Question

所以，我的问题是。我有餐桌付款和餐桌收费，因此，一笔付款有很多收费，但一笔收费有一笔付款。我需要选择所有付款，其中一笔付款的所有费用总额！=一笔付款的所有费用。

代码示例

4:2
9:8

Answer 1

您也许可以不用加入就可以做到这一点，只需选择一个子项即可获得payment_charge的总和，并将其与每次付款的charge_summ进行比较：

select payment_account, payment_agreement, payment_pay, payment_charge
from payment 
where payment.charge_summ <> (select sum(payment_charge)
                        from charge 
                        where payment.payment_id = charge_payment_id)

编辑您可以使用JOIN进行操作，而无需内部选择：

select payment_id,payment_account, payment_agreement, payment_pay, charge_summ,sum(payment_charge) 
from payment join
     charge
     on payment_id = charge_payment_id
group by payment_id,payment_account, payment_agreement, payment_pay,charge_summ
having charge_summ <> sum(payment_charge)

Answer 2

在任何情况下，表结构都会有用，请尝试以下查询：

select payment_account, payment_agreement, payment_pay, payment_charge
from payment join
(select charge_payment_id, sum(payment_charge) as totally from charge group by 
charge_payment_id) A 
on payment_id = charge_payment_id and payment_charge != totally;

在这里，您首先要汇总费用并与结果进行比较。

一对多内部连接

2 个答案: