排序合併聯接

Question

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

「user_likes」（用戶的所有喜歡）表「user_following」（所有用戶按照潛艇）與user_id(varchar(15)) and post_sub(varchar(15))

和「user_post」與user_id(int(10)) post_txt(text) sub_id(varchar(30) parallel to follow_sub) post_unix_time(float)

我有2種方式來訂購的帖子：

1）通過最新的職位潛艇，用戶如下：

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

2）通過算法計算每篇文章的喜歡，但所有的SUBS（即使用戶不跟隨他們）：

 $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-兩者結合起來用LIKES ALGORITHM將帖子呈現給他所關注的潛在用戶。

這是我試圖做到這一點，但它告訴我錯誤 - 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"; 

我怎樣才能使它發揮作用？

Answer 1

它給人一種布爾像MySQL的任何味道或mysqli的您正在使用返回false，而不是因爲你有一個聲明的對象無效SQL

FROM userpost.*

應該

FROM userpost