Default browser xml stylesheet when referenced xsl stylesheet is not found?

Question

I would like the raw XML to be displayed when my stylesheet is not found by the browser, is it possible?

Below I have put the header of my XML document.

<?xml version="1.0" encoding="UTF-8" standalone="no"?><?xml-stylesheet type="text/xsl" href="./my_sheet.xsl" title="relativeRef" alternate="YES"?>

When the stylesheet is not found on my server I return a 404 error but this makes the XML parser of browser stop and displaying an error message. What should I do? Here is the JAVA code of the servlet which serves the stylesheet.

@Override
protected void doGet(HttpServletRequest req, HttpServletResponse resp)
        throws ServletException, IOException {      

    edamisUtils.log("Serving my_sheet.xsl");

    //If we could not load my_sheet we throw a 404 error.
    if(my_sheet== null){
        edamisUtils.debug(5, "Did not find the my_sheet.xsl, sending a 404 status response.");
        resp.sendError(HttpServletResponse.SC_NOT_FOUND);
        return;
    }

    //Set the MIME type of the response to text/xsl
    resp.setContentType("text/xsl");
    OutputStream os = resp.getOutputStream();
    os.write(my_sheet.array());

}

Thanks all and have a nice day!

Answer 1

Instead of returning a 404, you could return a 200 and a "copy-all" XSL.

Something like

if(my_sheet == null){
   copy the copy_all XSL in the response output stream;
}

A copy-all XSL example can be found here