如何根据用户是否登录显示或隐藏信息

Question

I am trying to develop a website and I am having issues with trying to hide information based on if a user is logged in.

I am using mysqli as a database and have registered a test user. Unfortunately, when using echo it doesn't change what is seen.

I have this in the header to create the session:

<?php
require 'includes/dbh.inc.php';
session_start();
 ?>

and then after that, I am using this in the index.php to try and show or hide based on if they are logged in:

<?php
if (!isset($_SESSION['userid'])) {
echo '<p class="login-status">You are logged in!</p>';
}

else {
echo '<p class="login-status">You are logged out!</p>';
}
 ?>

Regardless of what I seem to do it always displays "you are logged in!"

Answer 1

You are using a different index in login.inc.php and index.php. Try this in your index.php and it should work:

<?php
require 'header.php';
 ?>

<main>
  <div class="wrapper-main">
<section class="section-default">

<?php
print_r($_SESSION);
if (isset($_SESSION['idusers'])) {
echo '<p class="login-status">You are logged in!</p>';
}

else {
echo '<p class="login-status">You are logged out!</p>';
}
 ?>

</section>
</div>
</main>

 <?php
require 'footer.php';
  ?>

I tried your code locally and it is working now.

EDIT: I copied my working code from your index.php. Just replace it in your index.php file.