[jQuery] Re: what's the best way to write this small function

[bingo]

hi Richard,

Thanks for your help..although I got it working but my approach
definetly seems to have bug...here is what is working for me

jQuery.fn.customLoad = function(url){
    $(this).ajaxStart(function(){
         $(this).html("<div align=\"center\"><img
src=\"/cake2/img/images/loadingAnimation.gif\" /></div>");
   });
   $(this).load(url);
   return;
}

your approach seems cleaner and without bug..I think I will go with
yours..

Thanks


On May 20, 9:50 pm, "Richard Worth" <[EMAIL PROTECTED]> wrote:
> On 5/20/07, bingo <[EMAIL PROTECTED]> wrote:
>
>
>
>
>
> > jQuery.fn.customLoad = function(url){
> >         return $(this).ajaxStart(function(){
> >                                 $(this).html("<div align=\"center\"><img
> > src=\"/cake2/img/images/
> > loadingAnimation.gif\" /></div>");
> >                                 $(this).load(url);
> >                         });
> > };
>
> > As you can see, I am just trying over write load function so that
> > whenever any div is being loaded, just attach a new div that display
> > loading animation.
>
> Try this:
>
> jQuery.fn.customLoad = function(url){
>     $(this).bind("ajaxStart", function(){
>         $(this).html("<div align=\"center\"><img
> src=\"/cake2/img/images/loadingAnimation.gif\" /></div>");
>     });
>     $(this).bind("ajaxStop", function(){
>         $(this).unbind("ajaxStart");
>     });
>     return $(this).load(url);
>
> };
>
> I found the unbind necessary to keep ajaxStart from being called a second
> time after the element is loaded. I think this may be a bug.
>
> - Richard D. Worth