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Home/ Questions/Q 761309
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Editorial Team
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Editorial Team
Asked: May 14, 20262026-05-14T15:49:20+00:00 2026-05-14T15:49:20+00:00

html <a href=# onclick=MyObj.startup()>click me</a> js code var MyObj = { startup : function()

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html

<a href="#" onclick="MyObj.startup()">click me</a>

js code

var MyObj = 
{
    startup : function()
    {
        var ajax = null;
        ajax = new XMLHttpRequest();
        ajax.open('GET', 'http://www.nasa.gov', true);
        ajax.onreadystatechange = function(evt)
        {
            if(ajax.readyState == 4)
            {
                if (ajax.status == 200)
                {
                    window.dump(":)\n");
                }
                else
                {
                    window.dump(":(\n");
                }
            }
        }
        ajax.send(null);
    }
}

ajax.status always returning 0, no matter which site it is, no matter what is the actual return code. I say actual, because ajax.statusText returning correct value, eg OK or Redirecting…

ajax.readyState also returns proper values and 4 at the end.

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  1. Editorial Team
    Editorial Team
    2026-05-14T15:49:21+00:00Added an answer on May 14, 2026 at 3:49 pm

    You can overcome this easily in a local environment by setting up a php proxy (xampp a server and pass a querystring for the url you want to grab). Have your php proxy wget the url and echo its contents. That way your local html file (when viewed as http://localhost/your.html) can send ajax requests out of domain all day. Just don’t expect the content to work as though it were local to that domain.

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