排序合并联接

Question

我有表user_id(int(10)) and follow_sub(varchar(30))

“user_likes”（用户的所有喜欢）表“user_following”（所有用户按照潜艇）与user_id(varchar(15)) and post_sub(varchar(15))

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

我有2种方式来订购的帖子：

1）通过最新的职位潜艇，用户如下：

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

2）通过算法计算每篇文章的喜欢，但所有的SUBS（即使用户不跟随他们）：

 $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joins-两者结合起来用LIKES ALGORITHM将帖子呈现给他所关注的潜在用户。

这是我试图做到这一点，但它告诉我错误 - 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"; 

我怎样才能使它发挥作用？

Answer 1

它给人一种布尔像MySQL的任何味道或mysqli的您正在使用返回false，而不是因为你有一个声明的对象无效SQL

FROM userpost.*

应该

FROM userpost