php mysql连接使用

Question

$user_res_get = do_mysql_query("SELECT /* forums.php */ userid, subject, " .
    " forumid FROM topics WHERE id=$topicid") or sqlerr(__FILE__, __LINE__);

和:

$user_res_get = do_mysql_query("SELECT /* forums.php */ id, name, FROM " .
    "forums WHERE id=$forumid") or sqlerr(__FILE__, __LINE__);

forumd 和 id 相同

如何“加入”他们？

我试过创建这样的东西:

$user_res_get = do_mysql_query("SELECT u.userid, u.subject, u.forumid, a.id, " .
    "a.name FROM topics AS u JOIN forums as a ON u.forumid = a.id WHERE " .
    "u.forumid=$topicid") or sqlerr(__FILE__, __LINE__);

但没有运气:/ 有什么建议 ？ 编辑: Ant 尝试打印:

$user_row_get = mysql_fetch_assoc($user_res_get);
$user_row_get['subject']
$user_row_get['name']

和其他人...

Answer 1

代替:

WHERE u.forumid=$topicid

尝试:

WHERE u.id=$topicid