从 MySQL 中的 2 个不同的表中创建一个别名列

Question

我有一个与 2 个不同表相关的数据库表，例如：

   === inventory ===
+------------+-----------+
|    code    |   total   |
+------------+-----------+
| el_pr_25   |     45    |
| el_pr_11   |     33    |
| mob_tp_x93 |     23    |
| mob_tp_t55 |     33    |
| el_pr_x73  |     25    |
| mob_tp_25  |     22    |
+------------+-----------+

= electricity =
+-----+-------+
|  id | name  |
+-----+-------+
|  25 | test1 |
|  11 | test2 |
| x73 | test3 |
+-----+-------+

  == mobile ==
+-----+----------+
|  id |   name   |
+-----+----------+
| x93 | test 66  |
| t55 | test 222 |
| 25  | test 323 |
+-----+----------+

我要选择的主表是库存表，库存表通过code列连接到另一个表，对于电力表有一个el_pr_前缀后跟电力表的id而对于移动表前缀是mob_tp_是前缀，我想从电力和移动表中选择名称列的库存表，例如结果将是这样的：

SELECT code,total, ... as name FROM inventory; 


         === inventory ===
+------------+-----------+----------+
|    code    |   total   |  name    |
+------------+-----------+----------+
| el_pr_25   |     45    | test1    |
| el_pr_11   |     33    | test2    |
| mob_tp_x93 |     23    | test 66  |
| mob_tp_t55 |     33    | test 22  |
| el_pr_x73  |     25    | test3    |
| mob_tp_25  |     22    | test 323 |
+------------+-----------+----------+

Answer 1

我们可以试试下面的连接查询：

SELECT
    i.code,
    i.total,
    COALESCE(e.name, m.name) AS name
FROM inventory i
LEFT JOIN electricity e
    ON i.code REGEXP CONCAT('el_.*_', e.id, '$')
LEFT JOIN mobile m
    ON i.code REGEXP CONCAT('mob_.*', m.id, '$');

Demo

上述查询使用COALESCE 技巧为每个项目选择正确的名称，它假定给定项目仅匹配electricity 或mobile 表。

但是，您的数据库设计并不理想。最好只有一个包含移动和电气（和其他）项目元数据的表。此外，您的表应该有适当的连接列，不需要需要复杂的子字符串或正则表达式操作来匹配。我建议如下：

inventory
+----+------------+-----------+
| id |    code    |   total   |
+----+------------+-----------+
| 1  | el_pr_25   |     45    |
| 2  | el_pr_11   |     33    |
| 3  | el_pr_x73  |     25    |
| 4  | mob_tp_x93 |     23    |
| 5  | mob_tp_t55 |     33    |
| 6  | mob_tp_25  |     22    |
+----+------------+-----------+

items
+--------------+----------+-------------+
| inventory_id | name     | type        |
+--------------+----------+-------------+
| 1            | test1    | electricity |
| 2            | test2    | electricity |
| 3            | test3    | electricity |
| 4            | test 66  | mobile      |
| 5            | test 222 | mobile      |
| 6            | test 323 | mobile      |
+--------------+----------+-------------+

Answer 2

您可以使用inner join 和union [all] 作为：

select i.*, e.name from inventory i 
                   inner join electricity e on ( i.code = concat('el_pr_',e.id) ) union all
select i.*, m.name from inventory i 
                   inner join mobile m on ( i.code = concat('mob_tp_',m.id) );

Rextester Demo

Answer 3

我建议将您的数据模型更改为单独的列，并通过同时使用它们形成独特的约束。

|code  |   total   |type
+------+-----------+
| 25   |     45    |el_pr_
| 11   |     33    |el_pr_
| x93  |     23    |mob_tp_
| t55  |     33    |mob_tp_
| x73  |     25    |el_pr_
| 25   |     22    |mob_tp_

并将名称来源合并到一个表中

|  id | name  |type
+-----+-------+
|  25 | test1 |el_pr_
|  11 | test2 |el_pr_
| x73 | test3 |el_pr_
| x93 | test 66  |mob_tp_
| t55 | test 222 |mob_tp_
| 25  | test 323 |mob_tp_

然后

select *
from inventory i
join tab_names n on i.type = n.type and i.code = n.id

这只是一个粗略的轮廓！