how to call a function $_SESSION[usr]->logout() inside <a> - php

Question

I have a class named User which has a function named logout(). I create an instance of this class in index.php and i pass it's value to $_SESSION[usr] before i call memberspage.php . In memberspage.php i have a link named logout which when clicked i want the logout() function to run and also send the user to index.php. For this purpose i've done something like this.
Log out
I know that -> causes the problem but i don't know how to fix it. thnx for your time.
The following code worked for me
Log out
but there is a problem. If i go to the page(memberspage.php) where the above code is and i press the back arrow (not logout link) the logOut() function will still be used(the session is destroyed and i will have to log in again to access memberpage.php) . I don't get it because i thought that the only way to call the logOut() function was to click on Log out link.

Answer 1

If $_SESSION[usr]->logout() is working for you as you said in your comment. I don't know how.
But here is just for calling a php function inside anchor tag.It's totally depend on your function response.
<?php
function usr(){
return "abc";
}
?>
Log out

Answer 2

First i suggest that you change your use of session you can create a page for example session.php where all your session is place, it can also be the re directory page of your login page.
like this one named login.php
create in your form make action redirect to session.php
i also suggest that all your php codes of login are inside the session.php then make this one.
<?php
session_start();
$host = "localhost";
$uname = "root";
$pass = "";
$db = "mydb;
//database connection
$conn = mysqli_connect($host, $uname, $pass, $db);
mysqli_select_db($conn, $db);
if(!$conn){
die("Connection failed: " . mysqli_connect_error());
}
if(isset($_POST['username'])){
$username = $_POST['username'];
$password = $_POST['password'];
$username = stripslashes($username);
$password = stripslashes($password);
//$username = mysqli_real_escape_string($username);
//$password = mysqli_real_escape_string($password);
$sql = "SELECT * FROM table WHERE username = '" .$username. "' AND password = '".$password."' LIMIT 1";
$res = mysqli_query($conn, $sql);
if(mysqli_num_rows($res) > 0){
if($data = mysqli_fetch_assoc($res))
{
$_SESSION['type'] = $data['type'];
if(isset($_SESSION["login_user"]))
{
if($data['type'] == 'admin'){
header('location: admin.php');
}
else if($data['type'] == 'customer'){
header('location: customerhome.php');
}//header('location: uservalidation.php');
}
}
}
else{
//header('location: #');
echo '<script>';
echo 'alert("Invalid no?")';
echo '</script>';
header('location: logind.php');
}
}
?>
then create another page which is logout.php
put this code inside:
<?php
session_start();
header('location: index.php');
session_destroy();
?>
then save put the a link your page for logout.php

Answer 3

Add file logout.php and put into them your logout implementation:
<?php
header('Content-Type: application/json');
$_SESSION[usr]->logout();
echo json_encode(['message' => 'ok']);
And call this file with AJAX:
<script>
function logout() {
$.ajax({
url: '/logout.php'
}).then(function (res) {
window.location.href = '/';
});
}
</script>
Log out