slideshow javascript - fade not working in chrome

the following slideshow works in all browsers as far as switching images,
but it is supposed to have a fade from image to image.

it works in IE8, but not Chrome, Safari, Opera, FireFox

anyone know how to fix that?

thanks

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">

<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>slideshow</title>

<SCRIPT LANGUAGE="JavaScript">
// Set slideShowSpeed (milliseconds)
var slideShowSpeed = 7000;
// Duration of crossfade (seconds)
var crossFadeDuration = 5000;
// Specify the image files
var Pic = new Array();
// to add more images, just continue
// the pattern, adding to the array below

Pic[0] = 'gallery/test1.jpeg'
Pic[1] = 'gallery/test2.jpeg'
Pic[2] = 'gallery/test3.jpeg'
Pic[3] = 'gallery/test4.jpeg'

// do not edit anything below this line
var t;
var j = 0;
var p = Pic.length;
var preLoad = new Array();
for (i = 0; i < p; i++) {
preLoad[i] = new Image();
preLoad[i].src = Pic[i];
}
function runSlideShow() {
if (document.all) {
document.images.SlideShow.style.filter="blendTrans(duration=100)";
document.images.SlideShow.style.filter="blendTrans(duration=crossFadeDuration)";
document.images.SlideShow.filters.blendTrans.Apply();
}
document.images.SlideShow.src = preLoad[j].src;
if (document.all) {
document.images.SlideShow.filters.blendTrans.Play();
}
j = j + 1;
if (j > (p - 1)) j = 0;
t = setTimeout('runSlideShow()', slideShowSpeed);
}
</script>

</head>

<body onLoad="runSlideShow()">

<img src="gallery/test1.jpeg" name='SlideShow'/>


</body>

</html>

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webdottAsked:
Who is Participating?
 
Dave BaldwinFixer of ProblemsCommented:
As far as I can tell, 'filter' is an IE only property.  Most slideshows use jQuery to create transitions in slideshows.  A lot of people use this one: http://www.dynamicdrive.com/dynamicindex14/fadeinslideshow.htm
0
 
webdottAuthor Commented:
thanks. that works for what i need.
0
 
Dave BaldwinFixer of ProblemsCommented:
You're welcome.
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