Mysql 如何检查JOIN的结果是否为NULL？

以下是我的表格结构：

-- categories
+----+-------------+
| id |     name    |
+----+-------------+
| 1  | political   |
| 2  | cultural    |
| 3  | social      |
| 4  | historical  |
+----+-------------+

-- tags
+----+-------------+-----------+-------------+-----------------+
| id |     name    | parent_id | category_id |  total_used_num |
+----+-------------+-----------+-------------+-----------------+
| 1  | president   | NULL      | 1           | 43              |
| 2  | society     | NULL      | 3           | 345             |
| 3  | Trump       | 1         | 1           | 5               |
| 4  | book        | 5         | 2           | 5473            |
| 5  | library     | NULL      | 2           | 433             |
+----+-------------+-----------+-------------+-----------------+

这是我的问题：

SELECT t1.name, 
       CONCAT(t2.name, ',', cat.name) t_p 
FROM   tags t1 
       LEFT JOIN tags t2 
              ON t1.parent_id = t2.id 
       INNER JOIN categories cat 
               ON t1.category_id = cat.id 
ORDER  BY total_used_num DESC, 
          t1.id 

/* output
+----+-------------+--------------------------+
| 1  | president   | NULL                     |
| 2  | society     | NULL                     |
| 3  | Trump       | president,political      |
| 4  | book        | library,cultural         |
| 5  | library     | NULL                     |
+----+-------------+--------------------------+

看到了吗？带有

NULL

的

CONCAT

的结果是

NULL

。不管怎样，我怎样才能避免呢？我还想把逗号设为有条件的。如果两个字段都不是空的，那么逗号应该在那里。这是预期结果：

+----+-------------+--------------------------+
| 1  | president   | political                |
| 2  | society     | social                   |
| 3  | Trump       | president,political      |
| 4  | book        | library,cultural         |
| 5  | library     | cultural                 |
+----+-------------+--------------------------+

我该怎么做呢？

只要使用

concat\u ws（）

：

---

您可以使用case语句。这是Oracle查询，您可以为MySQL修改：

SELECT   t1.name, 
            (case when t2.name is not null then ( t2.name || ', ') else '' end)  || cat.name 
FROM   tags t1 
       LEFT JOIN tags t2 
              ON t1.parent_id = t2.id 
       INNER JOIN categories cat 
               ON t1.category_id = cat.id  
ORDER  BY t1.total_used_num DESC, 
          t1.id ;

---

只要使用CASE WHEN语句，就可以得到预期的输出。 请参阅下面的代码

SELECT t1.name,  (CASE 
                WHEN  (t2.name IS NULL) 
                THEN  cat.name
                ELSE (CONCAT(t2.name, ',', cat.name))
                END) AS t_p 
FROM   tags t1 
       LEFT JOIN tags t2 
              ON t1.parent_id = t2.id 
       INNER JOIN categories cat 
               ON t1.category_id = cat.id

---

你到底是谁？你有没有在不到一分钟的时间里读过我的问题？！你真的很棒。我印象深刻。

SELECT t1.name,  (CASE 
                WHEN  (t2.name IS NULL) 
                THEN  cat.name
                ELSE (CONCAT(t2.name, ',', cat.name))
                END) AS t_p 
FROM   tags t1 
       LEFT JOIN tags t2 
              ON t1.parent_id = t2.id 
       INNER JOIN categories cat 
               ON t1.category_id = cat.id