Php session error undefined index

Question

session_start();

if (isset($_POST['username'] , $_POST['password'])) {

    $extractabout = $db->prepare("SELECT * FROM user WHERE username = :username && password = :password");

    $extractabout->execute([
        'username' => $_POST['username'],
        'password' => $_POST['password']
    ]);

    $infos = $extractabout->rowCount() ? $extractabout : [] ;

    foreach ($infos as $info) {
        if(!empty($info['username']) && !empty($info['password']) && !empty($info['id']) && !empty('role')) {
            $_SESSION['username'] == $info['username'];
            $_SESSION['password'] == $info['password'];         
        }

    }

    if(isset($_SESSION['username'], $_SESSION['password'], $_SESSION['user_id'], $_SESSION['role'])) {
        header("Location: test.php");
    }   

}

?>

I have an error when I give $_SESSION['username'] the value of $info['username'] called :

Undefined index: username
Undifined index: password

Multiple errors spotted. == is different from =. Remember to exit() after header("Location: $url"); call — Raptor
– Raptor, Commented Jan 8, 2015 at 11:23

Rizier123 · Accepted Answer · 2015-01-08 11:22:11Z

2

You don't assign the values, you only have to use one = so use this:

$_SESSION['username'] = $info['username'];
$_SESSION['password'] = $info['password'];

Also i would put a exit(); after each header so that you are sure the script stops to be executed!

header("Location: test.php");
exit();

answered Jan 8, 2015 at 11:22

Rizier123

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Comments

Rakesh Sharma · Accepted Answer · 2015-01-08 11:29:54Z

0

use = for assign values in session not ==(compare).

Also what is !empty('role') i think it would be !empty($info['role'])

if(!empty($info['username']) && !empty($info['password']) && !empty($info['id']) && !empty($info['role'])) {
        $_SESSION['username'] = $info['username'];
        $_SESSION['password'] = $info['password'];                     
}

edited Jan 8, 2015 at 11:29

answered Jan 8, 2015 at 11:23

Rakesh Sharma

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Comments

Needhi Agrawal · Accepted Answer · 2015-01-08 11:33:15Z

-1

You have syntax errors in your code as,corrected one is:

if(!empty($info['username']) && !empty($info['password']) && !empty($info['id']) && !empty($info['role'])) {
    $_SESSION['username'] = $info['username'];
    $_SESSION['password'] = $info['password']; 
}

edited Jan 8, 2015 at 11:33

answered Jan 8, 2015 at 11:24

Needhi Agrawal

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3 Comments

Raptor Over a year ago

Why separate isset() calls? That's not necessary.

Needhi Agrawal Over a year ago

This isset($_SESSION['username'], $_SESSION['password'], $_SESSION['user_id'], $_SESSION['role'])) is wrong, i just correct the syntax.

Rakesh Sharma Over a year ago

@NeedhiAgrawal you can use multiple check with comma separate in isset() follow stackoverflow.com/questions/14476609/…

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