“字段列表”中的未知列“parent.username”在 2 个表上执行 LEFT JOIN

Question

我正在构建一个嵌套的 cmets 功能。我有submissions_comments，看起来像：

+---------------+------------------+------+-----+---------+----------------+
| Field         | Type             | Null | Key | Default | Extra          |
+---------------+------------------+------+-----+---------+----------------+
| id            | int(10) unsigned | NO   | PRI | NULL    | auto_increment |
| user_id       | int(10) unsigned | NO   | MUL | NULL    |                |
| submission_id | int(11)          | NO   | MUL | NULL    |                |
| comment       | text             | NO   |     | NULL    |                |
| parent_id     | int(10) unsigned | YES  | MUL | NULL    |                |
| created       | datetime         | NO   | MUL | NULL    |                |
| created_ip    | int(11)          | NO   |     | NULL    |                |
| helpful_count | int(11)          | NO   | MUL | NULL    |                |
| deleted       | tinyint(4)       | NO   | MUL | 0       |                |
+---------------+------------------+------+-----+---------+----------------+'

和看起来像这样的用户：

  +----------------+------------------------+------+-----+-------------------+-----------------------------+
| Field          | Type                   | Null | Key | Default           | Extra                       |
+----------------+------------------------+------+-----+-------------------+-----------------------------+
| id             | int(10) unsigned       | NO   | PRI | NULL              | auto_increment              |
| email          | varchar(128)           | NO   | MUL | NULL              |                             |
| username       | varchar(23)            | NO   |     | NULL              |                             |
| name           | varchar(32)            | NO   |     | NULL              |                             |
| about          | varchar(255)           | NO   |     | NULL              |                             |
+----------------+------------------------+------+-----+-------------------+-----------------------------+

我正在使用 GROUP_CONCAT 和 LEFT JOIN 为我提供所有子 cmets（在我的 submissions_comments 表中），并使用此查询：

SELECT parent.id, MAX(parent.comment) as pcomm,
       GROUP_CONCAT(child.id ORDER BY child.id) as children,
       GROUP_CONCAT(child.comment ORDER BY child.id) as childrenComments
FROM   submissions_comments AS parent
 LEFT  JOIN submissions_comments AS child
   ON  child.parent_id = parent.id
WHERE  parent.parent_id IS NULL
GROUP  BY parent.id
ORDER  BY parent.id;

这给了我：

+----+-------------------------------+----------+--------------------------------------------------------+
| id | pcomm                         | siblings | siblingComments                                        |
+----+-------------------------------+----------+--------------------------------------------------------+
|  1 | This is a parent              | 2,4      | This is a child comment,This is a second child comment |
|  3 | I don't have any children | NULL     | NULL                                                   |
|  5 | Testing one two three         | NULL     | NULL                                                   |
|  6 | adsdfsasdf                    | NULL     | NULL                                                   |
|  7 | asdfadsfdsaf                  | NULL     | NULL                                                   |
|  8 | asdfasdsadfsadf               | NULL     | NULL                                                   |
|  9 | asdfsdafsdafdaafds            | NULL     | NULL                                                   |
+----+-------------------------------+----------+--------------------------------------------------------+

我的问题：我正在尝试在submissions_comments 上加入users u，这样我就可以获得username、email、name、about 对于每个家长评论和子 cmets。

我的查询：

SELECT parent.id, MAX(parent.comment) as pcomm,
       parent.username, 
       GROUP_CONCAT(child.id ORDER BY child.id) as children,
       GROUP_CONCAT(child.comment ORDER BY child.id) as childrenComments,
       GROUP_CONCAT(child.username ORDER BY child.id) as childrenUsernames
FROM   submissions_comments AS parent
 LEFT  JOIN submissions_comments AS child
   ON  child.parent_id = parent.id
LEFT JOIN users u on parent.user_id = u.id
GROUP  BY parent.id
ORDER  BY parent.id;

我想要什么：

+----+-------------------------------+----------+--------------------------------------------------------+-----------------------+
| id | pcomm                         | children | childrenComments                                       |childrenUsernames      |
+----+-------------------------------+----------+--------------------------------------------------------+-----------------------+
|  1 | This is a parent              | 2,4      | This is a child comment,This is a second child comment |blahbster, user123     |
|  3 | I don't have any children | NULL     | NULL                                                   |                       |
|  5 | Testing one two three         | NULL     | NULL                                                   |                       |
|  6 | adsdfsasdf                    | NULL     | NULL                                                   |                       |
|  7 | asdfadsfdsaf                  | NULL     | NULL                                                   |                       |
|  8 | asdfasdsadfsadf               | NULL     | NULL                                                   |                       |
|  9 | asdfsdafsdafdaafds            | NULL     | NULL                                                   |                       |
+----+-------------------------------+----------+--------------------------------------------------------+-----------------------+

我不断收到以下错误：错误 1054 (42S22)：“字段列表”中的未知列“parent.username”和错误 1054 (42S22)：“字段列表”中的未知列“child.username”。

Answer 1

如果您需要parent 和child 的用户名，您将不得不去获取它们；它们不在 submits_cmets 表中，而是链接到该表。

SELECT parent.id, MAX(parent.comment) as pcomm,
       pu.username, 

       /* display child comments in order of child USER id */
       GROUP_CONCAT(cu.id ORDER BY cu.id) as children,
       GROUP_CONCAT(child.comment ORDER BY cu.id) as childrenComments,
       GROUP_CONCAT(cu.username ORDER BY cu.id) as childrenUsernames

  /* handle parent submissions, with user identity */
  FROM submissions_comments AS parent
  LEFT JOIN users AS pu ON parent.user_id = u.id   /* get parents' user info */

  /* handle child submissions, with user identity */
  LEFT JOIN submissions_comments AS child  ON  child.parent_id = parent.id
  LEFT JOIN users AS cu ON child.user_id = cu.id   /* get children's user info */

GROUP  BY parent.id, pu.username   /* avoid stupid nonstandard MySQL GROUP BY */
ORDER  BY parent.id, pu.username

如您所见，您可以对查询进行一些调整以获得您需要的内容。

Answer 2

parent 是submissions_comments 的别名，它没有username 这样的列。
只有users 表有一个名为username 的列。

变化：

select ...
parent.username,
...

收件人：

select ...
u.username,
...

---

您的 group by 子句中还有一个错误：您必须将 username 添加到它以获得正确的行为：

GROUP BY parent.id, u.username