Question

我的查询

select count(*) as cnt,
       EXTRACT(day FROM current_date - min(txdate))::int as days,

       sum (Select opening from acledgerbal l 
              where acname='Arv' 
              union all 
              Select sum(v2.debit-v2.credit) as opening from acvoucher2 v2 where
              txdate<='05/03/2014') as opening

from acduebills acb,acledger l 

where (acb.opening+acb.debit-acb.credit) > 0 
   and acb.unitname='Sales' 
   and l.acname='Arv' 
   and l.acno=acb.acno

此处显示more than one row returned by a subquery used as an expression错误。如何在子查询中使用sum。

我正在使用postgresql 9.1

编辑：

我想得到acduebills表中的行数，即（acb.opening + acb.debit-acb.credit）＆gt; 0和acb.unitname ='销售'。在那之后我想得到相同条件下最小日期的一天的差异。在那之后，我想要开放，它来自两个表：acledgerbal和acvoucher2。 acvoucher条件检查txdate表格。

如何在单个查询中获取这些细节？如何在多个模式中获取相同的详细信息？

Answer 1

以下陈述中的问题：

   sum (  Select opening from acledgerbal l 
          where acname='Arv' 
          union all 
          Select sum(v2.debit-v2.credit) as opening from acvoucher2 v2,
          txdate<='05/03/2014' )

您使用UNION，因此此子查询至少返回2行。因此，您得到一个错误，即子查询不能返回多行：“用作表达式的子查询返回的多行”

尝试将其更改为：

   (Select SUM(opening) from acledgerbal l WHERE acname='Arv')
   +  
   (Select SUM(v2.debit-v2.credit) as opening from acvoucher2 v2 
                                           WHERE txdate<='05/03/2014')

Answer 2

这样的事情：

SELECT count(*) AS cnt
     , current_date - min(txdate)::date AS days   -- subtract dates directly

     , (SELECT round(sum(opening)::numeric, 2)
        FROM  (
           SELECT opening
           FROM   acledgerbal
           WHERE  acname = 'Arv' 

           UNION ALL
           SELECT debit - credit
           FROM   acvoucher2
           WHERE  txdate <= '2014-05-03'
           ) sub
       ) AS opening
FROM   acduebills b
JOIN   acledger   l USING (acno)
WHERE ((b.opening + b.debit) - b.credit) > 0 
AND    b.unitname ='Sales' 
AND    l.acname = 'Arv';

round()到小数位只适用于numeric类型，所以我投了总和。

在子查询中求和

2 个答案: