Specify a single logout case

Question

I'm trying to override all logout situations (timeout-sessions, or simply any method that refers to [logout.php] page!) by specifying a single logout button in the index page. Is this possible? And what did I miss? Thanks in advance.

The code used in the [index.php]:

<?php
session_start();
if(isset($_POST['logout'])) {
    $_SESSION['logout_command'] = 1;
    header("Location: logout.php");
}
?>

<form action="" method="post">
    <input type="submit" name="logout" value="LOGOUT!!!">
</form>

And the code in the [logout.php]:

<?php
session_start();
if(isset($_SESSION['logout_command'])) {
    $check = $_SESSION['logout_command'];
    if($check = 1) { // AND ONLY IF!
        session_destroy();
        header("Location: login.php");
    } else { // BACK OFF!
        header("Location: index.php");
    }
} else { // ALSO BACK OFF!
    $_SESSION['logout_command'] = 0;
    header("Location: index.php");
}
?>

UPDATE: So, I missed the comparing (==) signs... Also, the [exit()] function after each [header()] in a sequence... Thanks to everyone.

After every `header('Location: /example.php');` you need to use the `exit();` method. — Jaquarh, Sep 04 '17 at 17:00
@Qirel It wasn't my purpose to be honest... I'm just concerned about restricting the logout process... — , Sep 04 '17 at 17:10
@Sky7ure It's alright if your question is flagged as a duplicate - its just that we gather frequently asked questions/issues into single threads with a deeper explanation, so they become more useful and easier to use for a larger userbase. — Qirel, Sep 04 '17 at 17:14

score 2 · Accepted Answer · answered Sep 04 '17 at 17:00

2

You're not doing a conditional, you're assigning the value in your first if:

if($check == 1) {

answered Sep 04 '17 at 17:00

ishegg

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Wow! Is that all?!^_^" – Sep 04 '17 at 17:01
1

It might be, everything else looks OK. Did you check? – ishegg Sep 04 '17 at 17:02

Specify a single logout case

1 Answers1