$sql="SELECT a.id,b.zhname zh1,c.zhname zh2,d.nickname zhiuserid,a.shzhlx,a.jiner,a.rq,a.beizhu FROM `".C('DB_PREFIX')."zb` a
left join `".C('DB_PREFIX')."zh` b on b.zhid=a.zh1
left join `".C('DB_PREFIX')."zh` c on c.zhid=a.zh2
left join `".C('DB_PREFIX')."member` d on d.uid=a.zhiuserid
where a.`zuid` like '".$info['zuid']."'";我有点搞不太明白，希望能得到大家的帮助 最佳答案
