Question

我对mySQL查询很新，并努力实现我需要的结果。我需要提供两个用户信息表。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 |
+----------+--------+--------+

我构建了以下用于连接表的查询：

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

查询产生以下结果：

+--------+--------+------------------+
| 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 |
+--------+--------+------------------+

我正在努力实现的目标是：

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

你能帮忙并指出我正确的方向吗？

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

如果您有许多详细信息，可以动态编写。

Answer 2

您可以两次加入同一个表，一次用于详细信息ID 1，另一个用于详细信息ID 2

SELECT id, d1.detail, d2.detail, email
FROM users u
LEFT JOIN user_detail d1
ON u.id = d1.userid
AND d1.detailid = 1
LEFT JOIN user_detail d2
ON u.id = d2.userid
AND d2.detailid = 2
ORDER BY ID

mySQL表加入 - 错误的答案

2 个答案: