Den er vist fra en af dine hadesider:
http://www.tizag.com/ajaxTutorial/ajax-javascript.php[code]
<html>
<body>
<script language="javascript" type="text/javascript">
<!-- 
//Browser Support Code
function ajaxFunction(){
    var ajaxRequest;  // The variable that makes Ajax possible!    
    try{
        // Opera 8.0+, Firefox, Safari
        ajaxRequest = new XMLHttpRequest();
    } catch (e){
        // Internet Explorer Browsers
        try{
            ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");
        } catch (e) {
            try{
                ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");
            } catch (e){
                // Something went wrong
                alert("Your browser broke!");
                return false;
            }
        }
    }
    // Create a function that will receive data sent from the server
    ajaxRequest.onreadystatechange = function(){
        if(ajaxRequest.readyState == 4){
            var ajaxDisplay = document.getElementById('ajaxDiv');
            ajaxDisplay.innerHTML = ajaxRequest.responseText;
        }
    }
    var age = document.getElementById('age').value;
    var wpm = document.getElementById('wpm').value;
    var sex = document.getElementById('sex').value;
    var queryString = "?age=" + age + "&wpm=" + wpm + "&sex=" + sex;
    ajaxRequest.open("GET", "ajax-example.php" + queryString, true);
    ajaxRequest.send(null); 
}
//-->
</script>
<form name='myForm'>
Max Age: <input type='text' id='age' /> <br />
Max WPM: <input type='text' id='wpm' />
<br />
Sex: <select id='sex'>
<option value='m'>m</option>
<option value='f'>f</option>
</select>
<input type='button' onclick='ajaxFunction()' value='Query MySQL' />
</form>
<div id='ajaxDiv'>Your result will display here</div>
</body>
</html>['/code']
Kom så med hjælpen, for jeg ved hvis der er en som kan hjælpe mig, så er det dig!