基于状态

Question

t_user 
----------------------------- 
uid | username | full_name 
----------------------------- 
1 | dodo  | Dodo Ash 
2 | jane  | Jane Shalimar 
---------------------------- 


t_join 

----------------------------- 
j_id | uid_fk | uid | status 
----------------------------- 
1 | 1  | 2 | joining 
2 | 2  | 1 | joining 
----------------------------- 

t_message 
----------------------------- 
msg_id | message | uid_fk 
----------------------------- 
1  | hi all | 1 
2  | nice trip | 2 
----------------------------- 

PHP代码PHP选择显示消息显示根据加入数据：

$query = mysql_query("SELECT M.msg_id, M.uid_fk, M.message, M.created, U.full_name, U.profile_pic, U.username, U.uid, F.status, F.uid FROM t_haps_wall M, t_users U, t_join_user F WHERE 
     M.uid_fk=U.uid AND F.uid=U.uid AND F.status='joining' order by M.msg_id desc ") or die(mysql_error()); 

我与PHP SELECT代码的麻烦。 重点是朋友的消息将显示状态“加入”。那么，我该如何设定才能得到它？

Answer 1

使用此查询它将工作。

$query = mysql_query("SELECT M.msg_id, M.uid_fk, M.message, M.created, U.full_name, U.profile_pic, U.username, U.uid, F.status, F.uid FROM t_haps_wall M, t_users U, t_join_user F WHERE 
    M.uid_fk=U.uid AND F.uid=U.uid AND M.uid_fk = F.uid AND F.status='joining' order by M.msg_id desc ") or die(mysql_error());