Question

我试图减去2个别名以创建另一个别名，但我得到了一个＆＃34;未知列＆＃34;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，我想通过从profit中减去revenue来创建expense别名。原因是我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。

我该如何做到这一点？

编辑 - 我已经尝试了下面的答案并且得到了一个＆＃34;每个派生的表应该有别名＆＃34; PHPStorm出错，尝试运行查询时出现语法错误。

下面是新查询：

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

您需要将查询放在子查询中。

SELECT 
t.*,
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

注意：

您只能在GROUP BY, ORDER BY, or HAVING子句中使用列别名。

标准SQL不允许您这样做引用WHERE中的列别名条款。这种限制是强加的因为当WHERE代码是执行后，列值可能尚未执行确定。

Reference

Answer 2

用另一个选择包装它，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

使用2个其他别名的数学结果创建别名

2 个答案: