AJAX open('GET',url,true) returns undefined value

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AJAX open('GET',url,true) returns undefined value

Nagarajan

I'm learning AJAX myself and my code is

<html>
<head>
<script>
function callAjax()
{
var xmlreq = new XMLHttpRequest(); // since i'm using Mozilla
var url = "http://localhost/naga/get_details.php?name=naga";
var ope;
ope = xmlreq.open('GET',url,true);
alert(ope); // here i'm getting as 'undefined'
xmlreq.onreadystatechange = myFunc();
function myFunc()
{
if (xmlreq.readyState==4)
{
document.getElementById('one').innerHTML=xmlreq.responseText;
}
}
xmlhttp.send();
}
</script>
</head>
<body>
<div id="one">Microsoft</div>
<input type="button" onclick="callAjax()" value="Change link">

</body>
</html>

Tcobb

So far as I know, the "open" method/function of an httpXMLrequest object has no return value. The question is whether it gets the value from the server for you.

rnd_me

Tcobb;1257633 wrote:
So far as I know, the "open" method/function of an httpXMLrequest object has no return value. The question is whether it gets the value from the server for you.

from spec:

 void open(ByteString method, DOMString url, optional boolean async = true, optional DOMString? user = null, optional DOMString? password = null);

so yeah, undefined is to be expected.

change xmlreq.onreadystatechange = myFunc(); to xmlreq.onreadystatechange = myFunc; if you want the code to work as expected.