6

我相信我可以通过使用左外连接的 case 语句来优化这个 sql 语句。

但是我一直很难设置案例，一个用于总结代码类型 AB、CD，另一个用于所有其余的。

感谢您对此提供的任何帮助或提示。 

update billing set payments = isnull(bd1.amount, payments)
, payments = case 
       when payments is null then 0 
       else payments 
     end
, charges = case 
        when bd2.amount is not null then charges 
        when charges is null then 0 
        else charges 
      end
, balance = round(charges + isnull(bd1.amount, bi.payments), 2) 
from billing bi 

left outer join (select inv, round(sum(bd1.bal), 2) amount 
                from "bill" bd1 
                where code_type = 'AB'
                 or code_type = 'CD' 
                group by inv) bd1 
                on bd1.inv = bi.inv 
left outer join (select invoice, round(sum(bd2.bal), 2) amount 
                from "bill" bd2 
                where code_type <> 'AB'
                 and code_type <> 'CD' 
                group by inv) bd2 
                on bd2.inv = bi.inv;
4

2 回答 2

5

您可以将其简化为使用单个查询而不是两个。您仍然需要它，因为 UPDATE 中的 GROUP BY 不起作用。

UPDATE bi 
SET    payments = bd.payments, 
       charges= bd.charges, 
       balance = bd.balance 
FROM   billing bi 
       LEFT JOIN (SELECT bd.inv, 
                         payments = Round(Sum(CASE 
                                                WHEN code_type IN ( 'AB' , 'CD' ) THEN 
                                                bd.bal 
                                                ELSE 0 
                                              END), 2), 
                         charges = Round(Sum(CASE 
                                               WHEN code_type NOT IN ( 'AB' , 'CD' ) THEN 
                                               bd.bal 
                                               ELSE 0 
                                             END), 2), 
                         balance = Round(Sum(bd.bal), 2) 
                  FROM   bill bd 
                  GROUP  BY bd.inv) bd 
         ON bd.inv = bi.inv
于 2012-04-30T14:57:09.547 回答
1

也许是这样的：

update billing set payments = isnull(bd1.amount, payments)
, payments = isnull(payments, 0)
, charges = isnull(bd2.amount, isnull(charges, 0))
, balance = round(charges + isnull(bd1.amount, bi.payments), 2)
from billing bi 

left outer join (select inv, round(sum(bd1.bal), 2) amount 
                from "bill" bd1 
                where code_type in ('AB', 'CD')
                group by inv) bd1 
                on bd1.inv = bi.inv 
left outer join (select invoice, round(sum(bd2.bal), 2) amount 
                from "bill" bd2 
                where code_type not in ('AB', 'CD')
                group by inv) bd2 
                on bd2.inv = bi.inv;

两个左连接不是问题！

于 2012-04-30T14:50:57.507 回答