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Home/ Questions/Q 3280322
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T19:41:28+00:00 2026-05-17T19:41:28+00:00

I have some code that if I’m at home; it will display at popup

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I have some code that if I’m at home; it will display at popup saying that if you are at home. If you are not at home it will display a popup saying you are not at home. As the script is located on a server to give an IP address, how can I get it to say if you are not connected to the internet, as it will fail to load so I would have though that ip!=”” would work but doesn’t what have I done wrong?
Thanks

<script type="application/javascript">
    function getip(json)
    {
        ip=json.ip
        if(ip!="")
        {
            if(ip=="123.123.123.123")
            {
                alert("You are at home");
            }
            else
            {
                alert("you are not at home");
            }
        }
        else
        {
            alert("You are not connected to the internet");
        }
    }
</script>
<script type="application/javascript" src="http://jsonip.appspot.com/?callback=getip">
</script>
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  1. Editorial Team
    Editorial Team
    2026-05-17T19:41:29+00:00Added an answer on May 17, 2026 at 7:41 pm

    The way JSONP works is to call that method with the response, your server response from http://jsonip.appspot.com/?callback=getip looks something like this:

    getip({ "ip": "123.123.123.123", "addres": "123.123.123.123" });
    

    But when you’re not connected that script include just errors, and the method isn’t ever called, that’s the issue. Your if check, and the entire getip() method, isn’t ever hit. You could do a very simple timeout, like this:

    var timer = setTimeout(function() { 
      alert("You are not connected to the internet");
    }, 10000); //10 seconds
    function getip(json){
      clearTimeout(timer);
      if(json.ip == "123.123.123.123") {
        alert("You are at home");
      } else {
        alert("you are not at home");
      }
    }
    
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