使用2个其他别名的数学结果创建别名

Question

我试图减去2个别名,以便创建另一个别名,但我得到一个"未知列"错误.

这是我的SQL:

select o.id, o.name,
     (select sum(l.source_expense)
        from `assignments` as a
        left join `leads` as l on (l.id = a.lead_id)
        where a.{$this->sql_column}=o.id
        and l.date_created between {$this->date_from} and {$this->date_to}
        and find_in_set(l.vertical_id, '".implode(',', $this->app_user->verticals)."')
     ) as `expense`,
     (select sum(a.buyer_revenue)
        from `assignments` as a
        left join `leads` as l on (l.id = a.lead_id)
        where a.refunded=0
        and a.{$this->sql_column}=o.id
        and l.date_created between {$this->date_from} and {$this->date_to}
        and find_in_set(l.vertical_id, '".implode(',', $this->app_user->verticals)."')
     ) as `revenue`,
     `revenue` - `expense` as `profit`
     from {$this->sql_table} as o

基本上,我想profit通过减去revenuefrom 来创建一个别名expense.原因是我正在使用数据表并希望列可以排序.我已经知道我可以用PHP轻松做到这一点.

我怎么能做到这一点？

编辑 - 我尝试了下面的答案,并从PHPStorm获得"每个派生表应该有别名"错误,并在尝试运行查询时出现语法错误.

继承人新的查询:

select t.id, t.name, t.expense, t.revenue, t.revenue - t.expense as profit
from(select o.id, o.name,
     (select sum(l.source_expense)
        from `assignments` as a
        left join `leads` as l on (l.id = a.lead_id)
        where a.{$this->sql_column}=o.id
        and l.date_created between {$this->date_from} and {$this->date_to}
        and find_in_set(l.vertical_id, '".implode(',', $this->app_user->verticals)."')
     ) as `expense`,
     (select sum(a.buyer_revenue)
        from `assignments` as a
        left join `leads` as l on (l.id = a.lead_id)
        where a.refunded=0
        and a.{$this->sql_column}=o.id
        and l.date_created between {$this->date_from} and {$this->date_to}
        and find_in_set(l.vertical_id, '".implode(',', $this->app_user->verticals)."')
     ) as `revenue`
     from {$this->sql_table} as o
 ) as t

Answer 1

只需用另一个 select 包裹它，然后别名将可用于数学计算：

SELECT t.id,o.name,t.expense,t.revenue,
       t.revenue -t.expense as `profit`
FROM (Your Query Here) t