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Use only javascript to cycle through images

Greetings,

I'm trying to get the following code to work, but I keep getting a ton of undefined errors when it comes to the array imageCount.

I'm trying to write this so it will be handle any number of images.

Please let me know where I'm going wrong and why.

<!DOCTYPE HTML>
<html>
<body onload="init()">

<div id="slideshow">
  <div class="image"><img src="myimage1.jpg"></div>
  <div class="image" style="display:none"><img src="src="myimage2.jpg"></div>
  <div class="image" style="display:none"><img src="src="myimage3.jpg"></div>
  <div class="image" style="display:none"><img src="src="myimage4.jpg"></div>
</div>

<script>
// all your Javascript code goes here //
function init() {
	setInterval("showImages()", 2000);
}
function showImages(){ 
var image = document.getElementsByTagName('img').length;
	var index = 0;
	var imageCount = new Array();
	for( var i = 0; i < image; i++){
		if(image != 0){
			imageCount.push(document.getElementsByTagName('img')[i]);
			imageCount[i].style.display = 'none';
			imageCount[index].style.display ='none';
			index = (index + 1)% imageCount[index].length;
			imageCount[i].style.display = 'block';	
		}
	}
}
</script>

</body>
</html>

Open in new window

Michel Plungjan

Not quite clear on what you want to do - show one image at a time?

This is for sure not correct
imageCount[index].length;

there is nothing with length in the array...

Here is the beginning of your script, somewhat optimised

function init() {
setInterval(showImages, 2000); // using quotes creates an implicit eval - not best practice
}

function showImages(){
var images = document.getElementsByTagName('img'); // or document.images
for (var i=0, n=images.length; i<n; i++) {

If the idea is to show one and hide the previous, then I would do - notice the usage of onload

<!DOCTYPE HTML>
<html>
<head>
<script>
var index=0,images; // global vars
window.onload=function() {
  images = document.getElementsByTagName('img'); // or document.images
  setInterval(showImages, 2000);
}
function showImages(){ 
   images[index].style.display='none'
   index++;
   if (index>=images.length) index=0;
   images[index].style.display = 'block';  
}
</script>
</head>
<body>

<div id="slideshow">
  <div class="image"><img src="myimage1.jpg"></div>
  <div class="image" style="display:none"><img src="src="myimage2.jpg"></div>
  <div class="image" style="display:none"><img src="src="myimage3.jpg"></div>
  <div class="image" style="display:none"><img src="src="myimage4.jpg"></div>
</div>



</body>
</html>

Open in new window

max7

ASKER

Not quite clear on what you want to do - show one image at a time?

Exactly what I want to do

This is for sure not correct
imageCount[index].length;

there is nothing with length in the array...

Correct, but I couldn't figure out why not since I thought the following code would put the img elements inside that array:

var imageCount = new Array();
	for( var i = 0; i < image; i++){
		if(image != 0){
			imageCount.push(document.getElementsByTagName('img')[i]);

Open in new window

Is the problem that I didn't correctly write my for loop? I ask because I see that you did:

 for (var i=0, n=images.length; i<n; i++)

Open in new window

at best, I think would have done i<images.length; but I don't think I ever would have thought of n here. I thought i would represent the image elements ... what am I missing?

Also, comparing your script to mine, is creating a new array and putting the image elements inside of it even necessary? I thought it was but it appears not.

I noticed too that in your script, you used global variables ... is that at least part of the reason I was getting undefined script errors?

If the idea is to show one and hide the previous, then I would do - notice the usage of onload

Two things:

a) unless I remove the surrounding div tags around the img tags, your script will only show the first image and none of the others. How can I keep those div tags and still have it work?

b) you pointed out onload ... is that a better choice than init and if so, why?

ASKER CERTIFIED SOLUTION

Michel Plungjan

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max7

ASKER

Thanks very much for your help, yet again. Good to know you're still on EE, have a wonderful day :)