How to resolve Undefined index error in PHP

Question

I would like to show or hide a given set of links in the navigation bar of a web application based on the login state of the user.

Below is a snippet of my code;

<nav>
    <a id="mainpage">Main Page</a>
    <?php if ($_SESSION['logged_in'] === false) { ?>
    <a href="login2.php">Login</a>
    <a href="register.php">Register</a>
    <?php } else { ?>
    <a href="post.php">Posting</a>
    <a href="#">Members posts</a>
    <a href="logout.php" class="outbutton">Logout</a>
    <?php } ?>
</nav>

Here, my login page script;

if ($_SERVER["REQUEST_METHOD"] == "POST") {
    $username = ($_POST['username']);
    $password = ($_POST['password']);
    $q = "SELECT * FROM users WHERE username='$username' AND pass='$password'";
    $x = $conn->query($q);

    if ($x->num_rows > 0) {
        while ($row = $x->fetch_assoc()) {
        $_SESSION['logged_in'] = true;
        header("location: welcome.php");
    }
} else {
    die("Username or Password is incorrect");
}

My login script works as expected but in the index.php page, I get the error below when the user is not logged in:

Notice: Undefined index: logged_in

On the other hand, the links get displayed in the navigation bar when a user logs in successfully.

I am using session_start() at the beginning of my PHP script before any other codes.

use `isset` function like `if(isset($_SESSION['logged_in']))` — Bilal Ahmed, Oct 04 '18 at 06:48
Possible duplicate of ["Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset" using PHP](https://stackoverflow.com/questions/4261133/notice-undefined-variable-notice-undefined-index-and-notice-undefined) — Tom Udding, Oct 04 '18 at 06:50
If you are getting the error even after the login in index page then use session_start() in index page and also isset to check the logged_in session variable. — Vikash Dhiman, Oct 04 '18 at 06:53
@VikashDhiman I only get the error when I'm not logged in and I used session start — Jack potato, Oct 04 '18 at 06:54

score 1 · Accepted Answer · answered Oct 04 '18 at 07:01

try below where you are displaying the menu

<nav>
<a id="mainpage">Main Page</a>
<?php if (!isset($_SESSION['logged_in'])) { ?>
    <a href="login2.php">Login</a>
    <a href="register.php">Register</a>
<?php } else { ?>
    <a href="post.php">Posting</a>
    <a href="#">Members posts</a>
    <a href="logout.php" class="outbutton">Logout</a>
<?php } ?>

Bilal Ahmed · Answer 2 · 2018-10-04T06:56:05.073

0

You should used isset function like

if(isset($_SESSION['logged_in'])){
  // your code
}

Detail: first time session is not set & user is logout and there is no logged_in index in $_SESSION that's why php through warning

also you can use @ for more details

PHP supports one error control operator: the at sign (@). When prepended to an expression in PHP, any error messages that might be generated by that expression will be ignored.

edited Oct 04 '18 at 06:56

answered Oct 04 '18 at 06:49

Bilal Ahmed

4,005
3
22
42

If I do this the error will be gone but after logging out the navbar doesn't change it just shows the link for logged in users – Jack potato Oct 04 '18 at 06:53
use else condition and add code into else condition – Bilal Ahmed Oct 04 '18 at 06:57
Is it possible to use @ also on my login page? because I get an error when the user wants to login although there is no problem. here for example if ($_SESSION['logged_in'] === true) { ?> logout – Jack potato Oct 04 '18 at 07:11
you can use `@` for undefined error but this is not a good way – Bilal Ahmed Oct 04 '18 at 07:15

score 0 · Answer 3 · answered Oct 04 '18 at 06:55

0

session_start()
if(isset($_SESSION['logged_in'])){
    echo "SHOW MY LOGGED IN SESSION ".$_SESSION['logged_in'];
}else

answered Oct 04 '18 at 06:55

Ripon Uddin

709
3
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How to resolve Undefined index error in PHP

3 Answers3