MySql自我JOIN查询

毛格说恢复莫妮卡：

我有这张桌子

mysql> describe skill_usage;
+----------+---------+------+-----+---------+-------+
| Field    | Type    | Null | Key | Default | Extra |
+----------+---------+------+-----+---------+-------+
| skill_id | int(11) | NO   | MUL | NULL    |       |
| job_id   | int(11) | NO   | MUL | NULL    |       |
+----------+---------+------+-----+---------+-------+

并且知道在我的数据中，有一个job_id（6）分别用于skill_id3和4：

mysql>  select * from skill_usage;
+----------+--------+
| skill_id | job_id |
+----------+--------+
|        1 |      1 |
|        2 |      2 |
|        3 |      3 |     <----  matches only one part of the AND clause
|        3 |      4 |     <----  matches only one part of the AND clause
|        2 |      5 |
|        3 |      6 |     <==== matches both parts of the AND clause
|        4 |      6 |     <====
|        2 |      7 |
+----------+--------+
8 rows in set (0.00 sec)

这是我尝试过的：

SELECT DISTINCT s1.job_id FROM skill_usage AS s1 
  INNER JOIN skill_usage AS s2 ON s1.job_id = s2.job_id
    WHERE s1.skill_id IN (3,4)
    AND   s2.skill_id IN (3,4)

我认为这意味着“找到job_id同时匹配skill_id3和skill_id4的所有内容”。

显然不是：

mysql> SELECT DISTINCT s1.job_id FROM skill_usage AS s1
    ->   INNER JOIN skill_usage AS s2 ON s1.job_id = s2.job_id
    ->     WHERE s1.skill_id IN (3,4)
    ->     AND   s2.skill_id IN (3,4);
+--------+
| job_id |
+--------+
|      3 |
|      4 |
|      6 |
+--------+
3 rows in set (0.00 sec)

我做错了什么？我的查询应如何阅读？我认为这是一本好书或Udemy课程的时间，但我所拥有的并没有掩盖自我。

我的查询正确找到job_id= 6，但错误地找到了（IMO），找到job_id3和4。我希望它们使AND子句失败。

阿布拉：

替代解决方案。
（请参阅db fiddle）

select s1.job_id
  from skill_usage s1
  where s1.skill_id = 3
    and s1.job_id in (
                       select s2.job_id
                         from skill_usage s2
                        where s2.skill_id = 4
                     )

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