MySQL：使用同一表中输出行中其他行的列数据

Question

我有一个包含类别和子类别的表，与“cat_parent”列链接在一起，其中包含父类别 ID。

当我搜索某些内容时，我希望结果包含其父级的类别 slug（如果没有父级，则为 NULL）作为虚拟列“parent_slug”。

这是我的桌子：

+-----+------------+------------+------------+
| id  | cat_name   | cat_slug   | cat_parent |
+-----+------------+------------+------------+
| 1   | Cars       | cars       | 0          |
+-----+------------+------------+------------+
| 2   | Planes     | planes     | 0          |
+-----+------------+------------+------------+
| 3   | Volvo      | volvo      | 1          |
+-----+------------+------------+------------+
| 4   | Alfa Romeo | alfa-romeo | 1          | 
+-----+------------+------------+------------+
| 5   | Boeing     | boeing     | 2          | 
+-----+------------+------------+------------+
| 6   | Mitsubishi | mitsubishi | 1          | 
+-----+------------+------------+------------+
| 7   | Mitsubishi | mitsubishi | 2          | 
+-----+------------+------------+------------+

当我搜索“volvo”时，我希望结果是这样的：

+-----+----------+----------+------------+-------------+
| id  | cat_name | cat_slug | cat_parent | parent_slug |
+-----+----------+----------+------------+-------------+
| 3   | Volvo    | volvo    | 1          | cars        |
+-----+----------+----------+------------+-------------+

或者搜索 mitsubishi，它会是这样的：

+-----+------------+------------+------------+-------------+
| id  | cat_name   | cat_slug   | cat_parent | parent_slug |
+-----+------------+------------+------------+-------------+
| 6   | Mitsubishi | mitsubishi | 1          | cars        |
+-----+------------+------------+------------+-------------+
| 7   | Mitsubishi | mitsubishi | 2          | planes      |
+-----+------------+------------+------------+-------------+

并且，想象一下我要搜索“s”（如“%s%”），它看起来像这样：

+-----+------------+------------+------------+-------------+
| id  | cat_name   | cat_slug   | cat_parent | parent_slug |
+-----+------------+------------+------------+-------------+
| 1   | Cars       | cars       | 0          | NULL        |
+-----+------------+------------+------------+-------------+
| 2   | Planes     | planes     | 0          | NULL        |
+-----+------------+------------+------------+-------------+
| 6   | Mitsubishi | mitsubishi | 1          | cars        |
+-----+------------+------------+------------+-------------+
| 7   | Mitsubishi | mitsubishi | 2          | planes      |
+-----+------------+------------+------------+-------------+

我希望这是有道理的。我不想更改表结构或添加关系表，因为它对于简单类别来说非常好且快速。

而且，是的，三菱确实制造飞机。 ：P

提前致谢！

Answer 1

SELECT c.*, p.cat_slug as parent_slug FROM YOUR_TABLE_NAME as c
Left Join YOUR_TABLE_NAME as p on c.cat_parent = p.id;

将 YOUR_TABLE_NAME 替换为正确的表名称。