從兩個表中總結插入如果不存在

Question

我有一種情況，我想總結兩個表之間的差異。問題是一行可能存在於第二個表中，然後我想將它作爲新行插入。

僞

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

看圖像做出解釋

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

你需要做一個完全外部聯接 - 再總結一下。如果只運行內部查詢，您將獲得每個可能的行組合（不包括在t1中，在t1和t2中存在，在t2中包含） - 然後將其組合在一起並執行總和。

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 

見SQLFiddle Demo

Answer 3

我想你需要一個FULL JOIN（除非你真正想要的行插入到第一個表）

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); 

如果你確實需要的行插入到表1你將需要做2次操作，先插入，然後選擇：

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); 

編輯

如果在每個表中的行不unqiue，你需要總結他們，那麼你需要做的款項在子查詢的JOIN將引入交叉加盟：

考慮這個數據

Table1 
Seller Code Amount 
VL  500  10 
VL  500  20 

Table2 
Seller Code Amount 
VL  500  30 
VL  500  5 

當你加入這個你會得到：

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 

差的總和，然後-10代替-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); 

EDIT 2

在Daniel's answer的UNION方法比FULL進行更好的JOIN：

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

像這樣的事情？你不告訴插入到「因此TableX的」

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

通知你把過濾器作爲SELECT語句的插入後，應用過濾器 你沒有指定你需要什麼類型的SQL的什麼表，所以我不能說這是否工作