$query = "select u.*, d.depcode.*, s.seccode
from user u
inner join department d , section s on d.dep_id = u.userid and s.sec_id = u.userid
where u.username='$username' and u.password='$password'";
$query = "select u.*, d.depcode.*, s.seccode
from user u
inner join department d on d.dep_id = u.userid
inner join section s on s.sec_id = u.userid
where u.username='$username' and u.password='$password'";
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\AppServ\www\iso\login.php on line 102
Unknown table 'department'
Load balance : Server 03
