3

这4个字段相互关联

朋友 - 帖子 - 用户 - 提要

我希望它输出为：在此处输入图像描述

在我的查询中：

SELECT users.firstname, users.lastname, users.screenname, posts.post_id, posts.user_id,
posts.post, posts.upload_name, posts.post_type, 
DATE_FORMAT(posts.date_posted, '%M %d, %Y %r') AS date, 
COUNT(NULLIF(feeds.user_id, ?)) AS everybody, SUM(feeds.user_id = ?) AS you,
GROUP_CONCAT(CASE WHEN NOT likes.user_id = ? THEN 
             CONCAT_WS(' ', likes.firstname, likes.lastname)
                    END
            ) as names
FROM website.users users
INNER JOIN website.posts posts ON (users.user_id = posts.user_id)
LEFT  JOIN website.feeds feeds ON (posts.post_id = feeds.post_id)
LEFT  JOIN website.users likes ON (feeds.user_id = likes.user_id)
GROUP BY posts.pid
ORDER BY posts.pid DESC

现在，我有一个问题应该加入friends表的哪一部分，我想显示来自friend_id或user_id的所有帖子以及来自当前登录的用户的帖子。如果朋友表上没有匹配的朋友，那么只需输出用户的所有帖子。请各位，我需要你的帮助。

Friends.friend_id = 当前用户的好友

Friends.user_id = 用户的当前好友

因此，friends.friend_id = posts.user_id 或 friends.user_id = posts.user_id

如果我的朋友表无法理解，请帮我更改它以使其更好。

4

2 回答 2

2

如果我很了解您想根据friends = user_id 加入friends 表，并且如果与friends 表的user_id 上的JOIN 不匹配，则可以尝试以下方法：

SELECT users.firstname, users.lastname, users.screenname, posts.post_id, posts.user_id,
posts.post, posts.upload_name, posts.post_type, 
DATE_FORMAT(posts.date_posted, '%M %d, %Y %r') AS date, 
COUNT(NULLIF(feeds.user_id, ?)) AS friends, SUM(feeds.user_id = ?) AS you,
GROUP_CONCAT(CASE WHEN NOT likes.user_id = ? THEN 
             CONCAT_WS(' ', likes.firstname, likes.lastname)
                    END
            ) as names
FROM website.users users
INNER JOIN website.posts posts ON (users.user_id = posts.user_id)
LEFT  JOIN website.feeds feeds ON (posts.post_id = feeds.post_id)
LEFT  JOIN website.users likes ON (feeds.user_id = likes.user_id)
LEFT  JOIN website.friends friends ON ((posts.user_id = friends.user_id) OR (posts.user_id = friends.friends_id) )
GROUP BY posts.pid
ORDER BY posts.pid DESC

我基本上已经在您似乎想要加入的两个字段上添加了一个带有 OR 的朋友表的 JOIN ...

于 2012-05-13T08:44:05.810 回答
2

您希望看到来自用户或他的朋友的帖子。因此，不要加入用户，而是加入子查询，如下所示：

SELECT users.firstname, users.lastname, users.screenname,
       posts.post_id, posts.user_id, posts.post, posts.upload_name,
       posts.post_type, DATE_FORMAT(posts.date_posted, '%M %d, %Y %r') AS date, 
       COUNT(NULLIF(feeds.user_id, ?)) AS everybody,
       SUM(feeds.user_id = ?) AS you,
       GROUP_CONCAT(CASE WHEN NOT likes.user_id = ? THEN 
             CONCAT_WS(' ', likes.firstname, likes.lastname) END) as names
  FROM (SELECT user_id FROM website.users WHERE user_id = ?
        UNION ALL
        SELECT user_id FROM website.friends WHERE friend_id = ?
        UNION ALL
        SELECT friend_id FROM website.friends WHERE user_id = ?) AS who
  JOIN website.users users ON users.user_id = who.user_id
  JOIN website.posts posts ON users.user_id = posts.user_id
  LEFT  JOIN website.feeds feeds ON posts.post_id = feeds.post_id
  LEFT  JOIN website.users likes ON feeds.user_id = likes.user_i)
 GROUP BY posts.pid
 ORDER BY posts.pid DESC;

在这里测试输出。

于 2012-05-13T09:36:26.607 回答