Hello, I need help with a an image pop feature I need on my website. I dont know js but I was able to find the following script online. The only problem is that instead of popping the image on or near the original thumbnail it does it towards the top of the page.

Code:
function get(eid) {
var d = document;
var r = d.getElementById(eid);
return r;
}

function popImg(open, iref) {
if (open) {
// top for popup image 10 pixels
// below corresponding thumb
var top = (50 +
iref.offsetHeight +
iref.offsetTop) +
'px';
// left for popup image is aligned
// with left of thumbnail
var left = iref.offsetLeft + 'px';
// use same source file for popup
// as thumbnail
var img = '<img src="' +
iref.src + '" />';
var d = document;
// if popup hasn't yet been added,
// create and append to body
if (null == get('popImg')) {
var pop = d.createElement('DIV');
pop.id = 'popImg';
pop.style.position = 'absolute';
d.body.appendChild(pop);
}
// get reference to popup image
// container div
var pop = get('popImg');
// set image element into div
pop.innerHTML = img;
// position relative to thumbnail
pop.style.top = top;
pop.style.left = left;
// show the div and its image
pop.style.display = 'block';
}
else {
// since request was for close,
// (open==false), hide the div -
// don't destroy it, since it can
// be recycled cheaper
var pop = get('popImg');
pop.style.display = 'none';
}
}
Code:
<img src="exampleimg.jpg" alt="test" 
width="95"
height="63"
hspace="5px"
vspace="5px"
align="right"
onmouseover="popImg(true, this);"
onmouseout="popImg(false);" />
The last 2 lines are what the person said to put in the img tag.

So if anyone could solve my problem in any way I would greatly appreciate it!!

thanks