Question

假设有一首歌曲和一首歌，我们想知道喜欢同一首歌的所有朋友（直接朋友和朋友）：

这就是我们这样做的方式：

public function get_song_mates($uid, $song_id, $current_only = true, $limit = PHP_INT_MAX)
    {

        $sql    = "SELECT * 
                    FROM (
                    (
                        SELECT users . * ,  '1st' AS  `level`, 1 AS `level_i` 
                        FROM songs
                        JOIN users ON users.id = songs.user_id
                        JOIN friends AS my_friend ON my_friend.fid = users.id 
                        AND my_friend.uid = '".$uid."'                  
                        WHERE song_id =  '".$song_id."')
                    UNION (

                    SELECT second_friend . * ,  '2nd' AS  `level`, 2 AS `level_i`
                    FROM songs
                    JOIN users AS second_friend ON second_friend.id = songs.user_id
                    JOIN friends ON friends.uid = second_friend.id
                    JOIN users AS first_friend ON first_friend.id = friends.fid 
                    JOIN friends AS my_friend ON my_friend.fid = first_friend.id 
                    AND my_friend.uid = '".$uid."'                  
                    WHERE song_id =  '".$song_id."'


                    AND second_friend.id NOT IN
                        (
                            SELECT users.id 
                              FROM songs
                              JOIN users ON users.id = songs.user_id
                              JOIN friends AS my_friend ON my_friend.fid = users.id 
                               AND my_friend.uid = '".$uid."'
                             WHERE song_id = '".$song_id."'
                        )
                    ) ORDER BY `level_i` ASC, firstname, lastname
                    ) AS friend                 
                    GROUP BY id ORDER BY `level_i`";

        $users  = $this->getFromSql($sql);  
        return $users;
    }

但请注意我们必须添加：

AND second_friend.id NOT IN...

防止显示与直接朋友和朋友的朋友相同的用户..我很确定我需要使用GROUP BY，但我无法使用正确的语法。这里有什么线索吗？

- 编辑 -

friends(uid, fid)
songs(id, user_id, song_id)
users(id, frist_name, last_name)

Answer 1

修正如下：

公共函数get_song_mates（$ uid，$ song_id，$ current_only = true，$ limit = PHP_INT_MAX） {

    $sql    = "SELECT DISTINCT * 
                FROM (
                (
                    SELECT users . * ,  '1st' AS  `level`, 1 AS `level_i` 
                    FROM songs
                    JOIN users ON users.id = songs.user_id
                    JOIN friends AS my_friend ON my_friend.fid = users.id 
                    AND my_friend.uid = '".$uid."'                  
                    WHERE song_id =  '".$song_id."')
                UNION (

                SELECT DISTINCT second_friend . * ,  '2nd' AS  `level`, 2 AS `level_i`
                FROM songs
                JOIN users AS second_friend ON second_friend.id = songs.user_id
                JOIN friends ON friends.uid = second_friend.id
                JOIN users AS first_friend ON first_friend.id = friends.fid 
                JOIN friends AS my_friend ON my_friend.fid = first_friend.id 
                AND my_friend.uid = '".$uid."'                  
                WHERE song_id =  '".$song_id."'";

    $users  = $this->getFromSql($sql);  
    return $users;
}

GROUP一级和二级朋友

1 个答案: