I'm a beginner in jQuery area and I have simple question like this :
I want to load (AJAX) MySQL result in array, let's say :
$row[0] = first name
$row[1] = last name
$row[2] = phone number
I have no problem with PHP part, but I have difficulties to display each of that array content on different id. because syntax I found loads everything processed by PHP :
<script type="text/javascript">
$(document).ready(function(){
$('#mysql-result').load('ajax.php');
});
</script>
how to get 'First Name', 'Last Name' and 'Phone Number' from PHP with only one time load and still I can put the result in different . thank you.
UPDATE
I give you real example about what I need. Here's my HTML file named ajax.html :
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Ajax Trial</title>
</head>
<body>
<div id = "fistname"><!-- ajax result goes here --></div>
<div id = "lastname"><!-- ajax result goes here --></div>
<div id = "phonenumber"><!-- ajax result goes here --></div>
</body>
</html>
and here's my PHP file, named ajax.php :
<?php
require_once 'config-min.php';
$con = mysql_connect($DbServer,$DbUser,$DbPassword);
mysql_select_db($DbName, $con);
$result = mysql_query("SELECT FirstName, LastName, PhoneNumber FROM User WHERE ID = '201' LIMIT 1");
$row = mysql_fetch_array($result);
echo $row[0];
echo $row[1];
echo $row[2];
mysql_close($con);
?>
now, my question still same... how to get this PHP result (3 echos), load once, then displayed in those 3 different divs
