Binding image lazy loading to new images inserted after ajax request

Question

I'm using Mika Tuupola's Lazy Load plugin http://www.appelsiini.net/projects/lazyload to delay loading images as you scroll down a long image gallery. The problem is after 10 images, I use infinite scrolling so I fetch the next 10 images, and append them via ajax. Lazy Loading no longer works on this next batch of appended images.

It's a pretty javascript-heavy image gallery, so for everything else (such as tooltips, modal overlays, etc) I've been using jQuery's delegate() to bind to ajax-inserted elements. The problem with the Lazy Load plugin is that I'm not sure what event to bind to.

So say I want to lazy load images with a class of "lazy". I would write this:

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

and it works for the first 10 images, but stops working after inserting more via ajax. The only thing I can think of is to use delegate on a load event, like so:

$(document).delegate("img.lazy", "load", function(event) {  
    $(this).lazyload({ 
         effect: "fadeIn" 
    });     
});

but that breaks everything. Thanks!

EDIT: The jQuery I use to load more records (this is a Rails app):

$(window).scroll(function() {
    var url;
    url = $(".pagination .next_page").attr("href");
    if (url && $(window).scrollTop() > $(document).height() - $(window).height() - 50) {
    $(".pagination").html("<p>loading more images...</p>");
    return $.getScript(url);
    }
});

$(window).scroll();

@rgin Again, with what event? `lazyload` isn't an event, and `.on` requires an event to bind to. — Kevin B, Apr 05 '12 at 21:23

score 18 · Accepted Answer · answered Apr 05 '12 at 21:16

18

I would use the ajaxStop method.

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

removeClass prevents double initialization.

answered Apr 05 '12 at 21:16

Kevin B

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Wow, thank you that worked like a charm. Now I just need to understand how exactly `ajaxStop()` solved the problem (I read this http://api.jquery.com/ajaxStop/ but still a bit confused). Is it performance-friendly? – DelPiero Apr 05 '12 at 23:20
3

It is performance friendly, however it would be even better if you instead placed it in the callback of the specific ajax request that is updating the page. – Kevin B Apr 06 '12 at 00:25
1

@DelPiero - extending Kevin B's comment. The deferreds object is your friend in this scenario. See: http://api.jquery.com/category/deferred-object/ – trickyzter Apr 06 '12 at 14:06

score 0 · Answer 2 · answered Oct 20 '14 at 20:55

In addition to Kevin B's answer if you want to avoid ajaxStop jQuery now defines a when() function for that purpose

// the following code will be executed when the ajax request resolve.
// the ajax function must return the value from calling the $.ajax() method.

    $.when(load_new_image(1), load_new_image(2), load_new_image(3)).done(function(i1, i2, i3){

            if($('img.lazy').length){

                 $('img.lazy').lazyload({ 

                          effect:'fadeIn'

                 }).removeClass('lazy').addClass('lazyloaded');

            }

        });

        function load_new_image(image_id) {

            return $.ajax({
                url: "someUrl",
                dataType: "json",
                data:  yourJsonData,            
                ...
            });
        }

from: http://api.jquery.com/jQuery.when/

Binding image lazy loading to new images inserted after ajax request

2 Answers2

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