jQuery: Remove each image after it loads

After the image loads, I want to remove it.  This does NOT work:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Demo</title>
<style type="text/css">
img {
width: 100px;
}
</style>
<script type="text/javascript" src="https://code.jquery.com/jquery-3.1.0.js"></script>
<script type="text/javascript">

$(document).ready(function() {

    var imgI=0;
    $('.showImage img').each(function() {
        var tmpImg = new Image() ;
        tmpImg.onload = function(){

          $(this).remove();

         if( imgI++ === ($('.showImage img').length-1) ) {
          alert('all images loaded!');
         }
        }
        tmpImg.src = $(this).attr('src');
    }) ;
}) ;

</script>

</head>
<body>

 <div class="showImage"><img src="http://www.space.com/images/i/000/056/377/original/fullmoonmultiple.JPG?interpolation=lanczos-none&fit=inside%7C660:*?z2z2322z" alt=""/></div>
 <div class="showImage"><img src="https://upload.wikimedia.org/wikipedia/commons/e/e1/FullMoon2010.jpg?z3z32z22" alt=""/></div>
 <div class="showImage"><img src="https://upload.wikimedia.org/wikipedia/commons/9/9a/Howling_at_the_Moon_in_Mississauga.jpg?3z23z2z" alt=""/></div>

</body>
</html>

Open in new window

LVL 10
skijAsked:
Who is Participating?

[Product update] Infrastructure Analysis Tool is now available with Business Accounts.Learn More

x
I wear a lot of hats...

"The solutions and answers provided on Experts Exchange have been extremely helpful to me over the last few years. I wear a lot of hats - Developer, Database Administrator, Help Desk, etc., so I know a lot of things but not a lot about one thing. Experts Exchange gives me answers from people who do know a lot about one thing, in a easy to use platform." -Todd S.

Marco GasiFreelancerCommented:
Why not doing just this?
$(document).ready(function() {
    $('.showImage img').each(function() {
        var tmpImg = new Image() ;
        tmpImg.src = $(this).attr('src');
        $(this).remove();
    }) ;
}) ;

Open in new window

0
Marco GasiFreelancerCommented:
Or this?
$(document).ready(function() {

    var imgI=0;
    $('.showImage img').each(function() {
      var $this = $(this);
        var tmpImg = new Image() ;
        tmpImg.onload = function(){
         if( imgI++ === ($('.showImage img').length-1) ) {
          alert('all images loaded!');
         }
         $this.remove();
        };
        tmpImg.src = $(this).attr('src');
    }) ;
}) ;

Open in new window

See it here: http://test.webintenerife.com/hideimages.html
0
Julian HansenCommented:
You can do it like this
  $('.showImage img').on('load', function() {
    $(this).remove();
  });

Open in new window

0

Experts Exchange Solution brought to you by

Your issues matter to us.

Facing a tech roadblock? Get the help and guidance you need from experienced professionals who care. Ask your question anytime, anywhere, with no hassle.

Start your 7-day free trial
Learn SQL Server Core 2016

This course will introduce you to SQL Server Core 2016, as well as teach you about SSMS, data tools, installation, server configuration, using Management Studio, and writing and executing queries.

Julian HansenCommented:
Not sure what you are doing with

tmpImg.src = $(this).attr('src');

Open in new window


Where are you using tmpImg ? Only the last image loaded will actually have a value but you won't be able to access it because tmpImg is declared locally within the event handler?
0
leakim971PluritechnicianCommented:
Test page : https://jsfiddle.net/fmbLvx0x/

<script type="text/javascript">
$(window).load(function() {
    alert('all images loaded!');
    $('.showImage img').remove();
});
</script>

Open in new window

0
Slick812Commented:
greetings  skij , , ,  You show JS code that is suppose to eliminate Images <img> as soon as they load into the page, , I do not see this as an effective use of the page structure? ? If you do not want or need these images then do not place the all of <img> into the page. You are trying to get all of the <img> web locations into a newly created "tmpImg"

But I do NOT understand what your code efforts are trying to accomplish? ? ? You seem to need the images (as new images with the identical src ) stored to be used later for something else?

Can you tell us what you are trying to do? And what problem you are trying to FIX with the code block you show, and what result you need (what the new images) to happen, to get the page as you need it? ?
0
It's more than this solution.Get answers and train to solve all your tech problems - anytime, anywhere.Try it for free Edge Out The Competitionfor your dream job with proven skills and certifications.Get started today Stand Outas the employee with proven skills.Start learning today for free Move Your Career Forwardwith certification training in the latest technologies.Start your trial today
jQuery

From novice to tech pro — start learning today.