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Sistem Login Gagal - PHP ke Mysql
Outputnya selalu ERROR, kira2 permasalahannya dimana ya? maaf ya masih belajar xD
loginsubmit.php
<?php
session_start();
if ( isset($_POST['login']) ){
$logemail=$_POST['logemail'];
$logpass=$_POST['logpass'];
include('koneksi.php');
$query="SELECT * FROM users_data WHERE user_email='$logemail'";
$hasil=mysqli_query($query,$koneksi) or die ("ERROR");
$data=mysqli_fetch_array($hasil);
$security = "15111999pwdc677";
if ( md5($security . md5($logpass) . md5($security)) == $data['user_password'])
{
$_SESSION['user_email']=$logemail;
echo 'BERHASIL';
}else {
echo 'GAGAL';
}
}
?>
koneksi.php
<?php
$db_host = "localhost";
$db_user = "root";
$db_pass = "1qaz2wsx";
$db_name = "wisataindo";
$koneksi = mysqli_connect($db_host, $db_user, $db_pass, $db_name);
if(mysqli_connect_errno()){
echo 'Gagal melakukan koneksi ke Database : '.mysqli_connect_error();
}
?>
login.php (formnya)
<div id="headerlogin">
<h2 style="font-weight: bold; color: #fff; margin-left:1.4em;">Login WisataIndo</h2>
</div>
<div style="background-image: url('img/bglogin.jpg');">
<div class="container">
<hr id="line"/>
<form action="functions/loginsubmit.php" method="POST">
<div class="form-group">
<label> Email</label>
<input type="email" class="form-control input-lg" name="logemail"/>
</div>
<div class="form-group">
<label> Password</label>
<div class="input-group">
<input type="password" class="form-control input-lg" id="logpass" name="logpass"/>
<a href="#" class="input-group-addon" id="showLogpass"><i id="show" class="fa fa-eye" aria-hidden="true"></i></span></a>
</div>
</div>
<button type="submit" name="login" class="btn btn-primary btn-lg btn-block" id="login"><span class="fa fa-sign-in fa-lg"> Login</button>
<button type="reset" name="reset" class="btn btn-default btn-lg btn-block"><span class="fa fa-refresh fa-lg"></span> Reset</button>
</form>
<br />
<hr id="line"/>
<br />
<div align="center" style="color: #fff;">
Belum punya akun WisataIndo? Silahkan daftar terlebih dahulu.
<br />
<button class="btn btn-success btn-lg btn-block" id="btn-daftar">Daftar</button>
</div>
<br />
<br />
</div>
</div>
0
3 Jawaban:
Coba gini gann..
<?php
session_start();
if ( isset($_POST['login']) ){
$logemail=$_POST['logemail'];
$logpass=$_POST['logpass'];
include('koneksi.php');
$query="SELECT * FROM users_data WHERE user_email='$logemail'";
$hasil=mysqli_query($query,$koneksi) or die ("ERROR");
if(mysqli_num_rows($hasil) > 0){
while ($row = mysqli_fetch_assoc($hasil)) {
$data['user_password'] = $_SESSION['password'];
}
}
$security = "15111999pwdc677";
if ( md5($security . md5($logpass) . md5($security)) == $_SESSION['password']) {
$_SESSION['user_email']=$logemail;
echo 'BERHASIL';
}else {
echo 'GAGAL';
}
}
?>
0
Tapii saya masihh bingung maksud agan di bagian ini..
$security = "15111999pwdc677";
if ( md5($security . md5($logpass) . md5($security)) == $_SESSION['password']) {
$_SESSION['user_email']=$logemail;
echo 'BERHASIL';
}else {
echo 'GAGAL';
}
0
Atauu coba pastikan apakah ini udah benarr..
$db_host = "localhost";
$db_user = "root";
$db_pass = "1qaz2wsx";
$db_name = "wisataindo";
$koneksi = mysqli_connect($db_host, $db_user, $db_pass, $db_name);
0