CODEWITHSUNDEEP
phphtmlformsisset
ABor
ABor

Reputation: 173

PHP isset('submit') is always returning FALSE

I'm using an external php file in the "action" field of my HTML form so that the actions mentioned in the php file are undertaken on clicking the Submit button.

For this I've used the isset() function in the php file.

However, I'm finding that the isset function is always returning FALSE, resulting in the execution of the else statement (as seen in the console log).

If I remove the isset() function (and hence the if-else statements), then the code is working wonderfully.

Can you please check the problem in my code?

Also I've seen in other posts that I need to use some other argument along with isset(), for example,

if(isset($_POST['submit']) && !empty($_POST["xyz"]))

is this at all required?

P.S.: I'm still in the initial stage of the page development and hence I request you to please ignore the security concerns of my code, which I acknowledge that it exists. :)

My Sub-codes:

My HTML Form:

<form id="info-form" method="POST" action="form-submit.php">
    <label for="Name">What is your Name? </label> 
    <input required type="text" name="name" placeholder="Enter your full name here." />

    <label for="Email">What is your email ID? </label>
    <input required type="email" name="email" placeholder="[email protected]" />

    <label for="mobile">What is your 10-Digit Mobile Number? </label>
    <input required type="text" name="mobile" maxlength="10" />

    <button name="submit-form" type="submit" class="btn btn-lg btn-success"><i class="fa fa-paper-plane" aria-hidden="true"></i>
        Submit
    </button>
    <button type="reset" class="btn btn-lg btn-warning"><i class="fa fa-undo" aria-hidden="true"></i>
        Reset
    </button>
</form>

My form-submit.php file:

<?php
    if(isset($_POST['submit-form']))
    {
        require("database-connect.php");

        $name = $_POST['name'];
        $email = $_POST['email'];
        $mobile = $_POST['mobile'];

        $sql = "INSERT INTO tbl_details ".
               "(name,email_id,mobile_number) ".
               "VALUES ".
               "('$name','$email','$mobile')";

        mysql_select_db('db_info');
        $return = mysql_query( $sql, $connect );

        if(! $return )
        {
            die('Could not enter data: ' . mysql_error());
        }

        echo "Entered data successfully\n";
        mysql_close($connect);
    }

    else
    {
        echo "Not Set\n";
    }
?>

Upvotes: 1

Views: 2550

Answers (3)

deceze
deceze

Reputation: 522015

You really shouldn't be checking for the existence of a submit button value to begin with. As you see, some slight cosmetic changes to the frontend, like what kind of button is used to submit a form, shouldn't have any repercussions on the backend. The submit button and its value are pretty irrelevant to processing a form.

What you really want to check on the server is either if the request was a POST request, or whether the values which you want to work with are set, or both:

if ($_SERVER['REQUEST_METHOD'] == 'POST') …
if (isset($_POST['name'], $_POST['email'], $_POST['mobile'])) …

The most reasonable thing would be:

if ($_SERVER['REQUEST_METHOD'] != 'POST') {
    header('HTTP/1.0 405 Method Not Allowed');
    exit;
}

or:

if ($_SERVER['REQUEST_METHOD'] != 'POST') {
    header('Location: myform.html');
    exit;
}

After you've asserted that you're dealing with the right HTTP method, process your data:

$data = filter_input_array(INPUT_POST, [
    'name'   => FILTER_DEFAULT,
    'email'  => FILTER_VALIDATE_EMAIL,
    'mobile' => FILTER_DEFAULT
]);

As you see, you don't even need to interact with $_POST directly at all. See http://php.net/filter_input_array.

Upvotes: 5

x13
x13

Reputation: 2227

You should use an input-element instead of a button.

<button name="submit-form" type="submit" class="btn btn-lg btn-success">
<i class="fa fa-paper-plane" aria-hidden="true"></i>Submit
</button>

Should be replaced by an input like this:

<input name="submit-form" value="ButtonText" type="submit" class="btn btn-lg btn-success" />

The value of an input is the text it displays to the user.

Also, in your question you mention isset('submit') and isset($_POST['submit']), while others are right and this is not an existing name in your HTML code, your PHP code does contain the correct isset($_POST['submit-form']). It causes a little confusion for some 'quick' readers...


As far as the extra !empty($_POST["xyz"])), for your purpose (checking if the form is a postback) that won't be necessary.

Upvotes: 0

Ragu Natarajan
Ragu Natarajan

Reputation: 739

Change your button tag like this. I hope its will help you!

<input name="submit-form" value="Submit" type="submit" class="btn btn-lg btn-success">

Thank you!

Upvotes: 1

Related Questions