Question

我有表＆＃34; user_following＆＃34;（用户遵循的所有潜行客户）user_id(int(10)) and follow_sub(varchar(30))

＆＃34; user_likes＆＃34;（用户的所有喜欢）表user_id(varchar(15)) and post_sub(varchar(15))

和＆＃34; user_post＆＃34;与user_id(int(10)) post_txt(text) sub_id(varchar(30) parallel to follow_sub) post_unix_time(float)

我有两种订购帖子的方法：

1）用户遵循的最新帖子中的潜艇：

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

2）通过计算每篇文章喜欢的算法，但是所有的SUBS（即使用户没有关注它们）：

        $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;";

我想结合两个连接 - 用LIKES ALGORITHM将帖子呈现给他所关注的子用户。

这是我尝试这样做的方式，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";

我怎样才能让它发挥作用？

Answer 1

它给出了一个布尔值，因为你使用的mysql或mysqli的任何风格都是返回false而不是语句对象，因为你有无效的sql

FROM userpost.*

应该是

FROM userpost

按联合顺序排序

1 个答案: