Proper way to reset a GIF animation with display:none on Chrome

Question

Title is self-explanatory, but I\'ll provide a step-by-step view on the matter. Hopefully I\'m not the first one to have noticed this (apparently) bug on Webkit/Chrome.

Answer 1

Chrome deals with style changes differently than other browsers.

In Chrome, when you call .show() with no argument, the element is not actually shown immediately right where you call it. Instead, Chrome queues the application of the new style for execution after evaluating the current chunk of JavaScript; whereas other browsers would apply the new style change immediately. .attr(), however, does not get queued. So you are effectively trying to set the src when the element is still not visible according to Chrome, and Chrome won't do anything about it when the original src and new src are the same.

Instead, what you need to do is to make sure jQuery sets the src after display:block is applied. You can make use of setTimeout to achieve this effect:

var src = 'http://i.imgur.com/JfkmXjG.gif';
$(document).ready(function(){
    var $img = $('img');
    $('#target').toggle(
        function(){
            var timeout = 0; // no delay
            $img.show();
            setTimeout(function() {
                $img.attr('src', src);
            }, timeout);
        },
        function(){
            $img.hide();
        }
    );
});

This ensures that src is set after display:block has been applied to the element.

The reason this works is because setTimeout queues the function for execution later (however long later is), so the function is no longer considered to be part of the current "chunk" of JavaScript, and it provides a gap for Chrome to render and apply the display:block first, thus making the element visible before its src attribute is set.

Demo: http://jsfiddle.net/F8Q44/19/

Thanks to shoky in #jquery of freenode IRC for providing a simpler answer.

---

Alternatively, you can force a redraw to flush the batched style changes. This can be done, for example, by accessing the element's offsetHeight property:

$('img').show().each(function() {
    this.offsetHeight;
}).prop('src', 'image src');

Demo: http://jsfiddle.net/F8Q44/266/