Getting variable jQuery -> PHP (not working)

Question

I'm having problems with passing variables from jQuery into PHP. I search for the solutions on the internet, and i came to the AJAX. Never used Ajax before, so i guess that the problem is known.

So i have the following code in "index.php"

$("#inviteForm").submit(function(e) {

e.preventDefault();
var emailVal = $("#email").val();

$.ajax({
 type: "POST",
 url: "processAjax.php",
 data: {email: emailVal},
 success: function(data) {
    alert(data);  
}
});
});

In form, i have one input box (for email) and a submit button (the method is POST).

In processAjax.php i have the following code

<?php
    $x = $_POST['email'];
    return $x;
?>

So if i'm correct, if the $.ajax function is OK, the alert box should pop up. But it doesn't. i've also tried alert(x); but it didn't work.

Any idea what i'm doing wrong

The absolute first thing I'd check is whether the AJAX request is actually being sent. Use Firebug (for Firefox) or the Developer Tools in other browsers to do that (all accessed by hitting the F12 key). If it's not, you're probably trying to bind the submit event handler too early (before the form exists); if that's the case take a look at the .ready() function. — Anthony Grist
– Anthony Grist, Commented Jan 24, 2013 at 15:01
You don't return the value, you output it; as if you were outputting it to the user. Everything that is output on the page will be accessible through the data variable. — EM-Creations
– EM-Creations, Commented Jan 24, 2013 at 16:06

Moseleyi · Accepted Answer · 2013-01-24 14:57:55Z

7

Try echo $x; instead of return $x;

answered Jan 24, 2013 at 14:57

Moseleyi

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Clifford James · Accepted Answer · 2013-01-24 14:59:05Z

2

Try this:

<?php
    $x = $_POST['email'];
    echo $x;
?>

answered Jan 24, 2013 at 14:59

Clifford James

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mallix · Accepted Answer · 2013-01-24 15:17:11Z

1

Try this for better data manipulation. Use json_encode from the server side and json datatype to your ajax calls. Then to alert server response, just alert the key of the array like alert(data.value):

$.ajax({
 type: "POST",
 url: "processAjax.php",
 data: {email: emailVal},
 dataType: 'json'
 success: function(data) {
    alert(data.value);  
}

processAjax.php

$result['value'] = $_POST['email'];
echo json_encode($result);

edited Jan 24, 2013 at 15:17

answered Jan 24, 2013 at 14:59

mallix

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