PHP mysql如何关联三个表显示来自不同表的输出

Question

我有这个图

我想做的是输出：

你如何设法查询这个？

我有这段代码

SELECT users.firstname, users.lastname, 
       users.screenname, posts.post_id, posts.user_id,
       posts.post, posts.upload_name, 
       posts.post_type, posts.date_posted
FROM website.users users
INNER JOIN website.posts posts ON (users.user_id = posts.user_id)
ORDER BY posts.pid DESC
//PROBLEM with this one is that it only views the post from all users.

//SO I added
SELECT COUNT(user_id) AS friends, SUM(user_id = ?) AS you, user_id
FROM feeds WHERE post_id = ?
//This one will give you two fields containing how many different users **feeds** the
post

请帮帮我们。实际上这个我只是关注 Facebook的“LIKE”状态 唯一的问题是我不是这种东西的业余爱好者，所以我很高兴听到你的所有答案。我真的需要你的帮助

Answer 1

如果我理解正确，你想要一个带有feeds表的外联接（为了保留所有posts，即使没有关联的feeds），那么{ {1}}为每个帖子合并所有此类Feed，并GROUP BY post.pid所需信息。

我使用MySQL的GROUP_CONCAT()函数来获取所有用户（最多group_concat_max_len）的逗号分隔列表 给定帖子的“提要”（如果需要，您可以使用SELECT修饰符更改分隔符。）

SEPARATOR

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<强>更新

要获取“喜欢”帖子的用户的全名，不包括自己，需要第二次加入SELECT users.firstname, users.lastname, users.screenname, posts.post_id, posts.user_id, posts.post, posts.upload_name, posts.post_type, posts.date_posted, COUNT(feeds.user_id) AS friends, -- number of "likes" SUM(feeds.user_id = ?) AS you, -- did I like this? GROUP_CONCAT(feeds.user_id) -- who likes it? 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) GROUP BY posts.pid ORDER BY posts.pid DESC 表：

users

Answer 2

如果您想为所有用户执行此操作并同时获取Feed，则必须加入此feed表：

SELECT u.firstname, u.lastname, 
   u.screenname, p.post_id, p.user_id,
   p.post, p.upload_name, 
   p.post_type, p.date_posted,
   COUNT(f.user_id) AS friends, SUM(f.user_id = ?) AS you
FROM website.users u
INNER JOIN website.posts p ON (u.user_id = p.user_id)
LEFT  JOIN website.feeds f ON (p.post_id = f.post_id)
GROUP BY p.pid
ORDER BY p.pid DESC

这个应该可以解决这个问题......