PHP Undefined index when submitting form with Pjax

Question

I've been experimenting with Pjax and love it so far. It works great, but I've run into an issue. When submitting a form (post) with pjax, everything works, but I'm getting a "PHP Notice: Undefined index: name" in the error log.

Here's my code for submitting the form:

$(document).on('submit', 'form[data-pjax]', function(event) {
    event.preventDefault();
    $.pjax.submit(event, '#containercontainer');
    return false;
});

I've got all the required .js files included but not here for simplicity.

Here's an abridged version of my form:

<form data-pjax method="POST" action="add-process.php" id="addform">
    <div class="row">
        <div class="col-sm-5">
            <div class="form-group">
                <label for="name">Name:</label>
                <input type="text" class="form-control" id="name" placeholder="Name" name="name" autocomplete="off" autofocus>
            </div>
        </div>
        ...
    </div>
    <button id="submit_btn" type="submit" class="btn btn-success btn-block" style="font-weight:bold;">ADD</button>
</form>

In the add-process.php file, I'm getting the variables with $name = $_POST['name']; but the error log says undefined index for these lines.

UPDATE:

I changed my PHP code to include the recommended isset() as seen below:

// GET INFO FROM FORM
if(isset($_POST['userID'])) {
    $userID = $_POST['userID'];
}
if(isset($_POST['reason'])) {
    $reason = $_POST['reason'];
}
if(isset($_POST['otherreason'])) {
    $otherreason = $_POST['otherreason'];
}

if($reason == 3) {
    $reason = $otherreason;
}

if(!is_numeric($userID)) {
    echo 'User ID not numeric!';
    exit();
}

That fixed several things, but I'm still getting errors in the log:

PHP Notice:  Undefined variable: reason
PHP Notice:  Undefined variable: userID

Again, Thank you for any help!

score 0 · Answer 1 · answered May 26 '15 at 23:50

0

You get

PHP Notice:  Undefined variable: reason
PHP Notice:  Undefined variable: userID

because your isset($_POST['userID']) AND isset($_POST['reason']) returns false, because they are don't exist. That is why your variables undefined. You can add "default" values.

You can trace what your form send to server-side. Try this:

print_r($_POST);

or

print_r($_REQUEST);

And about form: check names of inputs. Are inputs with names 'userID' AND 'reason' exists?

answered May 26 '15 at 23:50

Mykola Vasilaki

416
6
11

Thank you! I commented out all the headers and put in echo statements to let me know how far it got and then tried print_r($_POST); and got this: "Array ( [userID] => 7 [reason] => 1 [otherreason] => ) Done!" Which I think is good! Would you mind helping me figure out what to do with this now? Thank you so much!!! – Matthew Greene May 27 '15 at 00:09
What you exactly want to say. ? explain a bit – Alive to die - Anant May 27 '15 at 00:16
I just want to be able to use these in php as regular variables... Like have them available for use as "echo $userID;" – Matthew Greene May 27 '15 at 00:34

score 0 · Answer 2 · answered May 27 '15 at 01:52

Any time that you set variables inside of conditionals, it is best to initialize them first:

$reason = null;
$userID = false;

/* .... your current code starts here   */

PHP throw the notices to help you avoid typo-induced bugs (ie later referencing uesrID, etc). Initializing them first makes it clear that you are in fact using the name you intended.

PHP Undefined index when submitting form with Pjax

2 Answers2