MySQL，需要帮助定义多个连接查询以返回完整信息

Question

所以，我修改了从this thread 学到的查询，但是当我在标签和猫之间进行过滤时，结果并不理想。过滤类别 5 将仅返回类别列表信息，标签将为空，而标签则相反。

    SELECT posts.id,time,title,
        GROUP_CONCAT(IFNULL(cats.id, '') ORDER BY cats.id DESC SEPARATOR '~') as catIdList,
        GROUP_CONCAT(IFNULL(cats.name, '') ORDER BY cats.id DESC SEPARATOR '~') as catNameList,
        GROUP_CONCAT(IFNULL(cats.slug, '') ORDER BY cats.id DESC SEPARATOR '~') as catSlugList,
        GROUP_CONCAT(IFNULL(cats.value, '') ORDER BY cats.id DESC SEPARATOR '~') as catValueList,
        GROUP_CONCAT(IFNULL(tags.id, '') ORDER BY tags.id DESC SEPARATOR '~') as tagIdList,
        GROUP_CONCAT(IFNULL(tags.name, '') ORDER BY tags.id DESC SEPARATOR '~') as tagNameList,
        GROUP_CONCAT(IFNULL(tags.slug, '') ORDER BY tags.id DESC SEPARATOR '~') as tagSlugList,
        GROUP_CONCAT(IFNULL(tags.value, '') ORDER BY tags.id DESC SEPARATOR '~') as tagValueList
    FROM posts
    LEFT JOIN termRelations ON ( posts.id = termRelations.postId )
    LEFT JOIN cats ON ( termRelations.termId = cats.id AND termRelations.termTypeId = 1 )
    LEFT JOIN tags ON ( termRelations.termId = tags.id AND termRelations.termTypeId = 0 )
    WHERE ( ( IFNULL(tags.id, '') = '4' ) )
    GROUP BY posts.id ORDER BY time DESC

IFNULL() 可以解决不存在的条目。上面的查询将返回：

    (
        [id] => 15
        [time] => 0
        [title] => post 15
        [catIdList] => 
        [catNameList] => 
        [catSlugList] => 
        [catValueList] => 
        [tagIdList] => 4
        [tagNameList] => tagname
        [tagSlugList] => tagname
        [tagValueList] => 
    )

    (
        [id] => 16
        [time] => 0
        [title] => post 16
        [catIdList] => 
        [catNameList] => 
        [catSlugList] => 
        [catValueList] => 
        [tagIdList] => 4
        [tagNameList] => tagname
        [tagSlugList] => tagname
        [tagValueList] => 
    )

如果没有WHERE ( ( IFNULL(tags.id, '') = '4' ) )，结果将是（当然还有所有其他帖子，因为它没有被过滤到这个标签中）：

    (
        [id] => 15
        [time] => 0
        [title] => post 15
        [catIdList] => 
        [catNameList] => 
        [catSlugList] => 
        [catValueList] => 
        [tagIdList] => 4
        [tagNameList] => tagname
        [tagSlugList] => tagname
        [tagValueList] => 
    )

    (
        [id] => 16
        [time] => 0
        [title] => post 16
        [catIdList] => 5~~
        [catNameList] => Movies~~
        [catSlugList] => movies~~
        [catValueList] => ~~
        [tagIdList] => 4~1~
        [tagNameList] => tagname~sand~
        [tagSlugList] => tagname~sand~
        [tagValueList] => ~~
    )

这当然是我想要的 - 所有相关信息！

这些表格是：

1. termRelations 包含 post ID 和 term ID，termTypeId 区分 cat 和 tags 表。
2. cats 包含术语 ID 和类别信息（名称、slug、parentId 等）
3. tags 包含术语 ID 和标签信息（名称、slug 等）
4. posts 包含帖子信息（标题、时间、正文等）

termRelations 的目的是将标签和类别绑定到帖子。此查询的目的是返回过滤结果（我希望用户能够查看具有特定标签和特定类别的帖子。）同时仍保留完整信息。

是否可以通过将cats 和tags 表合并到terms 中来解决这个问题？

我希望我知道怎么做，但是在这一点上，我几乎碰到了心理障碍。所以，一点帮助:)？谢谢！！

Answer 1

使用子查询将WHERE 更改为EXIST：

SELECT posts.id,time,title,
    GROUP_CONCAT(IFNULL(cats.id, '') ORDER BY cats.id DESC SEPARATOR '~')
      AS catIdList,
    GROUP_CONCAT(IFNULL(cats.name, '') ORDER BY cats.id DESC SEPARATOR '~') 
      AS catNameList,
    GROUP_CONCAT(IFNULL(cats.slug, '') ORDER BY cats.id DESC SEPARATOR '~') 
      AS catSlugList,
    GROUP_CONCAT(IFNULL(cats.value, '') ORDER BY cats.id DESC SEPARATOR '~') 
      AS catValueList,
    GROUP_CONCAT(IFNULL(tags.id, '') ORDER BY tags.id DESC SEPARATOR '~') 
      AS tagIdList,
    GROUP_CONCAT(IFNULL(tags.name, '') ORDER BY tags.id DESC SEPARATOR '~') 
      AS tagNameList,
    GROUP_CONCAT(IFNULL(tags.slug, '') ORDER BY tags.id DESC SEPARATOR '~') 
      AS tagSlugList,
    GROUP_CONCAT(IFNULL(tags.value, '') ORDER BY tags.id DESC SEPARATOR '~') 
      AS tagValueList
FROM posts
LEFT JOIN termRelations ON ( posts.id = termRelations.postId )
LEFT JOIN cats 
  ON ( termRelations.termId = cats.id AND termRelations.termTypeId = 1 )
LEFT JOIN tags 
  ON ( termRelations.termId = tags.id AND termRelations.termTypeId = 0 )
WHERE EXISTS
      ( SELECT *
        FROM termRelations 
        WHERE termRelations.termId = '4'
          AND termRelations.termTypeId = 0
          AND posts.id = termRelations.postId
      )
GROUP BY posts.id 
ORDER BY time DESC