从连接表中获取所有数据都存在且不存在

Question

连接表中的记录，存在且不存在

表格文章;

+------------+-------+--------+-
| id| cat-id |aut-id |content |
+------------+-------+--------+
| 1 |   1    |    3  |eeeeeee |
| 2 |   2    |    4  |fffffff |
| 3 |   8    |    5  |ggggggg |
| 4 |   9    |    2  |jjjjjjj |
+------------+-------+--------+

表类别;

+--------------+
| id| cat-name |
+--------------+
| 1 |   phy    |
| 2 |   che    | 
| 3 |   bio    |
+--------------+

表作者;

+--------------+
| id| aut-name |
+--------------+
| 4 |   joe    |
| 5 |   sam    | 
| 6 |   kim    |
+--------------+

我想要这个结果

+------------+-------+--------+--------+--------+
| id| cat-id |aut-id |cat-name|aut-name|content |
+------------+-------+--------+--------+--------+
| 1 |   1    |    3  |  phy   |unknown |eeeeeee |
| 2 |   2    |    4  |  che   |  joe   |fffffff |
| 3 |   8    |    5  |  uncat |  sam   |ggggggg |
| 4 |   9    |    2  |  uncat | unknown|jjjjjjj |
+------------+-------+--------+--------+--------+

Answer 1

只是LEFT JOIN数据，所以你不会过滤掉缺失的数据

QUERY：

SELECT 
    a.`id`, 
    a.`cat-id` , 
    a.`aut-id` , 
    COALESCE(c.`cat-name`, 'uncat') as 'cat-name',
    COALESCE(a1.`aut-name`, 'unknown') as 'aut-name' ,
    a.`content` 
FROM `article` a
LEFT JOIN `categories` c on c.`id` = a.`cat-id`
LEFT JOIN `authors` a1 on a1.`id` = a.`aut-id`

使用COALESCE处理空值并输入您想要的结果

DEMO

输出：

+------------+-------+--------+--------+--------+
| id| cat-id |aut-id |cat-name|aut-name|content |
+------------+-------+--------+--------+--------+
| 1 |   1    |    3  |  phy   |unknown |eeeeeee |
| 2 |   2    |    4  |  che   |  joe   |fffffff |
| 3 |   8    |    5  |  uncat |  sam   |ggggggg |
| 4 |   9    |    2  |  uncat | unknown|jjjjjjj |
+------------+-------+--------+--------+--------+