jQuery form data not sending to PHP page

Question

I'm very new to anything outside of PHP, so forgive me if this comes off as simple.

I'm trying to create a simple registration form, which then calls a PHP function, which inserts the data.

The problem is, the data doesn't seem to be going to the page.

My form:

<!DOCTYPE html>
<html>
<head>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"></script>
<script>
$(function () {
$('#register').on('submit', function (e) {
    $.ajax({
        type: 'POST',
        url: 'http://warofman.com/actions.php?type=register',
        data: $('#register').serialize(),
        success: function () {
            alert('Form was submitted.');
            console.log(e);
        }
        });
    e.preventDefault();
    });
});
</script>
</head>
<body>
    <form id="register">
        Username: <input type="text" name="login"><br>
        Email: <input type="text" name="email"><br>
        Password: <input type="password" name="password"><br>
        Confirm Password: <input type="password" name="confpass"><br>
        <input type="submit" name="submit" value="Register">
    </form>
</body>
</html>

And then, the PHP:

function register()
{
    $login = $_POST['login'];
    $email = $_POST['email'];
    $password = $_POST['password'];
    $confpass = $_POST['confpass'];

    do_reg($login, $email, $password, $confpass);
}

Where do_reg() calls the process of registering.

Show how you're inserting the information into the database because if that is how you're trying, it's not... correct. — iBrazilian2
– iBrazilian2, Commented Dec 11, 2013 at 0:09
What are you passing through serialize? the problem must be there. Try alerting the serialized data or try to log it.. — briosheje
– briosheje, Commented Dec 11, 2013 at 0:12
if you do in php "print_r($_POST);" it's receiving correctly the POST data? If so.. it's do_reg() problems — Wagner Leonardi
– Wagner Leonardi, Commented Dec 11, 2013 at 0:12
As it turns out, I am completely simple. The problem was in how I was calling my database class, so the code was not actually running. (Way to go PHP for silently failing!) I accepted Rao's answer for the best practice advice. :) — Seker
– Seker, Commented Dec 11, 2013 at 1:35

Nagendra Rao · Accepted Answer · 2013-12-11 00:25:21Z

Make the following changes, (follow good practice)

<form id="register" action="" method="post">
        <label for="login">Username:</label>
        <input id="login" type="text" name="login"><br>

        <label for="email">Email:</label>
        <input type="email" name="email" id="email" /><br>

        <label for="password">Password:</label> 
        <input type="password" name="password" id="password"><br>

        <label for="confpass">Confirm Password:</label>
        <input type="password" name="confpass" id="confpass"><br>

        <input type="submit" name="submit" value="Register">
    </form>

On php side:

You are writing everything inside a function(register) and not calling it, instead do something like this,

if(isset($_POST['submit'])){
  $login = $_POST['login'];
  $email = $_POST['email'];
  $password = $_POST['password'];
  $confpass = $_POST['confpass'];
  do_reg($login, $email, $password, $confpass);
}

thobens · Accepted Answer · 2013-12-11 00:44:11Z

0

You can't pass a query string in the URL using POST. Add the "type=register" to your data:

$.ajax({
    type: 'POST',
    url: 'http://warofman.com/actions.php',
    data: "type=register&" + $('#register').serialize(),
    success: function () {
        alert('Form was submitted.');
        console.log(e);
    }
    });
e.preventDefault();
});

answered Dec 11, 2013 at 0:44

thobens

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