mySQL 表连接 - 错误答案

Question

我对 mySQL 查询还很陌生，并且努力获得我需要的结果。 我有两个需要显示的用户信息表。表格如下:

users
+----+------------------+
| id | email            | 
+----+------------------+
|  1 |  joe@hotmail.com |
|  2 | john@hotmail.com |
|  3 | fred@hatmail.com |
+----+------------------+

user_detail
+----------+--------+--------+
| detailid | userid | detail | 
+----------+--------+--------+
|    1     |   1    | Joe    |
|    2     |   1    | Soap   |
|    1     |   2    | John   |
|    2     |   2    | Doe    |
|    1     |   3    | Fred   |
|    2     |   3    | Bloggs |
+----------+--------+--------+     

我构建了以下连接表的查询:

SELECT id, detail , email
FROM users
LEFT JOIN user_detail
ON users.id=user_detail.userid
ORDER by id

查询产生这个结果:

+--------+--------+------------------+
| userid | detail |    email         |
+--------+--------+------------------+
|   1    | Joe    |  joe@hotmail.com |
|   1    | Soap   |  joe@hotmail.com |
|   2    | John   | john@hotmail.com |
|   2    | Doe    | john@hotmail.com |
|   3    | Fred   | fred@hatmail.com |
|   3    | Bloggs | fred@hatmail.com |
+--------+--------+------------------+     

我正在努力实现的是:

+--------+---------+---------+------------------+
| userid | detail1 | detail2 |    email         |
+--------+---------+---------+------------------+
|   1    |  Joe    | Soap    | joe@hotmail.com  |
|   2    |  John   | Doe     | john@hotmail.com |
|   3    |  Fred   | Bloggs  | fred@hatmail.com |
+--------+---------+---------+------------------+    

你能帮我指出正确的方向吗？

Answer 1

SELECT id,
MAX(CASE WHEN detaildid=1 THEN detail END) as detail1,
MAX(CASE WHEN detaildid=2 THEN detail END) as detail2,
email
FROM users
LEFT JOIN user_detail
ON users.id=user_detail.userid
GROUP BY id,email
ORDER by id

如果你有很多 detailids，这可以动态编写。