Question

我有两张桌子。

a） ai_account

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b） ai_order_product

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我想为特定的supplier_id做一些计算。

1，totalAmount，我想做点什么   SUM（ai_order_product.quantity * ai_order_product.cost）

2，amountPaid，这是供应商支付的总金额   像SUM（ai_account.amount）这样的东西，参考supplier_id。

3）余额，这将由SUM计算（ai_order_product.quantity *   ai_order_product.cost） - SUM（ai_invoice.amount）

4）lastPayment日期，即MAX（ai_account.addDate）。

我尝试过这样的事情。

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

它无法正常工作，它会获取一些意外的值，而上述预期的结果是，

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

我想通过supplier_id获取一行。

P.S：我刚刚输入了一些虚拟值，这就是为什么计算主要是负数，而在应用中，计算值将是正数。

Answer 1

这是因为每个“ai_order_product”行都被多次计数（ai_account表中每行存在一次）。

试试这个：

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

这可能略有偏差，但这种一般逻辑应该有效。

Answer 2

在 SELECT 语句中使用 SUM 等聚合函数时，必须使用 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

当您选择多个supplier_id时，group by条款会启动。

Answer 4

请注意，您的预期值和实际值都加倍了。输入样本数据并运行查询后，我得到（对于supplier_id = 42）

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

那是因为每个表中有两行符合连接条件，导致结果加倍。

Answer 5

我认为您需要首先获取其中一个表的子总计，这样您才能返回1行。因此，内部选择返回最大值和总和，所以当你加入它时，你只得到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

按Ac.addDate分组，ac.amount，op.supplier_ID - 只是在哪里停止子句。

使用SUM计算，查询不起作用？

5 个答案: