Sorry for double posting!

 

 

Hi all,

 

Ok so I have a piee of Ajax in my site thats baffling me. Im totally new to Ajax so any help will be much appreciated!

 

 

 

1) I have a form that passes a variable to the ajax function and works perfectly (as below).

 

<form name='myForm'>

<input type="hidden" id='id' name='id' value='<? echo $id; ?>' />

<input type='button' onclick='ajaxFunction()' value='Query MySQL' />

</form>

 

 

 

2) The function passes the variable to the desired page (includes/process_ajax_gallery_request.php) and works perfectly.

 

 

3) In the file (includes/process_ajax_gallery_request.php) i have a php function that querys the database and outputs some images.

 

 

 

4) Heres the prob: the output happens and is sent to "ajaxDiv" (as you will see below) but I cannot see anything on screen! I checked the source code of the HTML page and i can see that its there (ie <img src=.......>) but I cant see anything on the screen.

 

I done a little research with no luck but is it something to do with the DOM or something? Totally new to this!

 

Heres the Ajax function that takes care of it all:

 

 

 

function ajaxFunction(){

var ajaxRequest;

 

try{

ajaxRequest = new XMLHttpRequest();

} catch (e){

try{

ajaxRequest = new ActiveXObject("Msxml2.XMLHTTP");

} catch (e) {

try{

ajaxRequest = new ActiveXObject("Microsoft.XMLHTTP");

} catch (e){

alert("Your browser broke!");

return false;

}

}

}

 

ajaxRequest.onreadystatechange = function(){

if(ajaxRequest.readyState == 4){

var ajaxDisplay = document.getElementById('ajaxdiv');

ajaxDisplay.innerHTML = ajaxRequest.responseText;

}

}

var id = document.getElementById('id').value;

var queryString = "?id=" + id;

ajaxRequest.open("GET", "includes/process_ajax_gallery_request.php" + queryString, true);

ajaxRequest.send(null);

}

 

 

 

Do I need to do something with this?

ajaxDisplay.innerHTML = ajaxRequest.responseText;

 

 

Any help is much appreciated.

 

Thank you in advance and for reading this far!

 

Wayne.

Hi!

 

@waynef

 

Instead of using innerHTML, try to create the element and add it as a child node within ajaxDisplay. Sometimes browser security can keep you from doing stuff, so try the same example by having the PHP script return a simple text string to make sure it works.