sorry sir i will attach my php and ajax file here ...
my html file
<html>
<head>
<script type="text/javascript">
function showHint(str)
{
var printWin = window.open("","printSpecial");
printWin.document.open();
printWin.document.write("inside the script");
printWin.document.close();
if (str.length==0)
{
document.getElementById("txtHint").innerHTML="";
return;
}
if (window.XMLHttpRequest)
{// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp=new XMLHttpRequest();
}
else
{// code for IE6, IE5
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange=function()
{
if (xmlhttp.readyState==4 && xmlhttp.status==200)
{
document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","http://localhost/ajax/gethint.php?q="+str,true);
xmlhttp.send();
var printWin = window.open("","printSpecial");
printWin.document.open();
printWin.document.write("req send to php");
printWin.document.close();
}
</script>
</head>
<body>
<p><b>Start typing a name in the input field below:</b></p>
<form>
First name: <input type="text" onkeyup="showHint(this.value)" size="20" />
</form>
<p>Suggestions: <span id="txtHint"></span></p>
</body>
</html>
my php file is :
<?php
// Fill up array with names
$a[]="Anna";
$a[]="Brittany";
$a[]="Cinderella";
$a[]="Diana";
$a[]="Eva";
$a[]="Fiona";
$a[]="Gunda";
$a[]="Hege";
$a[]="Inga";
$a[]="Johanna";
$a[]="Kitty";
$a[]="Linda";
$a[]="Nina";
$a[]="Ophelia";
$a[]="Petunia";
$a[]="Amanda";
$a[]="Raquel";
$a[]="Cindy";
$a[]="Doris";
$a[]="Eve";
$a[]="Evita";
$a[]="Sunniva";
$a[]="Tove";
$a[]="Unni";
$a[]="Violet";
$a[]="Liza";
$a[]="Elizabeth";
$a[]="Ellen";
$a[]="Wenche";
$a[]="Vicky";
//get the q parameter from URL
$q=$_GET["q"];
$hint="";
//lookup all hints from array if length of q>0
if (strlen($q) > 0)
{
$hint="";
for($i=0; $i<count($a); $i++)
{
if (strtolower($q)==strtolower(substr($a[$i],0,strlen($q))))
{
if ($hint=="")
{
$hint=$a[$i];
}
else
{
$hint=$hint." , ".$a[$i];
}
}
}
}
// Set output to "no suggestion" if no hint were found
// or to the correct values
if ($hint == "")
{
$response="no suggestion";
}
else
{
$response=$hint;
}
//output the response
echo $response;
?>
but when i type names i cannot see the suggestion .. can u find whats wrong
in this ?
On Fri, Mar 23, 2012 at 6:00 PM, Jeremy Geerdes <jrgeer...@gmail.com> wrote:
> It's tough to say exactly what's going on without being able to see the
> page that's causing trouble. However, I would note that using AJAX does not
> change the actual HTML document. The changes are done dynamically in the
> browser. So you should be able to see the effect in the browser itself, or
> in the browser's developer tools (e.g., Firebug), but you won't see it when
> you view source.
>
> Jeremy R. Geerdes
> Generally Cool Guy
> Des Moines, IA
>
> For more information or a project quote:
> jrgeer...@gmail.com
>
> If you're in the Des Moines, IA, area, check out Debra Heights Wesleyan
> Church!
>
> On Mar 23, 2012, at 7:01 AM, naveen raj wrote:
>
> > i am using ajax + php in xampp but i cannot see any change in my html
> > document even when the event is made ( option selected ) .. can any
> > one help me
> >
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--
naveen raj , ceg
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