加入MySQL之前的聚合表（GROUP BY）答案

【问题标题】：Aggregate table before join MySQL (GROUP BY)加入MySQL之前的聚合表（GROUP BY）
【发布时间】：2012-01-07 04:01:21
【问题描述】：

我需要先聚合一个表，然后再将其与其他表连接。

wp_postmeta GROUP BY wp_postmeta.post_id

有可能吗？我应该把它放在哪里？这是我的代码：

SELECT wp_posts.post_content, wp_posts.ID, wp_terms.slug
FROM wp_posts
JOIN wp_postmeta ON wp_posts.ID = wp_postmeta.post_id
JOIN wp_term_relationships ON wp_posts.ID = wp_term_relationships.object_id
JOIN wp_term_taxonomy ON wp_term_relationships.term_taxonomy_id = wp_term_taxonomy.term_taxonomy_id
JOIN wp_terms ON wp_term_taxonomy.term_id = wp_terms.term_id


WHERE wp_posts.post_type = 'my-type'
   AND wp_posts.post_status = 'publish'
   AND wp_terms.slug IN
      ('field1', 'field2', 'field3')
   AND
   (
        wp_postmeta.meta_key = 'exclude' AND wp_postmeta.meta_value <> '111'
        OR wp_postmeta.meta_key = 'include' AND wp_postmeta.meta_value = '22'
        OR wp_postmeta.meta_key <> 'include' AND wp_postmeta.meta_key <> 'exclude'
   )

在测试 ruak 示例时发现了一种新方法。我不知道它为什么有效，但它确实有效：

SELECT wp_posts.post_content, wp_posts.ID, wp_terms.slug
FROM wp_posts
JOIN wp_term_relationships ON wp_posts.ID = wp_term_relationships.object_id
JOIN wp_term_taxonomy ON wp_term_relationships.term_taxonomy_id = wp_term_taxonomy.term_taxonomy_id
JOIN wp_terms ON wp_term_taxonomy.term_id = wp_terms.term_id
WHERE wp_posts.post_type = 'my-type'
   AND wp_posts.post_status = 'publish'
   AND wp_terms.slug IN
      ('field1', 'field2', 'field3')
   AND NOT wp_posts.id IN                # ADDED NOT
   (
      SELECT wp_postmeta.post_id
        FROM wp_postmeta
        WHERE wp_postmeta.meta_key = 'exclude' AND wp_postmeta.meta_value = '111'  #SINCE NOT ABOVE I CHANGED <> to =
           OR wp_postmeta.meta_key = 'include' AND wp_postmeta.meta_value <> '22'  #SINCE NOT ABOVE I CHANGED = to <>
           #I DELETED THIS LINE
   )

现在，如果未设置包含或排除，则查询会返回数据。如果是，它会检查 ID。

我的零钱有什么现金吗？

更多改进：

SELECT wp_posts.post_content, wp_posts.ID, wp_terms.slug
FROM wp_posts
JOIN wp_term_relationships ON wp_posts.ID = wp_term_relationships.object_id
JOIN wp_term_taxonomy ON wp_term_relationships.term_taxonomy_id = wp_term_taxonomy.term_taxonomy_id
JOIN wp_terms ON wp_term_taxonomy.term_id = wp_terms.term_id

WHERE wp_posts.post_type = 'my-type'
AND wp_posts.post_status = 'publish'
AND wp_terms.slug IN
('field1', 'field2', 'field3')

AND wp_posts.id NOT IN
(
SELECT wp_postmeta.post_id
FROM wp_postmeta

WHERE wp_postmeta.meta_key = 'exclude'
AND wp_postmeta.meta_value IN
('$id', '$id,%', '%,$id,%', '%,$id')

OR wp_postmeta.meta_key = 'include' 
AND wp_postmeta.meta_value NOT IN
('$id', '$id,%', '%,$id,%', '%,$id')

wp_postmeta 表的外观示例：

场景 1 - 只有 id 为 18 的页面才能获取数据：

meta_id     post_id     meta_key    meta_value
1           30          include     18
2           30          _edit_lock  1322225789:1
3           30          _edit_last  1

Senario 2 - id 为 18 的页面不应获取数据：

meta_id     post_id     meta_key    meta_value
1           30          exclude     18
2           30          _edit_lock  1322225789:1
3           30          _edit_last  1

场景 1 - 所有页面都应获取数据：

meta_id     post_id     meta_key    meta_value
2           30          _edit_lock  1322225789:1
3           30          _edit_last  1

【问题讨论】：

标签： mysql join group-by where

【解决方案1】：

我认为你根本不应该真正加入wp_postmeta，因为你实际上并不需要它的任何东西；相反，您应该使用IN 或EXISTS 子句执行“半连接”。例如：

SELECT wp_posts.post_content, wp_posts.ID, wp_terms.slug
FROM wp_posts
JOIN wp_term_relationships ON wp_posts.ID = wp_term_relationships.object_id
JOIN wp_term_taxonomy ON wp_term_relationships.term_taxonomy_id = wp_term_taxonomy.term_taxonomy_id
JOIN wp_terms ON wp_term_taxonomy.term_id = wp_terms.term_id
WHERE wp_posts.post_type = 'my-type'
   AND wp_posts.post_status = 'publish'
   AND wp_terms.slug IN
      ('field1', 'field2', 'field3')
   AND wp_posts.id IN
   (
      SELECT DISTINCT wp_postmeta.post_id
        FROM wp_postmeta
        WHERE wp_postmeta.meta_key = 'exclude' AND wp_postmeta.meta_value <> '111'
           OR wp_postmeta.meta_key = 'include' AND wp_postmeta.meta_value = '22'
           OR wp_postmeta.meta_key <> 'include' AND wp_postmeta.meta_key <> 'exclude'
   )

【讨论】：

ruakh，我已经更新了我的问题。您对我的新示例有任何 cmets 吗？
@Hakan：如果wp_postmeta.post_id 是唯一的——也就是说，如果两个不同的wp_postmeta 记录总是具有不同的post_id 值——那么你的新示例对我来说看起来很完美。我唯一的评论是写“___ NOT IN (...)”比写“NOT ___ IN (...)”更常见；但那里没有真正的区别，所以你的方式是 100% 好的。
我现在开始掌握这个窍门（我想）。我再次更新了这个问题。艰难的 post_id 将列在表 wp_postmeta 的多行中。我将在我的问题中添加示例。

【解决方案2】：

您可以使用子查询： http://dev.mysql.com/doc/refman/5.0/en/from-clause-subqueries.html

【讨论】：