Jquery Ajax Loading image

Description

You should do this using jQuery.ajaxStart and jQuery.ajaxStop.

1. Create a div with your image
2. Make it visible in jQuery.ajaxStart
3. Hide it in jQuery.ajaxStop

Sample

<div id="loading" style="display:none">Your Image</div>

<script src="../../Scripts/jquery-1.5.1.min.js" type="text/javascript"></script>
<script>
    $(function () {
        var loading = $("#loading");
        $(document).ajaxStart(function () {
            loading.show();
        });

        $(document).ajaxStop(function () {
            loading.hide();
        });

        $("#startAjaxRequest").click(function () {
            $.ajax({
                url: "http://www.google.com",
                // ... 
            });
        });
    });
</script>

<button id="startAjaxRequest">Start</button>

More Information

• jQuery.ajaxStart()
• jQuery.ajaxStop()

Try something like this:

<div id="LoadingImage" style="display: none">
  <img src="" />
</div>

<script>
  function ajaxCall(){
    $("#LoadingImage").show();
      $.ajax({ 
        type: "GET", 
        url: surl, 
        dataType: "jsonp", 
        cache : false, 
        jsonp : "onJSONPLoad", 
        jsonpCallback: "newarticlescallback", 
        crossDomain: "true", 
        success: function(response) { 
          $("#LoadingImage").hide();
          alert("Success"); 
        }, 
        error: function (xhr, status) {  
          $("#LoadingImage").hide();
          alert('Unknown error ' + status); 
        }    
      });  
    }
</script>

Its a bit late but if you don't want to use a div specifically, I usually do it like this...

var ajax_image = "<img src='/images/Loading.gif' alt='Loading...' />";
$('#ReplaceDiv').html(ajax_image);

ReplaceDiv is the div that the Ajax inserts too. So when it arrives, the image is replaced.