Php 需要帮助时从表中获取朋友详细信息

这是我的表格结构（趣味用户）

我的友谊桌是

id  user_id friend_id   status  date_request    from_ip

下面是im用来获取登录用户好友详细信息的查询

SELECT `fun_friends`.`id` as fid, `fun_users`.`id` as uid, `fun_users`.`first_name`, `fun_users`.`profile_img`, `fun_users`.`profession`, `fun_users`.`place` FROM (`fun_friends`) JOIN `fun_users` ON `fun_users`.`id`=`fun_friends`.`friend_id` WHERE (`fun_friends`.user_id= '".$_SESSION['user_row_id']."' AND  `fun_friends`.`status` =1) OR (`fun_friends`.friend_id= '".$_SESSION['user_row_id']."' AND  `fun_friends`.`status` =1)

结果是

fid uid first_name       profile_img                       profession                             place
11  47  Agnii   thumbs/2013-03-311364721555.jpg       Software engineer                                 somewhere

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查询返回loggedin用户的详细信息，而不是他的朋友详细信息。有人能帮我吗？

下面的查询使用一个子查询，该子查询获取特定用户的所有好友。表

fun\u users

对子查询联接了两次，因为子查询上有两列依赖于它

SELECT  a.id AS FID,
        IF(a.user_ID = 'user_row_id_HERE', c.id, b.id) AS UID,
        IF(a.user_ID = 'user_row_id_HERE', c.first_name, b.first_name) AS first_name,
        IF(a.user_ID = 'user_row_id_HERE', c.last_name, b.last_name) AS last_name,
        IF(a.user_ID = 'user_row_id_HERE', c.profile_img, b.profile_img) AS profile_img,
        IF(a.user_ID = 'user_row_id_HERE', c.profession, b.profession) AS profession,
        IF(a.user_ID = 'user_row_id_HERE', c.place, b.place) AS place
FROM
        (
            SELECT  id, user_ID, friend_ID
            FROM    friendship
            WHERE   'user_row_id_HERE' IN (user_ID, friend_ID) AND
                    status = 1
        ) a
        INNER JOIN fun_users b
            ON a.user_ID = b.id
        INNER JOIN fun_users c
            ON a.friend_ID = c.ID

所以问题来了，这条线上发生了什么

IF(a.user_ID = 'user_row_id_HERE', c.id, b.id) AS UID

基本上，它从子查询中测试

用户ID

的值是否等于当前用户。如果恰好相等，则返回表

fun\u users c

中的列，反之亦然

要进一步了解加入的更多信息，请访问以下链接：

IF(a.user_ID = 'user_row_id_HERE', c.id, b.id) AS UID