Send single input with ajax/jquery/php - php

Question

I'm new to jQuery / AJAX.
I'm trying to send single input with jquery/ajax/php.
LIVE EXAMPLE
But, after pressing submit nothing is happening, where is my error?
Any help much appreciated.
HTML:
<form action="submit.php">
<input id="number" name="number" type="text" />
<input id="submit" name="submit" type="submit" />
</form>
JQUERY / AJAX:
$(document).ready(function(e) {
$('input#submit').click(function() {
var number = $('input[name=number]');
var data = 'number=' + number.val();
$.ajax({
url: "submit.php",
type: "GET",
data: data,
cache: false,
success: function(html) {
if (html == 1) {
alert('wyslane');
}
else {
alert('error');
}
}
});
return false;
});
});
PHP:
<?php
$mailTo = 'email#gmail.com';
$mailFrom = 'email#gmail.com';
$subject = 'Call Back';
$number = ($_GET['number']) ? $_GET['number'] : $_POST['number'];
mail($mailTo, $subject, $number, "From: ".$mailFrom);
?>

Answer 1

HTML:
<form id=submit action="">
<input id="number" name="number" type="text" />
<input name="submit" type="submit" />
</form>
The action URL is irrelevant as you want to submit your data via AJAX. Add the submit id to the form and override the default submit behavior, instead of overriding the onclick handler of the submit button. I'll explain in the JS section.
JS:
var number = $('input[name="number"]');
Quotes were missing.
$(document).ready(function(e) {
$('#submit').submit(function() {
var number = $('input[name=number]');
var data = 'number=' + number.val();
$.ajax({
url: "submit.php",
type: "GET",
data: data,
cache: false,
success: function(html) {
if (html == 1) {
alert('wyslane');
}
else {
alert('error');
}
}
});
return false;
});
});
I don't really understand your success callback, why do you expect that html should be equal to 1?

Answer 2

Atleast I got 404 error when pressed your submit button:
Not Found
The requested URL /index.php was not found on this server.
Additionally, a 404 Not Found error was encountered while trying to use an ErrorDocument to handle the request.
When you get it to work, remember to add mysql_real_escape_string function to avoid SQL injections http://php.net/manual/en/function.mysql-real-escape-string.php
Since you are also using ID for number, you could just use: var data = 'number=' + $('#number').val()
Also if you add ID to your form, you can use:
$('#formId').submit(function(){
});
instead of that click. This function will launch when that form is submitted. This is better way because users can submit the form with other ways aswell than just clicking the submit button (enter).

Answer 3

var number = $('input[name=number]');
is wrong. It's
var number = $('input[name="number"]');