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Editorial Team
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Asked: 2026-06-11T23:03:21+00:00

I know there are very tutorials out there and same questions as well, but

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I know there are very tutorials out there and same questions as well, but I have tried many times, and ajax didn’t work. plz correct my script:
here is index.php

<?php
echo'
<script type="text/javascript">
function ajax()
{
var xmlhttp;

xmlhttp=new XMLHttpRequest();
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("result").innerHTML=xmlhttp.responseText;
    }
  }
  xmlhttp.open("POST","ajax.php",true);
  xmlhttp.send();

    }
</script>

</head>

<body>
<p>&nbsp;</p>
<form id="form1" name="form1" method="post" action="" onsubmit="return ajax()">
  <p>
    <label for="num2">number 1</label>
    <input type="text" name="num1" id="num2" />
    *
  <label for="num3">number 2</label>
  <input type="text" name="num2" id="num3" />
    =
  <label for="result">Result</label>
  <input type="text" name="result" id="result" />
  </p>
  <p>
    <input type="submit"  value="Submit" />
  </p>
</form>';
?>

and here is ajax.php that is taking two variables and multiplying them and echoing the result, but my page refreses and didn’t see anything

    <?php
    $num1=$_POST["num1"];
    $num2=$_POST["num2"];

    $result=$num1*$num2;
    echo $result;
?>
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1 Answer

  1. Editorial Team

    This happens because function ajax() does not return anything so the form continues to submit.

    Add before function ends return false;

    Correct code:

    <html>
<head>

<script type="text/javascript">
function ajax(){
    var xmlhttp;
    xmlhttp=new XMLHttpRequest();
    xmlhttp.onreadystatechange=function(){
        if (xmlhttp.readyState==4 && xmlhttp.status==200){
            document.getElementById("result").value=xmlhttp.responseText;
        }
    }
    var params = "num1="+document.getElementById('num2').value+"&num2="+document.getElementById('num3').value;
    xmlhttp.open("POST","ajax.php",true);
    xmlhttp.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    xmlhttp.setRequestHeader("Content-length", params.length);
    xmlhttp.setRequestHeader("Connection", "close");
    xmlhttp.send(params);
    return false;
}
</script>

</head>

<body>
    <p>&nbsp;</p>
    <form id="form1" name="form1" method="post" action="" onsubmit="return ajax()">
        <p>
            <label for="num2">number 1</label>
            <input type="text" name="num1" id="num2" />
            *
            <label for="num3">number 2</label>
            <input type="text" name="num2" id="num3" />
            =
            <label for="result">Result</label>
            <input type="text" name="result" id="result" />
            </p>
            <p>
            <input type="submit"  value="Submit" />
        </p>
    </form>
</body>
</html>
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