mysql相关字段

Question

这4个字段彼此相关

我希望它输出为：

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，我遇到了一个问题，我应该在哪个部分加入朋友表，   我想显示来自friend_id或user_id 的所有帖子以及当前登录用户的帖子。如果没有朋友在朋友表上匹配，则只输出所有来自用户的帖子。请伙计们，我需要你的帮助。

friends.friend_id =当前用户的朋友

friends.user_id =用户的当前朋友

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

如果我的朋友表不可理解，请帮助我改变它以使其更好。

Answer 1

如果我很好理解你想基于friends = user_id加入好友表，如果不匹配好友表的user_id上的JOIN，那么你可以尝试这样的事情：

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

我基本上在好友表中添加了一个JOIN，并在你想要加入的两个字段上添加了一个OR ...

Answer 2

您希望查看来自用户或其朋友的帖子。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;

测试输出here。