hello everyone, im sort of new to jquery and ajax. i was getting familiar with using javascript and ajax when all of the sudden i started to do things and all directions pointed towards using jquery. so i had to learn jquery.. it was the best thing i did. its so much easy. this is a very simple way to show content from a php file using ajax

so with that in mind, i am going to show you a simple script which you can use to practice and see how it works.

im a PHP programmer so i will be performing the server side script using PHP.

i am using a Windows 7 Computer so i will be using notepad to create the files.

the first file we are going to create is the HTML file, you can call it jquery-simple-ajax.php or jquery-simple-ajax.html whatever you want. i am going to call it jquery-simple-ajax.php

so open a new notepad, copy and paste this code and save it as jquery-simple-ajax.php

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Simple JQuery Ajax Example by Webune.com</title>
<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js" type="text/javascript"></script>
<script type="text/javascript">
// This functions starts the ajax to show the output from post.php
function StartAjax(ResultsId){
	  type: "GET",
	  url: "post.php",
	  cache: false,
	  data: "name=John&location=Boston",
	  success: function(html){
<a href="#" onclick="StartAjax('ResultsId');">Click Here to see content from post.php</a>
<div id="ResultsId"></div>

next we are going to create the server side file called post.php

open another notepad, copy and paste the following code and save it as post.php

# The followin code will be sent to the ResultsId div in the jquery-simple-ajax.php file
echo 'This is the Data Received:<br><PRE>';print_r($_REQUEST);echo '</PRE>';