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Editorial Team
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Asked: 2026-05-29T10:03:47+00:00

It’s supposed to display 172-ajax-include.php when the user clicks the submit button. But it

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It’s supposed to display 172-ajax-include.php when the user clicks the submit button. But it doesn’t work. I think I did it right. Can’t figure out what I did wrong.

<html>
<head>
    <script type="text/javascript">
    function load()
    { /* start function load */
        if (window.XMLHttpRequest)
        { /* 1 if start */

       xmlhttp = new XMLHttpRequest();

        } /* 1 if end */
        else
        {  /* 1 else start */  
         xmlhttp = new ActiveXObject('Microsoft.XMLHTTP');   
        } /* 1 else end */

        xmlhttp.onreadystatechange = function() {
        if (xmlhttp.readyState == 4 && xmlhttp.status == 200 /* the 200  makes sure that it's not empty*/)
        { /* 2 if start */
            document.getElementById('adiv'),innerHTML =  xmlhttp.responseText;
        } /* 2 if end */

        xmlhttp.open('GET', '172-ajax-include.php', true);
        xmlhttp.send();
        }



    } /* end function load */
    </script>

</head>
<body>

    <input type="submit" onclick="load();">
    <div id="adiv"></div>
</body>
</html>
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1 Answer

  1. Editorial Team

    open() and send() must be placed outside the onreadystatechange handler.

     xmlhttp.onreadystatechange = function() 
    {
      if (xmlhttp.readyState == 4 && xmlhttp.status == 200)
        { /* 2 if start */
            document.getElementById('adiv'),innerHTML =  xmlhttp.responseText;
        } /* 2 if end */
     };
   xmlhttp.open('GET', '172-ajax-include.php', true);
   xmlhttp.send();

    EDIT:

    Change html button

    <input type"button" onclick="load()" value="Show"/>
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