Question

我在我的数据库中有这个表，我试图得到2组值并将它们放在一个联合中。这是表格。我的工会基本上有两个条款，其中interest_payment＆gt; 0和interest_paid＆gt; 0，如果interest_payment大于零，它将占用该行并将其与下一行相加，如果下一行具有interest_payment＆gt; 0它将该行添加到该行，但是如果interest_payment＆lt; 0，它应该在同一行中添加interest_paid。但是现在当我运行查询时，我明白了。基本上netInterest不应该显示Null，因为我使用的是sum，所以它应该只返回一行。

这是所有这些的sql。

Select sum(investedMoney) as investMoney , sum(estimatedEarning) as estimatedEarning , sum(NetInterest) as NetInterest, Eurosymbol
                from
                (
                select  sum(round(a.Amount * e.average_rate / f.average_rate, 2)) as investedMoney ,
                        sum(round((d.interest_payment + d.overdue_payment) * e.average_rate / f.average_rate, 2)) as estimatedEarning,
                        0 as NetInterest,
                        c.symbol as Eurosymbol
                from   investment a
                inner join money_offer b
                on a.ORIG_ID = b.investment_orig_id and b.UPDATE_DT is null
                inner   join payment_plan d 
                on d.offer_orig_id = b.ORIG_ID and d.interest_payment > 0 and d.UPDATE_DT is null
                inner   join currency c 
                        on  d.currency = c.ID 
                inner join exchange_rates e 
                        on  e.currency_id = a.Currency
                inner join exchange_rates f 
                        on  f.currency_id = a.Currency
                where   a.Owner = 533  and
                                a.UPDATE_DT is null

                        union 


                select  0 as investedMoney ,
                        0 as estimatedEarning,
                        sum(round((d.interest_paid+ d.overdue_paid) * e.average_rate / f.average_rate, 2)) as NetInterest,
                        c.symbol as Eurosymbol
                from   investment a
                inner join money_offer b
                on a.ORIG_ID = b.investment_orig_id and b.UPDATE_DT is null
                inner   join payment_plan d 
                on d.offer_orig_id = b.ORIG_ID and d.interest_paid > 0 and d.UPDATE_DT is null
                inner   join currency c 
                        on  d.currency = c.ID 
                inner join exchange_rates e 
                        on  e.currency_id = a.Currency
                inner join exchange_rates f 
                        on  f.currency_id = a.Currency
                where   a.Owner = 533  and
                                a.UPDATE_DT is null
                )tmptbl
                group by Eurosymbol;

Answer 1

在d.interest_payment > 0或d.interest_paid > 0的情况下，您在payment_plan表中存储不同的货币。我的猜测是你有虚拟ID来表示＆＃34;没有货币＆＃34;。假设您在currency.symbol中为该特定虚拟货币存储了NULL值，您可以在整个查询中使用max。

另外，我认为出于您的目的，您应该将union个查询合并为一个。这样，您可以处理同一记录中两个条件都为真的情况：它们将不会被复制。两种类型的数字之间的区别可以使用case when构造，如下所示：

  select      sum(case when d.interest_payment > 0 
                    then round(a.Amount * e.average_rate / f.average_rate, 2) 
                    else 0 
                  end) as investedMoney,
              sum(case when d.interest_payment > 0 
                    then round((d.interest_payment + d.overdue_payment) 
                               * e.average_rate / f.average_rate, 2)
                    else 0 
                  end) as estimatedEarning,
              sum(case when d.interest_paid > 0 
                    then round((d.interest_paid + d.overdue_paid) 
                               * e.average_rate / f.average_rate, 2)
                    else 0
                  end) as NetInterest,
              max(c.symbol) as Eurosymbol
  from        investment a
  inner join  money_offer b
          on  a.ORIG_ID = b.investment_orig_id and b.UPDATE_DT is null
  inner join  payment_plan d 
          on  d.offer_orig_id = b.ORIG_ID and d.UPDATE_DT is null
          and (d.interest_payment > 0 or d.interest_paid > 0)
  inner join  currency c 
          on  d.currency = c.ID 
  inner join  exchange_rates e 
          on  e.currency_id = a.Currency
  inner join  exchange_rates f 
          on  f.currency_id = a.Currency
  where       a.Owner = 533  
      and     a.UPDATE_DT is null

您还应该检查为什么从exchange_rates（e和f）两次加入完全相同的记录，这使得表达式e.average_rate / f.average_rate始终评估为1（假设费率永远不会为零。）

union在表中返回2行而不是1行

1 个答案: