东西？你不知道要插入“因此tableX”中的表格

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，在插入应用过滤器后，将过滤器设置为select语句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

来自两个表的总和如果不存在则插入

4 个答案: