MySQL - 如何将子查询中的列别名用于另一个子查询？

Posted 2023-05-09

【中文标题】MySQL - 如何将子查询中的列别名用于另一个子查询？

【英文标题】：MySQL - How to use the columns alias from subqueries for another sub query?

【发布时间】：2020-11-10 04:30:32

【问题描述】：

select name, 
 (select sum(balance) from customers group by name having balance>0 and `type of contract`!="loan") as holdings, 
 (select sum(balance) from customers group by name having balance<0 or `type of contract`="loan") as borrowings, 
 (select case when holdings-borrowings>0 then "positive" else "negative" end from customers) as `positive/negative`, 
 holdings-borrowings as total 
 from customers 
 group by name 
 order by name;

错误代码：1054。“字段列表”中的“馆藏”列未知。

表定义为，name varchar，type of contractvarchar，balance int。 我知道错误出在哪里，我无法使用子查询中的列别名，但我不知道如何在另一种方法中执行查询。

【问题讨论】：

正确的解决方案 - 将您的查询转换为 CTE 或子查询。姑息疗法 - 使用用户定义的变量。

【参考方案1】：

您可以使用conditional aggregation尝试以下操作

注意：列别名不能用作投影列表中的参考，这是您收到错误的原因

select name, 
sum(case when balance>0 and `type of contract`!="loan" then balance else 0 end) as holdings,
sum(case when balance<0 and `type of contract`="loan" then balance else 0 end) as borrowings,
case when (sum(case when balance>0 and `type of contract`!="loan" then balance else 0 end)-
sum(case when balance<0 and `type of contract`="loan" then balance else 0 end))>0 then 'positive' else 'negative' end as `positive/negative`,
sum(case when balance>0 and `type of contract`!="loan" then balance else 0 end)-
sum(case when balance<0 and `type of contract`="loan" then balance else 0 end) as total
from customers   
group by name 
order by name

或 -

select name, holdings,borrowings,case when holdings-borrowings>0 then 'Postivie' else 'Negative' end as `positive/negative`,holdings-borrowings as total
from
(
select name, 
    sum(case when balance>0 and `type of contract`!="loan" then balance else 0 end) as holdings,
    sum(case when balance<0 and `type of contract`="loan" then balance else 0 end) as borrowings
    from customers   
    group by name
)A order by name

【讨论】：

但在持有和借款时，我需要余额的总和，而不仅仅是余额。

@NithinGowda，你执行了查询吗？

抱歉过早发表评论。它工作正常。谢谢。

@NithinGowda，请不要忘记将其标记为已接受的答案 :)