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Home/ Questions/Q 4052066
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Editorial Team
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Editorial Team
Asked: May 20, 20262026-05-20T14:16:24+00:00 2026-05-20T14:16:24+00:00

What’s the fastest way to check if my server is online via JavaScript? I’ve

  • 0

What’s the fastest way to check if my server is online via JavaScript?

I’ve tried the following AJAX:

function isonline() {
    var uri = 'MYURL'
    var xhr = new XMLHttpRequest();
    xhr.open("GET",uri,false);
    xhr.send(null);
    if(xhr.status == 200) {
        //is online
        return xhr.responseText;
    }
    else {
        //is offline
        return null;
    }   
}

The problem is, it never returns if the server is offline. How can I set a timeout so that if it isn’t returning after a certain amount of time, I can assume it is offline?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-20T14:16:25+00:00Added an answer on May 20, 2026 at 2:16 pm

    XMLHttpRequest does not work cross-domain. Instead, I’d load a tiny <img> that you expect to come back quickly and watch the onload event:

    function checkServerStatus()
    {
        setServerStatus("unknown");
        var img = document.body.appendChild(document.createElement("img"));
        img.onload = function()
        {
            setServerStatus("online");
        };
        img.onerror = function()
        {
            setServerStatus("offline");
        };
        img.src = "http://myserver.com/ping.gif";
    }
    

    Edit: Cleaning up my answer. An XMLHttpRequest solution is possible on the same domain, but if you just want to test to see if the server is online, the img load solution is simplest. There’s no need to mess with timeouts. If you want to make the code look like it’s synchronous, here’s some syntactic sugar for you:

    function ifServerOnline(ifOnline, ifOffline)
    {
        var img = document.body.appendChild(document.createElement("img"));
        img.onload = function()
        {
            ifOnline && ifOnline.constructor == Function && ifOnline();
        };
        img.onerror = function()
        {
            ifOffline && ifOffline.constructor == Function && ifOffline();
        };
        img.src = "http://myserver.com/ping.gif";        
    }
    
    ifServerOnline(function()
    {
        //  server online code here
    },
    function ()
    {
        //  server offline code here
    });
    
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