Question

我正在使用数据库hyperfile client / serveur与windev合作。

我有一个名为Operation with colums（accountNumber，date，amount，operationType）的表。

operationType可以采用两个值：“付款”和“取款”。

我想选择帐户中完成的操作列表，我的列表应显示5个列：日期，帐户编号，金额，操作类型和余额。

最后一栏（余额）应该是当前日期之前完成的所有操作与“付款”类型之间的差额，以及当前日期之前完成的所有操作与“撤销”类型之间的差额

我尝试以下sql代码

SELECT accountNumber, date as dateOpe, amount, operationType, (SUM (SELECT Operation.amount 
                                                                   FROM Operation 
                                                                   WHERE Operation.date<=dateOpe AND Operation.operationType='payment')
                                                             -SUM (SELECT Operation.amount
                                                                   FROM Operation
                                                                   WHERE Operation.date<=dateOpe AND Operation.operationType='withdrawal')) as balance
 FROM Operation

但我总是有一个错误告诉我，我没有权利在SUM

中添加一个选择

请有人帮助我。如何编写这样的SQL查询。

提前致谢

Answer 1

请试试这个：

SELECT accountNumber, date, amount, operationType, 
       (SELECT SUM(amount) AS balance  
          FROM operation WHERE operationType='payment' and date=date) 
      -(SELECT SUM(amount) AS balance  
          FROM operation WHERE operationType='withdrawal' and date=date) 
       as balance  
       FROM operation

Answer 2

尝试使用子查询查找SUM，如下所示。可能有更优雅的解决方案，但这应该有效。

SELECT date, accountNumber,
    operationType, deposits - withdrawals AS balance
FROM Operations o INNER JOIN (
    SELECT accountNumber, SUM(amount) AS withdrawals
    FROM Operaions
    WHERE operationType = 'withdrawal'
    GROUP BY accountNumber
) a ON  o.accountNumber = a.accountNumber
INNER JOIN (
    SELECT accountNumber, SUM(amount) AS deposits
    FROM Operations
    WHERE operationType = 'deposit'
    GROUP BY accountNumber
)b ON o.accountNumber = b.accountNumber

Answer 3

我会使用逻辑表达式，如果为false则为0，如果为true，则为1。因此，如果表达式为假，Operation.amount * <logical expression>将产生0，如果表达式为真，则Operation.amount将产生SELECT accountNumber, date as dateOpe, amount, operationType, SUM (Operation.amount * (Operation.date<=dateOpe AND Operation.operationType='payment')) - SUM (Operation.amount * (Operation.date<=dateOpe AND Operation.operationType='withdrawal')) as balance FROM Operation。因此，以下查询应该接近解决方案（未经测试）：

std::thread

根据其他选择使用其中一列进行选择

3 个答案: