Postingan lainnya
Cara memisahkan akses login!
Saya punya sebuah kasus dimana ketika si admin login dan user tidak login ke aplikasi, namun status admin dan user sama2 online statusnya di tabel databasenya. Nah, bagaimana caranya agar kondisi si user tidak sama statusnya.. supaya dibaca terpisah user yang lain dapat status online dan offline jika hanya jika mereka login atau logout saja... jadi tidak semua rata sama.. 1 admin login, semua user statusnya ikut-ikutan jadi online. Terimakasih
Berikut kode dan screenshootnya :
<?php
session_start();
require_once 'dbconnect.php';
if (isset($_SESSION['userSession'])!="") {
header("Location: home.php");
exit;
}
if (isset($_POST['btn-login'])) {
$email = strip_tags($_POST['email']);
$password = strip_tags($_POST['password']);
$email = $DBcon->real_escape_string($email);
$password = $DBcon->real_escape_string($password);
$query = $DBcon->query("SELECT user_id, email, password FROM tbl_users WHERE email='$email'");
$row=$query->fetch_array();
$count = $query->num_rows; // if email/password are correct returns must be 1 row
if (password_verify($password, $row['password']) && $count==1) {
$_SESSION['userSession'] = $row['user_id'];
header("Location: home.php");
$update = ("UPDATE tbl_users SET status='Online'");
$DBcon->query($update);
} else {
$msg = "<div class='alert alert-danger'>
<span class='glyphicon glyphicon-info-sign'></span> Invalid Username or Password !
</div>";
}
$DBcon->close();
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Coding Cage - Login & Registration System</title>
<link href="bootstrap/css/bootstrap.min.css" rel="stylesheet" media="screen">
<link href="bootstrap/css/bootstrap-theme.min.css" rel="stylesheet" media="screen">
<link rel="stylesheet" href="style.css" type="text/css" />
</head>
<body>
<div class="signin-form">
<div class="container">
<form class="form-signin" method="post" id="login-form">
<h2 class="form-signin-heading">Sign In.</h2><hr />
<?php
if(isset($msg)){
echo $msg;
}
?>
<div class="form-group">
<input type="email" class="form-control" placeholder="Email address" name="email" required />
<span id="check-e"></span>
</div>
<div class="form-group">
<input type="password" class="form-control" placeholder="Password" name="password" required />
</div>
<hr />
<div class="form-group">
<button type="submit" class="btn btn-default" name="btn-login" id="btn-login">
<span class="glyphicon glyphicon-log-in"></span> Sign In
</button>
<a href="register.php" class="btn btn-default" style="float:right;">Sign UP Here</a>
</div>
</form>
</div>
</div>
</body>
</html>
4 Jawaban:
Jawaban Terpilih
$update = ("UPDATE tbl_users SET status='Online'"); tambahin where gan kayak script dibawah ..
UPDATE tbl_users SET status = 'online' WHERE user_id = 3;
follow ig ini ya gan @fredricopohan
Thankyou bwt responya @HandiNursan Udah Solved gan. Barusan ane tambahin filter berdasarkan email bukan id.
Begini gan:
$update = ("UPDATE tbl_users SET status='Online' WHERE email='$email' ");
$DBcon->query($update);