如何根据返回的行 SELECT 多列？

Question

如果我运行 SELECT 查询并返回包含冗余信息的行，我如何在一行中显示该信息？

目前我的查询是这样的：

SELECT email, first_name, group FROM membership

这会返回一个结果集，如下所示：

------------------------------------
|    email    | first_name | group |
------------------------------------
|  bob@co.uk  |     bob    |   g1  |
|  jim@co.uk  |     jim    |   g1  |
|  jim@co.uk  |     jim    |   g2  |
|  jane@co.uk |    henry   |   g2  |
------------------------------------

相反，我需要结果集如下所示：

--------------------------------------------
|    email    | first_name | group | group |
--------------------------------------------
|  bob@co.uk  |     bob    |   g1  |       |
|  jim@co.uk  |     jim    |   g1  |   g2  |
|  jane@co.uk |    henry   |   g2  |       |
--------------------------------------------

重要的是，列名“组”尽管是多余的，但不会以不同的方式命名。

编辑：为清楚起见，组的数量可能会改变。 edit2：我使用的完整代码，用于初始查询，是：

SELECT 
    persons.mail AS email,
    persons.fname AS 'first_name',
persons.lname AS 'last_name',
groups.name as add_list
FROM group_membership
LEFT JOIN groups ON group_membership.gid = groups.gid
LEFT JOIN persons ON group_membership.uid = persons.uid
WHERE persons.mail != ""
ORDER BY last_name

Answer 1

你所要求的是不可能的。 如果有四行的值：

|  jim@co.uk  |     jim    |   g2  |

您是否希望结果有 4 个名为group的列，而不是您当前显示的两个group列？

你可能想要的是：

select email, first_name, group_concat(`group`) group by email, first_name

Answer 2

对于您的示例数据，这将执行以下操作：

select email, first_name,
  min(`group`) `group`, 
  case when max(`group`) <> min(`group`) then max(`group`) end `group`
from membership
group by email, first_name 

请参阅演示。
结果：

> email      | first_name | group | group
> :--------- | :--------- | :---- | :----
> bob@co.uk  | bob        | g1    | 
> jane@co.uk | henry      | g2    | 
> jim@co.uk  | jim        | g1    | g2

Answer 3

根据上面的示例，以下将起作用

     SELECT email, first_name, max(case 
        when 
      group ='g1' then group else null end) 
       as groupg1, 
      max(case 
        when 
      group ='g2' then group else null end) 
       as groupg2
      FROM membership group by email, 
      first_name