Handling exceptions in XMLHttpRequest.open() method

Here is a snippet of code I am working with:

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

While this code functions properly by loading a default xml file if the specified file is not found, a 404 error appears only in the console. The error does not display anywhere else on the page.

My inquiry pertains to how I can handle this exception and whether it is necessary to include additional code to verify the existence of the file, especially when the current code works without it.

Answer №1

To retrieve the HTTP response code, use xhttp.status; a 200 (OK) or 304 (Not Modified) typically indicate 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);
    }
}

Ensure that you properly declare your variables using var to avoid implicit globals, which can lead to issues.

It's highly recommended to make asynchronous XHR requests instead of synchronous ones to prevent the browser from locking up during the request.

As for the second part of your query, there should be no issue as long as your application can handle the exception, which it appears to be doing.

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