php - MySQL WHERE 通过 AS 语句

Question

我的 table

Table1 :
___________________________________
| ID | UserID | Amount | Date     |
___________________________________
| 1  | 1      | 10     | 2014-12-10
| 2  | 1      | 5      | 2014-12-12
| 3  | 2      | 50     | 2014-12-12

Table2 :
_________________________________________________
| ID | UserID | Amount | Description  | Date     |
_________________________________________________
| 1  | 2      | 50     | Test Payment | 2014-12-10

我希望用户计算余额。我使用这个 MySQL 代码

SELECT
Distinct(UserID) As User,
(SELECT COALESCE(SUM(Amount),0) FROM Table1 WHERE UserID=User) AS Credit,
(SELECT COALESCE(SUM(Amount),0) FROM Table2 WHERE UserID=User) AS Received,
((SELECT COALESCE(SUM(Amount),0) FROM Table1 WHERE UserID=User) - (SELECT COALESCE(SUM(Amount),0) FROM Table2 WHERE UserID=User) As Balance
FROM Table1 WHERE Balance>0

但我无法通过 as 声明。

Answer 1

您无法访问aliases in WHERE 。您应该在那里放置完整的表达式或使用 HAVING:

SELECT
Distinct(UserID) As User,
(SELECT COALESCE(SUM(Amount),0) FROM Table1 WHERE UserID=User) AS Credit,
(SELECT COALESCE(SUM(Amount),0) FROM Table2 WHERE UserID=User) AS Received,
((SELECT COALESCE(SUM(Amount),0) FROM Table1 WHERE UserID=User) - (SELECT COALESCE(SUM(Amount),0) FROM Table2 WHERE UserID=User) As Balance
FROM Table1 
HAVING Balance>0

-在大多数情况下，在 WHERE 中使用完整表达式是一件奇怪的事情，因为它会导致计算两次。