First off, this group is really not geared toward general AJAX help. That said,
it would also be much more effective for you to provide a link to your page,
working or not. If we can see the live page, even if it's not working, we can
utilize a number of debugging tools against your code, rather than relying on
our eyes to find errors.
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 9:18 AM, Naveen Raj wrote:
> 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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