Question

我只需要在付款日期相同的情况下对列的数据求和。

以下是返回的数据：

judgment    PaymentDate    interestrate  current1  paymentamount  principalamount
2/1/2008    7/31/2010       9.00          5781.04   -315.07        -270.07
2/1/2008    7/31/2010       9.00          5781.04    272.59         227.59

以下是查询：

select tb1.judgment, fulldate as PaymentDate, m.interestrate, m.current1, paymentamount, principalamount from master m
            inner join AARS_JudgmentsWithPA tb1 on tb1.jmacct = m.number
            inner join courtcases cc on cc.accountid = m.number
            where m.lastinterest != '2099-01-01 00:00:00.000'
            and tb1.fulldate > tb1.judgment
            and cast(tb1.PrincipalAmount as money) != 0
            and tb1.judgment != ''
            and m.number = 568463

对于提供的示例数据，我需要合并这些行，因为付款日期是相同的。因此，付款金额应缩减为（-270.07 + 227.59），即-42.48。

我尝试按照全额付款和付款金额进行分组，但是我希望我对每个列进行分组，然后再次返回这两行。

我错过了什么？

Answer 1

您需要为未分组的列指定聚合。对于重复MIN()的数字列，MAX()或AVG()是合适的。对于字符串列MIN()或MAX()可行。

select tb1.judgment, 
       fulldate as PaymentDate, 
       MIN(m.interestrate) AS interestrate, 
       MIN(m.current1) AS current1,
       SUM(paymentamount) AS paymentamount, 
       SUM(principalamount) AS principalamount
from master m
inner join AARS_JudgmentsWithPA tb1 on tb1.jmacct = m.number
inner join courtcases cc on cc.accountid = m.number
where m.lastinterest != '2099-01-01 00:00:00.000'
    and tb1.fulldate > tb1.judgment
        and cast(tb1.PrincipalAmount as money) != 0
        and tb1.judgment != ''
        and m.number = 568463
group by tbl.judgment, fulldate

另一种选择是对重复和聚合其余列的列进行分组：

select tb1.judgment, 
       fulldate as PaymentDate, 
       m.interestrate, 
       m.current1, 
       SUM(paymentamount) AS paymentamount, 
       SUM(principalamount) AS principalamount
from master m
inner join AARS_JudgmentsWithPA tb1 on tb1.jmacct = m.number
inner join courtcases cc on cc.accountid = m.number
where m.lastinterest != '2099-01-01 00:00:00.000'
    and tb1.fulldate > tb1.judgment
        and cast(tb1.PrincipalAmount as money) != 0
        and tb1.judgment != ''
        and m.number = 568463
group by tbl.judgment, fulldate, m.interestrate, m.current1

Answer 2

当您使用聚合函数时，您没有遗漏任何内容，在本例中为sum（），您必须按每个其他非聚合列进行分组。前几个句子以一种非常详细的方式解释：http://dev.mysql.com/doc/refman/5.0/en/group-by-extensions.html以及需要您按以下分组的函数列表：http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html

总结一个特定的列

2 个答案: