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Editorial Team
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Asked: 2026-06-17T16:33:46+00:00

This code wont execute. I think there is something wrong with the xhttp.send() function.

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This code wont execute. I think there is something wrong with the xhttp.send() function. Because the alert function before it executes but the alert function after it doesn’t execute:

<html>
<head>
    <title>PAGE OUTPUT</title>
    <script type="text/javascript">
    function aa()
    {   
        var xhttp;
        alert("hi welcome"); 
        if(window.XMLHttpRequest)
        {
            xhttp=new XMLHttpRequest();
            alert("hi");
        }
        else
        {
            xhttp=new ActiveXObject("Microsoft.XMLHTTP");
        }
        xhttp.open("GET","vehicle.xml",false);
        alert("OPEN EXECUTED");
        xhttp.send();
        alert("SEND EXECUTED");
        xmlDoc=xhttp.responseXML;
        alert("HI I HAVE REACHED OVER HERE");
        var vehicle=xmlDoc.documentElement;
        var car=vehicle.firstChild.nodeValue;
        var price=car.firstChild.nodeValue;
        alert(price);
    }
    </script> 
</head>
<body>
<center><input type="submit" onclick="aa()"></center>
</body>

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1 Answer

  1. Editorial Team

    You’d better to check if the URL is like file:///path/to/file the browser will fail to send any XHR from the page, for security reasons (nowadays that seems unreasonable).

    You should launch a server for this case, and visit that page in http://localhost/path/to/file. The easiest way I know is to launch a Python SimpleHTTPServer. You might install python yourself if you are using MS Windows, or it will be available in Mac/any Linux distributions. Then follow these steps

    • launch a terminal (on MS Windows it’s called a command line prompt, using Win+R then type cmd to launch one)
    • cd to the path where the files locate
    • execute python -m SimpleHTTPServer
    • visit http://localhost:8000/html_file in your browser
    • click on that button as usual

    Now you will see those alerts. Additionally, you might have to change xmlDoc=xhttp.responseXML to xmlDoc = xhttp.responseText if necessary.

    Then try jQuery, that’s much easier for ajax.

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