Question

Ok folks,

basically I need to change the value of the input (which is the upload button in this case) clicked on after the upload begins and finishes etc, which you can see in my code. The problem is there could be a number of these fields ( makes sense in the context of what its for ) created on the page, so I have to use $(this) so only the element clicked (#up for instance) get the call back changes. I cant seem to figure this out at all.

Would greatly appreciate the help.

Im using this plugin, http://valums.com/ajax-upload/, to do this and heres my code :

new AjaxUpload('#up', {
    name:'uploadedfile', 
    action: 'image-upload.php', 
    onSubmit: function(file, extension) {
        $(this).val("uploading...");    
    },

    onComplete: function(file, response){

        $(this).val(response).removeClass('unactive');

        }


    });

Thanks!

Answer 1

A work around:

var up = $("#up");
new AjaxUpload(up, {
   ...
   onSubmit: function(file, ext) {
       up.val("uploading...");
   }, 
   ...
});

Using how scoping and closures work you can just look up the chain for up. This also caches your selector.

This kind of thing is common practice with that = this or self = this