使用SUM進行計算，查詢不起作用？

Question

我有兩個表。

一） ai_account

B） ai_order_product

我想要做特定supplier_id一些計算。

1，總金額，我想這樣做 SUM（ai_order_product.quantity * ai_order_product.cost）

2，amountPaid，這是由供應商支付的總金額，這將是 像SUM （ai_account.amount）參考supplier_id。

3）平衡，這將通過SUM來計算（ai_order_product.quantity * ai_order_product.cost） - SUM（ai_invoice.amount）

4）lastPayment日期，這將是MAX（ai_account.addDate）。

我試着做這樣的事情。

SELECT SUM(op.quantity * op.cost) as totalAmount, 
     SUM(ac.amount) as amountPaid, 
     SUM(op.quantity * op.cost) - SUM(ac.amount) as balance, 
     MAX(ac.addDate) as lastPayment 
    FROM ai_order_product op 
     LEFT JOIN ai_account ac 
      ON (op.supplier_id = ac.trader_id) 
    WHERE op.supplier_id = 42 

它不能正常工作，它獲取一些意想不到的值，而對於被預計上述結果是，

for supplier_id = 42, 
1) totalAmount = 1375, 
2) amountPaid = 7000, 
3) balance = -5625, 
4) lastPayment = 2011-11-23 

and for supplier_id = 35, 
1) totalAmount = 1500, 
2) amountPaid = 43221, 
3) balance = -41721, 
4) lastPayment = 2011-11-28 

and for supplier_id = 41 
1) totalAmount = 0 
2) amountPaid = 3000, 
3) balance = -3000, 
4) lastPayment = 2011-11-09 

我想supplier_id獲取一行。

P.S：我剛剛輸入了一些虛擬值，這就是爲什麼計算大多是負值，而在應用中計算值是正值。

Answer 1

那是因爲每個「ai_order_product」行都會被多次計數（ai_account表中每行存在一次）。

試試這個：

SELECT 
    op.totalAmount as totalAmount 
    , SUM(ac.amount) as amountPaid 
    , op.totalAmount - SUM(ac.amount) as balance 
    , MAX(ac.addDate) as lastPayment 
FROM (
    select supplier_id, sum(quantity * cost) as totalAmount 
    from ai_order_product 
    group by supplier_id) op 
LEFT JOIN ai_account ac ON (op.supplier_id = ac.trader_id) 
WHERE op.supplier_id = 42 

這可能是稍微偏離，但這個一般的邏輯應該工作。

Answer 2

你有使用聚合函數時，像SUM SELECT語句使用國家GROUP BY。

SELECT op.supplier_id as supplierId, 
     SUM(op.quantity * op.cost) as totalAmount, 
     SUM(ac.amount) as amountPaid, 
     SUM(op.quantity * op.cost) - SUM(ac.amount) as balance, 
     MAX(ac.addDate) as lastPayment 
    FROM ai_order_product op 
     LEFT JOIN ai_account ac 
      ON (op.supplier_id = ac.trader_id) 
    GROUP BY op.supplier_id 
    HAVING supplierId = 42 

Answer 3

SELECT 
    SUM(op.quantity * op.cost) as totalAmount 
    , ac2.amountPaid 
    , SUM(op.quantity * op.cost) - ac2.balance 
    , ac2.lastPayment 
FROM ai_order_product op 
LEFT JOIN (SELECT 
      ac.supplier_id 
      , MAX(ac.addDate) as lastPayment 
      , SUM(ac.amount) as balance 
      FROM ai_account ac 
      WHERE (op.supplier_id = ac.supplier_id) 
      GROUP BY ac.supplier_id) ac2 ON (ac2.supplier_id = op.supplier_id) 
WHERE op.supplier_id = 42 
GROUP BY op.supplier_id 

當您選擇多個supplier_id時，group by子句將啓動。

Answer 4

請注意，您的預期值和實際值會加倍。輸入您的樣本數據和運行查詢後，我得到（對於supplier_id = 42）

+-------------+------------+---------+-------------+ 
| totalAmount | amountPaid | balance | lastPayment | 
+-------------+------------+---------+-------------+ 
|  2750 |  14000 | -11250 | 2011-11-23 | 
+-------------+------------+---------+-------------+ 

那是因爲你已經得到了在符合加入條件的每個表2行，造成的結果增加了一倍。

Answer 5

我認爲你需要首先得到子總數爲一個表，讓你只有1行回來。因此這裏的內部選擇返回最大值和總和，所以當你加入它時你只能得到1行。

SELECT SUM(op.quantity * op.cost) as totalAmount, 
     ac.addDate as lastPayment, 
     ac.amount as amountPaid, 
     SUM(op.quantity * op.cost) - SUM(ac.amount) as balance 
    FROM ai_order_product op 
     INNER JOIN (
      SELECT max(IaiaddDate) as addDate, sum(iai.amount) as Amount, iai.supplier_ID 
      FROM ai_account iai 
      Group by supplier_ID) ac 
     ON AC.Supplier_ID = Op.Supplier_ID 
    WHERE op.supplier_id = 42 

集團通過Ac.addDate，ac.amount，op.supplier_ID --just櫃面where子句都不放過。