Linking page in a div with html forms and php - php

Question

So I have this html code
<div id="wrap">
<div id="header">
</div>
<div id="content">
<form method="POST" action="code.php">
Name:
<input type="text" name="name" size="50">
<input type=submit value="Get Code">
</form>
</div>
<div id="footer">
</div>
Is it possible to load the code.php after the user clicks submit into the #content div?
Essentially, what I want is when the user clicks the submit button, the code.php after processing is loaded onto the same #content div.
So let say in my code.php, after processing the inputted data, I come up with this lilne of code,
<?php
some processing code here;
$name = 'john';
echo $name;
?>
So then after hitting submit, user would see
<div id="content">
john
</div>
Hope I didn't complicate my question by repeating myself, please let me know if this is possible with javascript, php or whatever.
Thanks for the read!

Answer 1

#JohnP yes, $.load is a good solution. However, you'll need to send the form data in the request:
UPDATED [3] for sending a POST with multiple fields and checkboxes:
$('form').submit(function(){
// create an object to send as a post
var form = this,
fields = form.elements,
el,
post = {};
for (var i = fields.length; i--; ) {
el = fields[i];
if (el.name) {
switch (el.type) {
case 'checkbox':
case 'radio':
post[el.name] = (el.checked) ? el.value : '';
break;
default:
post[el.name] = el.value;
}
}
}
// send the form data in the load request...
$('#content').load(this.action, post);
return false;
});
This will send the data as a POST.

Answer 2

Since you've tagged jQuery, I'll use a jQuery example
$(document).ready(function(){
$('form').submit(function(){
$('#content').load('code.php');
return false;
})
})
This makes a couple of assumptions here
This assumes that code.php is in the same path that you are in now.
There is only one form in the page.
As #johnhunter points out, this example obviously won't work with post. You can send the post data along with the method. See here for usage : http://api.jquery.com/load
EDIT
Here's a fiddle example : http://jsfiddle.net/jomanlk/J4Txg/
It replaces the form area with the content from jsfiddle/net/echo/html (which is an empty string).
NOTE 2 Make sure to include the code in $(document).ready() or include it at the bottom of the page. It goes without saying you need jQuery in your page to run this.

Answer 3

You might want to check out jquery form plugin http://jquery.malsup.com/form/#

Answer 4

in simple way use
<div id="content">
if(isset($_POST['submit'] && !empty($_POST) )
{
// do your all post process
$name ='John';
echo $name;
}
else {
<form method="POST" action="$_SERVER['PHP_SELF']">
<label for="uname" >Name:</label><input type="text" name="uname" id="uname" size="50">
<input type=submit value="Get Code" name="submit">
</form>
}
</div>