Image height always returned as 0

Question

I'm trying to follow this SO answer on how to tint an image, but when i do it, $("#myselector").height(); keeps been returned as 0. What am I doing wrong? Heres my code:

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="utf-8">
    <title>Test</title>
    <link rel="stylesheet" href="style.css">
    <!--[if IE]>
        <script src="http://html5shiv.googlecode.com/svn/trunk/html5.js"></script>
    <![endif]-->
</head>
<script src="http://codeorigin.jquery.com/jquery-1.10.2.min.js"></script>

<script>
$(document).ready(function() {
  overlay = $("#overlay");
img = $("#myimg");
overlay.width($("#myimg").width());
alert($("#myimg").height());
overlay.height("100");
overlay.css("top", img.offset().top + "px");
overlay.css("left", img.offset().left + "px");

});


</script>

<body id="home">

<div id="overlay" class="overlay"></div>
<img id="myimg" src="building.png" />

</body>
</html>

on your source , he use overlay.height(img.css("height")); – Janith Chinthana Aug 12 '13 at 12:19 — Janith Chinthana, Aug 12 '13 at 12:19

score 5 · Answer 1 · edited May 23 '17 at 12:22

5

My best guess is that the image isn't loaded yet.

Your script runs onready. If you change it to onload you'll probably get a better result.

Change

$(document).ready(function() {

to

$(window).load(function() {

More: window.onload vs $(document).ready()

edited May 23 '17 at 12:22

Community

1
1

answered Aug 12 '13 at 12:16

Halcyon

57,230
10
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score 4 · Accepted Answer · answered Aug 12 '13 at 12:15

4

Use image' load event:

$(document).ready(function() {
  overlay = $("#overlay");
  img = $("#myimg");
  img.load( function(){
    overlay.width($("#myimg").width());
    alert($("#myimg").height());
    overlay.height("100");
    overlay.css("top", img.offset().top + "px");
    overlay.css("left", img.offset().left + "px");
  });
});

answered Aug 12 '13 at 12:15

Ivan Chernykh

41,617
13
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You are a beautiful person. Thanks! – MissCoder87 Aug 12 '13 at 12:16

score 1 · Answer 3 · answered Aug 12 '13 at 12:16

1

Replace:

$(document).ready(function() {

on:

$(window).load(function() {

answered Aug 12 '13 at 12:16

Michael

15,386
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In an answer like this explaining what effect the change has is important. Why do you think `$(window).load` is superior to `$(document).ready` in this case? – Jason Aller May 01 '14 at 03:06

Image height always returned as 0

3 Answers3