Mysql Unknown column in where 子句 union all

Posted 2023-04-12

【中文标题】Mysql Unknown column in where 子句 union all

【英文标题】：Mysql Unknown column in where clause union all

【发布时间】：2020-03-07 08:07:02

【问题描述】：

仍然与我的Previous Question 相关，有一个像这样的表（tb_data）

+---------+---------------------+---------------------+---------------------+---------------------+-------+ 
| Disease | Additional_Disease1 | Additional_Disease2 | Additional_Disease3 | Additional_Disease4 | Room  |
+---------+---------------------+---------------------+---------------------+---------------------+-------+
| A01     | A03                 | A02                 |                     |                     | Man   |
| A03     | A02                 |                     |                     |                     | Woman |
| A03     | A05                 |                     |                     |                     | Child |
| A03     | A05                 |                     |                     |                     | Man   |
| A02     | A05                 | A01                 | A03                 |                     | UGD   |
+---------+---------------------+---------------------+---------------------+---------------------+-------+ 

我的问题是如何做到这一点

+---------+-------+ 
| Disease | Total |
+---------+-------+
| A03     | 2     |
| A02     | 1     |
| A01     | 1     |
| A05     | 1     |
+---------+-------+

这是我的代码尝试

    select Disease, count(*) total
    from (
    select Disease from tb_data
    union all select Additional_Disease1 from tb_data
    union all select Additional_Disease2 from tb_data
    union all select Additional_Disease3 from tb_data
    union all select Additional_Disease4 from tb_data
) t
where Disease is not Null
and Room = 'Man'
group by Disease
order by total desc, Disease

导致错误的原因

Error Code: 1054. Unknown column 'Room' in 'where clause'

【问题讨论】：

任何时候你发现自己有枚举列名（上面说'2'），你应该考虑你的架构设计是否真的是最优的

最佳？你能详细说明/举个例子吗？我对编程有点陌生，所以我仍然不太了解......

您的设计应该类似于 Nick 的子查询的输出

【参考方案1】：

您的问题是您的派生表不包含 Room 列。您可以在派生表中按Room 过滤：

select Disease, count(*) total
from (
    select Disease from tb_data where Room = 'Man'
    union all select Additional_Disease1 from tb_data where Room = 'Man'
    union all select Additional_Disease2 from tb_data where Room = 'Man'
    union all select Additional_Disease3 from tb_data where Room = 'Man'
    union all select Additional_Disease4 from tb_data where Room = 'Man'
) t
where Disease is not Null
group by Disease
order by total desc, Disease

或者在派生表中包含Room 列：

select Disease, count(*) total
from (
    select Room, Disease from tb_data
    union all select Room, Additional_Disease1 from tb_data
    union all select Room, Additional_Disease2 from tb_data
    union all select Room, Additional_Disease3 from tb_data
    union all select Room, Additional_Disease4 from tb_data
) t
where Disease is not Null
and Room = 'Man'
group by Disease
order by total desc, Disease

【讨论】：

@KanienNotiros 不用担心 - 我很高兴能提供帮助。

【参考方案2】：

如果tb_data 很大或由子查询生成，那么我不推荐使用union all 方法来取消透视它。而是：

select (case n.n
           when 1 then Disease
           when 2 then Additional_Disease1
           when 3 then Additional_Disease2
           when 4 then Additional_Disease3
           when 5 then Additional_Disease4
        end) as the_disease, count(*)
from tb_data d cross join
     (select 1 as n union all
      select 2 as n union all
      select 3 as n union all
      select 4 as n union all
      select 5 as n
     ) n
where Room = 'Man'
group by the_disease
having the_disease is not null