Somehow this array is not an array?

Question

The following code return some results that are baffling me...

if(is_array($loginUser)) {         
    $_SESSION['name'] = $loginUser['firstname'];
    $_SESSION['id'] = $loginUser['id'];
    print_r($_SESSION['name']);
    print_r($loginUser);
    var_dump($loginUser[1]);
    exit();
    header("Location: ../index.php?page=home");
}

That returns:

Notice: Undefined index: firstname in wwwroot/includes/userhandler.php on line 124

Notice: Undefined index: id in wwwroot/includes/userhandler.php on line 125

Array ( [0] => Array ( [0] => 4 [id] => 4 [1] => Johnny [firstname] => Johnny [2] => Appleseed [lastname] => Appleseed [3] => [email protected] [email] => [email protected] [4] => johnny'shashedpassword [password] => johnny'shashedpassword ) )

Notice: Undefined offset: 1 in wwwroot/includes/userhandler.php on line 129

NULL

Do a $loginUser = current($loginUser); to fix your problem — Peter
– Peter, Commented Sep 15, 2015 at 10:24
@MohammadAlabed Because something is obvious to you, that doesn't mean it is obvious to others as well. — Peter
– Peter, Commented Sep 15, 2015 at 10:30

AgeDeO · Accepted Answer · 2015-09-15 09:39:47Z

1

The id and firstName are stored in an array with key 0

change

$_SESSION['name'] = $loginUser['firstname'];
$_SESSION['id'] = $loginUser['id'];

to

$_SESSION['name'] = $loginUser[0]['firstname'];
$_SESSION['id'] = $loginUser[0]['id'];

but I see no reason why the you have an extra dimension in your $loginUser array. Try to refactor the way $loginUser is building its array

Later in the code you are calling var_dump($loginUser[1]); This returns your second error message because there is only a $loginUser[0]

answered Sep 15, 2015 at 9:39

AgeDeO

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2 Comments

Calvin Raines Over a year ago

Thank you for the explanation. I'm looking at it now to figure out why there are two arrays. I have never run into this before, and it certainly was not intentional.

AgeDeO Over a year ago

no problem, we where all beginners once don't let people like Mohammad Alabed put you down. Keep programming and you will be a professional one day

Peter · Accepted Answer · 2015-09-15 10:27:43Z

0

Your array is in another array. You can solve this by using the following line of code:

$loginUser = current($loginUser);

Furthermore, when using var_dump I always find it easy to use <pre /> tags. This makes it easier for you to see how the array is built up.

Example:

echo "<pre>";
var_dump($loginUser);
echo "</pre>";
exit;

answered Sep 15, 2015 at 10:27

Peter

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