Question

这是mytable

   customerID  invoice     payment
   1            101            300
   1            101            300
   3            102            200
   4            103            200
   1            104            200

我想做的是，首先：分组发票，然后对分组发票的付款进行总计（300 + 200）

 select customerID  , sum(payment) as totalpaid from mytable WHERE
 customerID  ='1' group by `invoice`

预期结果是500。但是使用上面的代码我得到800。如何正确做到这一点？

Answer 1

如果要排除重复项，请使用DISTINCT：

SELECT 
  customerID, 
  SUM(payment) as totalpaid 
FROM 
   (SELECT DISTINCT customerID, invoice, payment FROM customers) 
GROUP BY customerID

对于customerID = 1，您可以

SELECT 
  customerID, 
  SUM(payment) as totalpaid 
FROM 
   (SELECT DISTINCT customerID, invoice, payment FROM customers) 
WHERE customerID = 1

无需GROUP BY。

SQLITE3 php中GROUP BY之后的总和

1 个答案: