jQuery fadeOut（）之后不要重新定位其余元素

Question

The following is a simplified example of a slightly less simple game I am working on as practice:

The HTML:

<div>
    <img class="small" src="http://vignette4.wikia.nocookie.net/rickandmorty/images/d/dd/Rick.png/revision/latest?cb=20131230003659">
    <img class="small" src="https://ih0.redbubble.net/image.111135319.1949/flat,800x800,075,f.u1.jpg">
    <img class="small" src="http://67.media.tumblr.com/c38fff03b8dd7aaf75037eb18619da57/tumblr_n436i3Falo1sndv3bo1_1280.png">
</div>

The Javascript:

$(".small").click(function() {
    $(this).fadeOut();
});

And here it is in a JSFiddle: https://jsfiddle.net/rzu1yvpa

My Problem:

When I click on one of the images, it fades out as expected, but depending on which image disappears, the remaining images are re-positioned to fill the empty space. How can I make the remaining images keep their positions after other images have faded out? I have been searching and experimenting with this for a while now, but I am new to all things web, and am not sure what I should be looking for.

I have tried changing the position and display properties, but nothing has worked so far. It looks like fadeOut() sets the display property of the faded out element to none - maybe this is why the remaining images are ignoring them and filling the space? I have also tried putting the images in a bootstrap grid, but haven't been able to get that to work either, and would prefer a non-bootstrap solution.

Answer 1

jQuery's .fadeOut() and .fadeIn() methods animates the opacity of the matched elements, and once the opacity reaches 0, the display style property is set to none , so the element no longer affects the layout of the page.

This is what is causing the issue, when the display is set to none , the image is removed from the "flow".

You could use jQuery's fadeTo instead, which just fades to a degree of opacity, and does not set the display.

$(".small").click(function() {
    $(this).fadeTo(600,0);
});

FIDDLE

Answer 2

you have atleast two options:

1. Just change opacity of clicked image with .fadeTo() ( https://api.jquery.com/fadeTo/ ) that is just change of opacity (css - opacity to 0 with animation)

2. or harder way if you want to remove image itself you can swap it with div container for example with same width, heigth etc. using for example jq function .replaceWith() ( http://api.jquery.com/replacewith/ )

Greetings

Answer 3

The answer of Adeneo explains perfectly what's happening. To further optimise your code I'd suggest to use JS and a bit CSS instead:

Javascript

$('.small').on('click',function(event){
  $(this).addClass('small--animation');
});

CSS

.small {
  opacity: 1;
  transition: opacity .2s ease-in-out;
}
.small--animation {
  opacity: 0;
}