如何在同一个选择查询中将一个表的值交换到另一个表的值？答案

【问题标题】：How can I swap a value from one table to that of another table in the same select query?如何在同一个选择查询中将一个表的值交换到另一个表的值？
【发布时间】：2018-07-31 08:15:19
【问题描述】：

我的数据库由两个表组成

用户表

+---------+----------+---------------+-----------+
| user_id | username | password_hash | user_type |
+---------+----------+---------------+-----------+
| 1       | Admin    | hash          | 0         |
| 2       | Student  | hash          | 1         |
| 3       | Teacher  | hash          | 2         |
+---------+----------+---------------+-----------+

门票表

+-----------+---------+-----------------+----------------+--------------------+---------------+-------------+
| ticket_id | user_id | ticket_category | ticket_summary | ticket_description | ticket_status | assignee_id |
+-----------+---------+-----------------+----------------+--------------------+---------------+-------------+
| 1         | 2       | Network         | Test Network   | Lorem Ipsum        | 1             | 1           |
| 2         | 3       | Printer         | Test Printer   | Lorem Ipsum        | 1             |             |
+-----------+---------+-----------------+----------------+--------------------+---------------+-------------+

tickets.user_id 和tickets.assignee_id 都是users.user_id 的外键

我正在使用以下 select 语句查询这些表

SELECT username AS `Opened By`, assignee_id AS `Assigned To`, ticket_category AS `Category`, ticket_summary AS `Summary`, ticket_description AS `Description`, IF(ticket_status = 1,'Open','Closed') AS `Status` FROM tickets INNER JOIN users ON tickets.user_id = users.user_id;

这样构建一个数据集

+-----------+-------------+----------+--------------+-------------+--------+
| Opened By | Assigned To | Category | Summary      | Description | Status |
+-----------+-------------+----------+--------------+-------------+--------+
| Student   | 1           | Network  | Test Network | Lorem Ipsum | Open   |
| Teacher   |             | Printer  | Test Printer | Lorem Ipsum | Open   |
+-----------+-------------+----------+--------------+-------------+--------+

在我的 select 语句中，我想将 assignee_id 编号替换为它在 users 表中也对应的用户名。

例如，第一张工单 (ticket_id = 1) 的 assignee_id 为 1，我希望“分配给”列显示用户名“Admin”而不是 1

如何修改我的选择查询来完成此操作？

【问题讨论】：

标签： mysql

【解决方案1】：

只需为assignee_id 再次加入用户表。请注意，它必须是LEFT JOIN，以防没有人分配给票证。您可能还想在u2.username 上使用COALESCE 来输出空白而不是(null)（即COALESCE(u2.username, '')）。 SQLFiddle.

SELECT u1.username AS `Opened By`, 
       u2.username AS `Assigned To`, 
       ticket_category AS `Category`, 
       ticket_summary AS `Summary`, 
       ticket_description AS `Description`, 
       IF(ticket_status = 1,'Open','Closed') AS `Status` 
FROM tickets 
JOIN users u1 ON tickets.user_id = u1.user_id
LEFT JOIN users u2 ON tickets.assignee_id = u2.user_id

样本数据的输出：

Opened By   Assigned To   Category   Summary       Description   Status
Student     Admin         Network    Test Network  Lorem Ipsum   Open 
Teacher     (null)        Printer    Test Printer  Lorem Ipsum   Open

【讨论】：

【解决方案2】：

select T.OpenedBy,u.username as AssignedTo,Category,Summary,Description,Status from 
(
 SELECT username AS `OpenedBy`, assignee_id , ticket_category AS `Category`, ticket_summary AS `Summary`, ticket_description AS `Description`, IF(ticket_status = 1,'Open','Closed') AS `Status` FROM tickets INNER JOIN users ON tickets.user_id = users.user_id
) as T

inner join
users u on T.assignee_id=u.user_id

【讨论】：