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!