Question

I use lazyload() on my ecommerce website. lazyload() works great. I use this code to do that:

$(function(){
  $("img.lazy").lazyload({ 
  effect : "fadeIn"
          });
});

Also there are some filter like color, price etc. which is run an ajax and show new result. When new result come, lazyload() doesn’t work. I know that I have to use lazyload() with live(), delegate() or on(). But I am a newbie on jquery and I couldn’t do that. Can anyone help me please? My jquery version is 1.7.2 and we can use on().

Answer 1

NOTE: At the time this answer was accepted, this worked. The lazy loading plugin changed and now it is broken. I cannot unaccept the answer nor can I delete it.

There is another Q&A that suggests handling it as part of the AJAX request responsible for inserting the additional DOM:

Binding image lazy loading to new images inserted after ajax request

However, this is how I would do it:

$(document).on('DOMNodeInserted', 'img.lazy', function() {
    $(this).lazyload({
        effect: 'fadeIn'
    });
});

Demo: jsfiddle