Submit a form and reload data on same page - php

Question

I got a left_column with a #form, when I sumbmit it should load the results on #content_div without refreshing the page.
Im using this:
<script type="text/javascript" src="http://code.jquery.com/jquery-1.4.2.min.js"></script>
<script type="text/javascript" src="http://ajax.microsoft.com/ajax/jquery.validate/1.7/jquery.validate.min.js"></script>
<script type="text/javascript">
$(function() {
$('#dateform').submit(function(evt) {
evt.preventDefault();
$.ajax({
url: "charts/client.php",
type: 'POST',
data: $(this).serialize(),
success: function(result) {
$('#content_div').html(result);
}
});
});
});
</script>
<div id="content_div">
Nothing seems to appear.
And firebug reports this:
ReferenceError: google is not defined
This charts/client.php is using google api, and yes i've declared it like this:
<script language="javascript" type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.0/jquery.min.js"></script>
<script type="text/javascript" src="https://www.google.com/jsapi"></script>
What am I doing wrong?
Thanks

Answer 1

use ajax form
<script>
// wait for the DOM to be loaded
$(document).ready(function()
{
// bind 'myForm' and provide a simple callback function
$("#tempForm").ajaxForm({
url:'../calling action or servlet',
type:'post',
beforeSend:function()
{
alert("perform action before making the ajax call like showing soinner image");
},
success:function(e){
alert("data is"+e);
alert("now do whatever you want with the data");
}
});
});
</script>
you can find the plugin here

Answer 2

It seems that the error you get is from the client.php, not from the actual jquery ajax script.
Probably you tried to call a google method before creating a new instance of the google object you used. I could help you out more if you post the client.php's code here.
For example, when i have worked with gmaps api:
trying to do:
geocoder.geocode( { 'address': target}, function(results, status) {...
before setting :
var geocoder = new google.maps.Geocoder();
will return "google is not defined";