我试图减去2个别名以创建另一个别名,但我得到了一个"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"每个派生的表应该有别名" 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
答案 0 :(得分:2)
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:强>
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。
答案 1 :(得分:1)
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