PHP mysql如何关联三个表，显示来自不同表的输出

Question

我有这张图表

我想做的是有这个输出：

你是怎么做到这个查询的？

我有这个代码

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

请帮助大家。其实这个我只是关注Facebook的“LIKE”状态 唯一的问题是我不是这类东西的业余爱好者，所以我很高兴听到你所有的答案。我真的需要你的帮助

Answer 1

如果我对您的理解正确，您需要与feeds 表进行外连接（以保留所有posts，即使没有关联的feeds），然后GROUP BY post.pid 以合并将每个帖子的所有此类提要汇总在一起，并SELECT 所需的信息。

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

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

---

更新

要获取“赞”过帖子的用户全名，不包括自己，需要第二次加入users表：

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(
         CASE WHEN NOT likes.user_id = ? THEN                -- exclude self
           CONCAT_WS(' ', likes.firstname, likes.lastname)   -- full names
         END
       )
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

Answer 2

如果您想为所有用户执行此操作并同时获取提要，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

这个应该可以解决问题...