Question

当其他表中的某些字段不为空时，我想从表中选择特定列。

换句话说，我有两个表，它们有共同的字段，所以当字段名为＆＃34; salesordrkey＆＃34;在表号1中不是null然后我需要从表号2获取公共字段，否则从表号1获取公共字段

这就是我要做的事情

Select slsordr.salesordrkey,
       whissue.warehissuekey, 
       issuppk.issueprodpackkey,
       (Case When  whissue.salesordrkey Is Not Null Then
           (Select slsordr.busipartnerkey,
                   slsordr.contractkey,
                   slsordr.salesmankey,
                   slsordr.customerkey)
        Else
           (Select whissue.busipartnerkey,
                   whissue.contractkey,
                   whissue.salesmankey,
                   slsordr.customerkey)
        End)
 From warehissues whissue 
     Inner Join issueprodpacks issuppk on whissue.warehissuekey = issuppk.warehissuekey
     Left Join slssalesordrs slsordr on whissue.salesordrkey = slsordr.salesordrkey 
 Where
      whissue.partitionkey = @prtnkey and 
      issuppk.prodpackkey = @prodpackkey

但是这没有用......有没有其他方法可以实现这一目标？

Answer 1

我假设你的逻辑是倒退的。而不是：

case when  whissue.salesordrkey is not null . . .

我想你打算：

case when  slsordr.salesordrkey is not null

第一个不检查left join是否有效。

因此，您可以使用coalesce()：

select slsordr.salesordrkey, whissue.warehissuekey, issuppk.issueprodpackkey,
       coalesce(slsordr.busipartnerkey, whissue.busipartnerkey) as busipartnerkey,
       coalesce(slsordr.contractkey, whissue.contractkey) as contractkey,
       coalesce(slsordr.salesmankey, whissue.salesmankey),
       coalesce(slsordr.customerkey, whissue.customerkey)
from warehissues whissue inner join
     issueprodpacks issuppk
     on whissue.warehissuekey = issuppk.warehissuekey left join 
     slssalesordrs slsordr
     on whissue.salesordrkey = slsordr.salesordrkey
where whissue.partitionkey = @prtnkey and 
      issuppk.prodpackkey = @prodpackkey

我应该注意，这不是100％完全相同的逻辑。某些匹配的记录可能在NULL表中具有slsordr个值。然后使用coalesce()将引入另一个表中的值，即使存在匹配项。但是，这通常是逻辑。

否则，您需要使用多个case语句：

       (case when slsorderkey is not null then slsordr.busipartnerkey
             else whissue.busipartnerkey
        end) as busipartnerkey,

case是一个表达式。它只返回一个值，而不是多个值。

SQL - 在case中选择语句

1 个答案: