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Home/ Questions/Q 7947215
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Editorial Team
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Editorial Team
Asked: June 4, 20262026-06-04T01:22:07+00:00 2026-06-04T01:22:07+00:00

I have the following piece of code (only relevant parts): xhttp=new XMLHttpRequest(); xhttp.open(GET,doc_name,false); xhttp.send();

  • 0

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?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-04T01:22:08+00:00Added an answer on June 4, 2026 at 1:22 am

    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)

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