Permisi gan, butuh bantuan.
Saya bikin form login dan ada error.
Berikut kodingannya:
Quote:
Code:
<?php
$username=$_POST["username"];
$password=md5($_POST["password"]);
require("koneksi.php");
$connection = mysql_connect($server,$user,$pass);
$db = mysql_select_db($db);
$query="select *,count (*) as cek from guru
INNER JOIN ms_jabatan
ON guru.level=ms_jabatan.level
where guru.kd_guru='".$username."' and guru.password='".$password."'";
$ambil_data=mysql_query($query);
while($guru = mysql_fetch_array($ambil_data)) //ERROR DI SINI
{
$cek = $guru['cek'];
$username = $guru['kd_guru'];
$nama = $guru['nama_guru'];
$level = $guru['level'];
$jabatan = $guru['jabatan'];
}
if ($cek == 1 && $level == "1")
{
session_start();
$_SESSION['Auser'] = $username;
$_SESSION['Anama'] = $nama;
$_SESSION['Alevel'] = $level;
$_SESSION['Ajabatan'] = $jabatan;
header("Location:tampkurikulum.php");
}
else if ($cek == 1 && $level == "2")
{
session_start();
$_SESSION['Kuser'] = $username;
$_SESSION['Knama'] = $nama;
$_SESSION['Klevel'] = $level;
$_SESSION['Kjabatan'] = $jabatan;
header("Location:view_guru.php");
}
/*else if ($cek == 1 && $level == "3")
{
session_start();
$_SESSION['Wuser'] = $username;
$_SESSION['Wnama'] = $nama;
$_SESSION['Wlevel'] = $level;
$_SESSION['Wjabatan'] = $jabatan;
header("Location:tampwali.php");
}
else if ($cek == 1 && $level == "4")
{
session_start();
$_SESSION['Guser'] = $username;
$_SESSION['Gnama'] = $nama;
$_SESSION['Glevel'] = $level;
$_SESSION['Gjabatan'] = $jabatan;
header("Location:tampguru.php");
}*/
else
{
echo "Kode atau Password Salah!";
}
?>
Saya juga kurang mengerti di query
Quote:
select *,count (*) as cek from guru
Error yang saya alami adalah muncul peringatan
Quote:
Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in
dan variabel cek yang tidak bisa terdefinisi,
Mungkin agan agan sekalian bisa memberikan solusi dan pencerahan