XMLHttpRequest将变量传递给php脚本

<script language="javascript" type="text/javascript">
    // Browser Support 
function Heatctrl(heatMode){
    var heatControlRequest;

    try{
        //Good Browsers
        heatControlRequest = new XMLHttpRequest();
    } catch (e) {
        //Internet Explorer
        try{
            heatControlRequest = new ActiveXObject("Msxml2.XMLHTTP");
        } catch (e) {
            try{
                heatControlRequest = new ActiveXObject("Microsoft.XMLHTTP");
            }catch (e) {
                alert("Brower not supported");
                return false;
            }
        }
    }


    // Function to receive data from server and store in variable
    heatControlRequest.onreadystatechange = function(){
        if (heatControlRequest.readystate == 4){
            alert(heatControlRequest.responseText);
            //document.getElementById("controlMode").innerHTML=heatControlRequest.responseText;
            //var heatControlResponse = heatControlRequest.responseText;
        }
    }
    var url = "heatControl.php?heatmode="+heatMode;
    // Send the request
    heatControlRequest.open("GET",url,true);
    heatControlRequest.send();
}   
</script>

HTML

<div id="boilerButtons">
    <button type="button" onclick="Heatctrl('On')">Heating On</button>
    <button type="button" onclick="Heatctrl('Off')">Heating Off</button>
    <button type="button" onclick="Heatctrl('Boost')">Boost</button>
</div>

和PHP

<?php
$controlMode = $_GET['heatmode'];
echo $controlMode; 
?>

我非常感谢任何帮助,我从事电子工作而不是编程工作,而且我现在已经为此困难了两天.