PHP会话在localhost上工作但不在服务器上工作

Question

感谢您加入此事，我在使用服务器上的会话变量登录时遇到了麻烦，而且它在localhost上工作得很好。这是我的代码。

我的目标网页

<?php
require_once 'lock.php';
?>
My HTML code will be here....

lock.php

<?php
session_start();
require_once 'includes/connect.php';
if(empty($_SESSION['logged_in'])){
    header('Location: login.php?action=not_yet_logged_in');
}
$action = $_GET['action'];
if($action=='logout'){
    session_destroy();
    header('Location: login.php?action=logout');
}
?>

的login.php

<?php
session_start();
require_once 'includes/connect.php';
if(isset($_SESSION['logged_in'])==true){
    header('Location: index.php?action=already_logged_in');
}
?>


<?php
if (isset($_GET['action'])) {
if($_GET['action']=='not_yet_logged_in'){
    echo "<div class='infoMesssage'>You cannot go to the index page because you are not yet logged in.</div>";
    }else if($_GET['action']=='logout'){
        echo "<div class='infoMesssage'>You have scuccessfully logout.</div>";
    }
}                   

 if($_POST){
    $enUsername = $_POST['username'];
    $enPassword = $_POST['password'];

    if(empty($_POST['username'])){
        $username = 'mockadmin';
        $password = 'mockadmin';
    }else {
        $chLogin = mysqli_query($con, 'select * from admin where user_name="'.$enUsername.'" AND password="'.$enPassword.'"');
        $chUser = mysqli_fetch_object($chLogin);

        $userId = $chUser->id;
        $username = $chUser->user_name;
        $password = $chUser->password;
    }

    if($_POST['username']==$username && $_POST['password']==$password){
      $_SESSION['logged_in'] = true;
      header('Location: index.php');                
   } else{
      echo "<div class='failedMessage'>Access denied. :(</div>";
   }
}
?>
<form class="m-t" method="post" role="form" action="login.php">
    <div class="form-group">
        <input type="text" name="username" class="form-control" placeholder="Username">
    </div>
    <div class="form-group">
         <input type="password" name="password" class="form-control" placeholder="Password">
    </div>
    <button type="submit" name="btnlogin" class="btn btn-primary block full-width m-b">Login</button>
</form>

我的数据库 connect.php

<?php  
$con = mysqli_connect('localhost', 'liveuser', 'livepass', 'mocktest');
if (mysqli_connect_errno()){
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>

感谢您的帮助。再次感谢。