按组合连接排序

Question

我有user_id(int(10)) and follow_sub(varchar(30)) 的表“user_following”（用户关注的所有订阅者）

“user_likes”（用户的所有喜欢）表与user_id(varchar(15)) and post_sub(varchar(15))

和user_id(int(10)) post_txt(text) sub_id(varchar(30) parallel to follow_sub) post_unix_time(float)的“user_post”

我有两种方式来排序帖子：

1) 通过用户关注的最新帖子，订阅者：

        $result11="SELECT * FROM user_post
        INNER JOIN users_following ON user_post.sub_id=users_following.follow_sub
        WHERE  users_following.user_id='$userid' 
        ORDER BY user_post.post_id DESC";

2) 通过计算每篇文章的点赞数的算法，但包含所有订阅者（即使用户不关注它们）：

        $result11="SELECT user_post.*, (LOG10(ABS(COUNT(user_likes.post_id)) + 1) * SIGN(COUNT(user_likes.post_id)))/user_post.post_unix_time AS like_count
                    FROM user_post LEFT JOIN user_likes 
                    ON user_post.post_id = user_likes.post_id
                    GROUP BY user_post.post_id
                    ORDER BY like_count desc;"; 

我想结合这两个连接 - 使用 LIKES ALGORITHM 将帖子呈现给他关注的订阅者的用户。

这是我尝试的方式，但它显示错误-mysql_num_rows() expects parameter 1 to be resource, boolean given in：

            $result11="SELECT user_post.*, (LOG10(ABS(COUNT(user_likes.post_id)) + 1) * SIGN(COUNT(user_likes.post_id)))/user_post.post_unix_time AS like_count
            FROM user_post.*
            LEFT JOIN user_likes ON user_post.post_id = user_likes.post_id
            INNER JOIN users_following ON user_post.sub_id=users_following.follow_sub
            WHERE  users_following.user_id='$userid' 
            GROUP BY user_post.post_id
            ORDER BY like_count DESC";

我怎样才能让它工作？

Answer 1

它给出一个布尔值，因为你使用的任何风格的 mysql 或 mysqli 都返回 false 而不是语句对象，因为你有无效的 sql

FROM userpost.*

应该是

FROM userpost