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SELECT count(*) as ini FROM `info_log`

$date = date("Y-m-d H:i:s");
$ip = isset($_SERVER['HTTP_X_FORWARDED_FOR']) ? $_SERVER['HTTP_X_FORWARDED_FOR']: $_SERVER['REMOTE_ADDR'];
$alamat_ip = $_SERVER['REMOTE_ADDR'];
mysql_query("INSERT INTO `datelrmg`.`data` (`id`, `nim`, `nama`, `alamat`, `hp`, `pesan`, `ip`, `alamat_ip`, `date`, `email`)
VALUES (NULL, NULL, 'Tamu melihat dosen MI', '', NULL, NULL, '$ip', '$alamat_ip', '$date', '');");

$query = mysql_query("SELECT a.id, a.city, a.aktif, a.firstname, a.lastname , b.contextid, b.roleid, c.instanceid, d.fullname   FROM  `info_user` a
left join info_role_assignments b on b.userid=a.id
left join info_context c on c.id=b.contextid
left join info_course d on d.id=c.instanceid
WHERE a.firstname LIKE 'dosen' and a.aktif='Y' and b.roleid='3' ORDER BY `c`.`instanceid` ASC");
$query = mysql_query("SELECT a.id, a.firstname, a.city, a.lastname, b.roleid, b.contextid, c.instanceid, e.name, e.parent  FROM `info_user` a 
left join info_role_assignments b on b.userid=a.id  
left join info_context c on c.id=b.contextid 
left join info_context d on d.id=c.instanceid
left join info_course_categories e on e.id=c.instanceid
WHERE a.firstname LIKE 'teknik' and b.roleid='3' and e.parent='3' ORDER BY `a`.`lastname` ASC");

while($row=mysql_fetch_array($query)){
?>

SELECT `username`, `password`, `firstname`, `lastname` FROM info_user where username like '13%'

ini untuk melihat nim mhasiswa angkatan 2013
klw ada kesalhan di tahun 2013 contoh terbalik tabel firstname cara ganti seperti ini .. beres dahhh

update `info-user` set firstname=username where username like '13%'

Belajar dulu Rating: 4.5 Diposkan Oleh: Catatanku