Question

我有一个名为问题的表，其中包含以下行：

questions.id | questions.target_username
   1         | every.one
   2         | every.one

我还有以下行的答案表：

answers.id | answers.username       
   1       | guy
   1       | maricela 
   2       | mikha

正如您所看到的，相同的问题ID可能会有来自不同用户的答案。我想在以下情况下只选择行：

a）问题的答案不包括（mikha）为answers.username，在这种情况下只显示一次问题ID

b）问题的答案用户名（mikha）为answers.username。

我使用以下查询：

SELECT questions.id, answers.username
FROM questions
LEFT JOIN answers ON ( questions.id = answers.id ) 
WHERE questions.target_username =  'every.one'
AND (
answers.username IS NOT NULL 
OR answers.username =  'mikha'
)
GROUP BY questions.id, answers.username

结果我期待：

questions.id | answers.usernme
   1         |  
   2         | mikha

我真正得到的是：

questions.id | answers.usernme
   1         | guy
   1         | maricela  
   2         | mikha

提前致谢问候 Micahel

Answer 1

您只需要将answers.username = 'mikha'条件移动到on子句，此时甚至可以删除条件中的冗余条件。

SELECT questions.id, answers.username
FROM questions
LEFT JOIN answers ON(questions.id = answers.id AND answers.username = 'mikha')
WHERE questions.target_username = 'every.one'
GROUP BY questions.id, answers.username;

编辑：我不确定这是否是一项要求，但如果您想省略尚未有人回答的问题，以下是您想要的：

SELECT questions.id, answers.username
FROM questions
LEFT JOIN answers ON(questions.id = answers.id AND answers.username = 'mikha')
LEFT JOIN answers others ON(questions.id = others.id AND others.username <> 'mikha')
WHERE questions.target_username = 'every.one'
AND (
    (
        answers.id IS NOT NULL
        AND others.id IS NULL
    )
    OR (
        answers.id IS NULL
        AND others.id IS NOT NULL
    )
)
GROUP BY questions.id, answers.username;

编辑2：这是我在测试表中的内容，以及查询＃2的结果：

SELECT * FROM questions;
+----+-----------------+
| id | target_username |
+----+-----------------+
|  1 | every.one       |
|  2 | every.one       |
|  3 | every.one       |
|  4 | every.one       |
+----+-----------------+
4 rows in set (0.00 sec)

SELECT * FROM answers;
+-----+----+----------+
| ida | id | username |
+-----+----+----------+
|   1 |  1 | guy      |
|   2 |  1 | maricela |
|   3 |  2 | mikha    |
+-----+----+----------+
3 rows in set (0.00 sec)

(Run query #2 above)
+----+----------+
| id | username |
+----+----------+
|  1 | NULL     |
|  2 | mikha    |
+----+----------+
2 rows in set (0.00 sec)

编辑3：以下是更新的表格数据和符合您在评论中添加的条件的更新查询：

SELECT * FROM questions;
+----+-----------------+
| id | target_username |
+----+-----------------+
|  1 | every.one       |
|  2 | every.one       |
|  3 | every.one       |
|  4 | every.one       |
+----+-----------------+
4 rows in set (0.00 sec)

SELECT * FROM answers;
+-----+----+----------+
| ida | id | username |
+-----+----+----------+
|   1 |  1 | guy      |
|   2 |  1 | maricela |
|   3 |  2 | mikha    |
|   7 |  4 | guy      |
|   8 |  4 | mikha    |
+-----+----+----------+
5 rows in set (0.00 sec)

新查询：

SELECT questions.id, answers.username
FROM questions
LEFT JOIN answers ON(questions.id = answers.id AND answers.username = 'mikha')
LEFT JOIN answers others ON(questions.id = others.id AND others.username <> 'mikha')
WHERE questions.target_username = 'every.one'
AND (
    answers.username = 'mikha'
    OR (
        answers.id IS NULL
        AND others.id IS NOT NULL
    )
)
GROUP BY questions.id, answers.username;

结果：

+----+----------+
| id | username |
+----+----------+
|  1 | NULL     |
|  2 | mikha    |
|  4 | mikha    |
+----+----------+
3 rows in set (0.00 sec)

Answer 2

你必须改变答案表，

更改答案。想问题。我添加answer.Id作为主键...

在你的目标不明确，因为你期望的结果有第2行，mikha（你说如果用户名是mikha，我只需要问题ID）？

Answer 3

如果你加入来回答id只有那些'mikha'的用户名你应该得到想要的结果。那些有答案而不是mikhas答案的id将获得null值。我不明白你在哪里需要问题表来获取这些信息：

SELECT answers.id, a2.username
FROM answers
LEFT JOIN (
   SELECT id, username
   FROM answers
   WHERE username = 'mikha'
) AS a2 ON a2.id = answers.id
GROUP BY answers.id

Answer 4

您获得此结果的原因是：

answers.username IS NOT NULL OR answers.username = 'mikha'状况不佳。第二种情况answers.username = 'mikha'已经是answers.username IS NOT NULL的情况 - 所以这是一个毫无意义的第二个条件。
LEFT JOIN加入此条件时，基本上只相当于INNER JOIN。

如果值等于，GROUP_BY仅显示第一次出现

4 个答案: