javascript - XMLHttpRequest.status always returning 0 - Stack Overflow

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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', '', 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.

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.

• javascript
• firefox
• xmlhttprequest

Share Improve this question Follow asked May 2, 2010 at 6:31 PabloPablo 29.5k3737 gold badges134134 silver badges226226 bronze badges

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2 Answers 2

Sorted by: Reset to default 4

You can overe 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.

Is your site part of http://www.nasa.gov/? Otherwise, XMLHttpRequest will fail due to Same Origin Policy.

Also, if the page is served as a non-HTTP request, the status can be 0. See https://developer.mozilla/En/Using_XMLHttpRequest#section_3.

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