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dhtmlajax sample program

1. test “dhtmlajax” sample program: send_request.html
<script>
function outputResponse(loader){
    if(loader.xmlDoc.responseXML!=null)
        alert("We Got Response\n\n"+loader.doSerialization())
    else
        alert("Response contains no XML")
}

function sendRequestGet(){
    if(!document.getElementById('sync').checked){
        dhtmlxAjax.get("process.php?"+document.getElementById('params').value,outputResponse);
        alert("Request Sent");
        }else{
        var loader = dhtmlxAjax.getSync("process.php?"+document.getElementById('params').value);
        alert("Request Sent");
        outputResponse(loader)
        }
}

function sendRequestPost(){
    if(!document.getElementById('sync').checked){
        dhtmlxAjax.post("process.php",document.getElementById('params').value,outputResponse);
        alert("Request Sent");
    }else{
        var loader = dhtmlxAjax.postSync("process.php",document.getElementById('params').value);
        alert("Request Sent");
        outputResponse(loader)
    }
                
}
</script>

2. change "process.php" to "process.aspx", they do same things

3. always get “We Got Response” message, even program name change to an not exist

With above code, "We Got Response" message will appear each time when response can be parsed as XML, so if 404 page in your case returns valid xHTML content - it will be parsed as XML. 
You can change your code as 


function outputResponse(loader){
  if(loader.xmlDoc.responseXML!=null && loader.xmlDoc.doXPath("/top_tag")) 
     alert("We Got Response\n\n"+loader.doSerialization()) 
else
     alert("Response contains no XML") 
}

where top_tag - name of top tag in XML