PHP Indonesia - Facebook: maaf gan para master2 saya ingin tanya, saya masih...

PHP Indonesia - Facebook PHP Indonesia is a community for everyone that loves PHP. Our focus is in the PHP world but our topics encompass the entire LAMP stack. Topics include PHP coding, to memcached handling, db optimizations, server stack, web server tuning, code deploying, hosting options and much much more. Youtube Channel : http://www.youtube.com/user/OurPHPIndonesia NOTE: Rules for Wall posting: https://www.facebook.com/groups/35688476100/doc/10151597056316101/ Daftar Keanggotaan : http://www.facebook.com/groups/35688476100/doc/10150671770741101/

maaf gan para master2 saya ingin tanya, saya masih... Jul 27th 2013, 18:27, by Fatima Wa Khilafah Fatima Wa Khilafah 6:27pm Jul 27 maaf gan para master2 saya ingin tanya, saya masih newbie.. program saya tdk bisa login ada warning seperti ini... solusinya gimana ya? Ðiky Ķaurika Herlambanģ 6:28pm Jul 27 cb tampilin scriptnya Fatima Wa Khilafah 6:30pm Jul 27 <?php include "koneksi.php"; function anti_injection($data){ $filter = mysql_real_escape_string(stripslashes(strip_tags(htmlspecialchars($data,ENT_QUOTES)))); return $filter; } $username = anti_injection($_POST['username']); $pass = anti_injection(($_POST['password'])); // pastikan username dan password adalah berupa huruf atau angka. if (!ctype_alnum($username) OR !ctype_alnum($pass)){ echo "Sekarang loginnya tidak bisa di injeksi lho."; } else{ $login=mysql_query("SELECT * FROM admin WHERE username='$username' AND password='$pass' AND blokir='N'"); $ketemu=mysql_num_rows($login); $r=mysql_fetch_array($login); // Apabila username dan password ditemukan if ($ketemu > 0){ session_start(); include "timeout.php"; $_SESSION[namauser] = $r[username]; $_SESSION[namalengkap] = $r[nama_lengkap]; $_SESSION[passuser] = $r[password]; $_SESSION[leveluser] = $r[level]; // session timeout $_SESSION[login] = 1; timer(); $sid_lama = session_id(); session_regenerate_id(); $sid_baru = session_id(); mysql_query("UPDATE admin SET id_session='$sid_baru' WHERE username='$username'"); header('location:media.php?module=home'); } else{ echo "<link href=../config/adminstyle.css rel=stylesheet type=text/css>"; echo "<center>LOGIN GAGAL! <br> </center>"; } } ?> Fatima Wa Khilafah 6:34pm Jul 27 warningnya di script bagian ini $ketemu=mysql_num_rows($login); $r=mysql_fetch_array($login); Wahyu Linkmu 6:34pm Jul 27 kayaknya si itu eror di perintah mysql_query("....'); ny amas,,, coba perintah yang di dalam kurung di cek lagi.. Fatima Wa Khilafah 6:36pm Jul 27 script yg ini $login=mysql_query("SELECT * FROM admin WHERE username='$username' AND password='$pass' AND blokir='N'"); sdh sy cek Wahyu Linkmu 6:37pm Jul 27 dak mbok kamu coba cek scrip SELECT * FROM admin WHERE username='$username' AND password='$pass' AND blokir='N' kamu jalanin di mysql..???username n passwordnya kamu isi sesuai yang ada di database kamu..? Fatima Wa Khilafah 6:38pm Jul 27 sudah Wahyu Linkmu 6:38pm Jul 27 hasilnya jadi..? Fatima Wa Khilafah 6:39pm Jul 27 g bisa hasilnya error kyk di atas itu pak.. Wahyu Linkmu 6:40pm Jul 27 coba kamu export databasenya dan kirim ke aku,, nanti kita koreksi bareng-bareng Ðiky Ķaurika Herlambanģ 6:46pm Jul 27 $login=mysql_query("SELECT * FROM admin WHERE username='$username' AND password='$pass' AND blokir='N' )or die(mysql_error()); <-- cb sqlnya tambahin mysql _error biar ketauan errornya Fatima Wa Khilafah 6:47pm Jul 27 sudah sy kirim di message Wahyu Linkmu 6:48pm Jul 27 ok Fatima Wa Khilafah 6:51pm Jul 27 sudah sy tambahkan mysql _error sama sj hasilnya kyk gt.. Fikri Handa Herriansyah 6:58pm Jul 27 Pastebin dong code nya :) mungkin bisa dibantu Gembel Berkelas 7:07pm Jul 27 ane mah biasanya kalo query nya rada2 ga beres, langsung test echo "SELECT * FROM admin WHERE username='$username' AND password='$pass' AND blokir='N'"; exit; liat output nya, kalo kira2 sesuai dgn harapan, test deh d phpmyadmin

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