XMLHttpRequest.open() exception handling

Question

I have the following piece of code (only relevant parts):

xhttp=new XMLHttpRequest();
xhttp.open("GET",doc_name,false);
xhttp.send();
xmlDoc=xhttp.responseXML;
if(xmlDoc==null)
{
   xmlDoc=loadXMLDoc(defaultXml);
}

This runs fine as I load a default xml file if the specified file does not exist but shows a 404 error only in the console if the file does not exist. (This error does not reflect anywhere in the page except the console).

My question is that how should I check for this exception & is it necessary to add an extra piece of code for checking the existence of the file when the code runs without it?

Answer 1

You can access the HTTP response code via xhttp.status; either a 200 (OK) or 304 (Not Modified) would normally be considered a successful request.

xhttp=new XMLHttpRequest();
xhttp.open("GET",doc_name,false);
xhttp.send();

if (xhttp.status === 200 || xhttp.status === 304) {
    xmlDoc=xhttp.responseXML;
    if(xmlDoc==null)
    {
       xmlDoc=loadXMLDoc(defaultXml);
    }
}

Make sure you're declaring your variables first using var, otherwise you'll have implicit globals, which are bad.

Also make sure you have a good reason for doing this synchronously; synchronous XHR's lock up the browser whilst the request is pending. Making it async is highly recommended.

For the second part of your question, there's no problem what-so-ever; as long as your app can handle the exception. (which is seems to do)