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>

avatar newsyahrul
@newsyahrul

16 Kontribusi 0 Poin

Diperbarui 7 tahun yang lalu

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';
  }
}
?>

avatar dodipsitorus
@dodipsitorus

412 Kontribusi 145 Poin

Dipost 7 tahun yang lalu

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';
  }

avatar dodipsitorus
@dodipsitorus

412 Kontribusi 145 Poin

Dipost 7 tahun yang lalu

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);

avatar dodipsitorus
@dodipsitorus

412 Kontribusi 145 Poin

Dipost 7 tahun yang lalu

Login untuk ikut Jawaban