Question

I have an image pop-up ability on my website, in order to show users the full resolution picture when they click on a smaller version on the page.

This is the current CSS that positions it:

div#enlargedImgWrapper {
    position: absolute;
    top: 30px;
    left: 55px;
    z-index: 999;
}

The problem now is that if I click on an image further down the page, the window still appears in the top left corner of the page, where I can’t see it until I scroll back up. I need it to appear relative to the window, whatever its current position relative to the document is.

Note: I don’t want to use position: fixed; as some images might be taller than the screen, so I want users to be able to scroll along the image as well.

My idea was to use JS to change the top value:

var scrollValue = ???;
document.getElementById('enlargedImgWrapper').style.top = scrollValue+30 + 'px';

How can I detect by how much the user has scrolled down the page (var scrollValue)?
Or is there a ‘better’ way to do this?

Edit: if possible I would like to do this without jQuery.

Answer 1

Pure JavaScript uses scrollTop and scrollLeft:

var scrollLeft = (window.pageXOffset !== undefined) ? window.pageXOffset : (document.documentElement || document.body.parentNode || document.body).scrollLeft;
var scrollTop = (window.pageYOffset !== undefined) ? window.pageYOffset : (document.documentElement || document.body.parentNode || document.body).scrollTop;

https://developer.mozilla.org/en-US/docs/Web/API/Element.scrollTop

jQuery version:

var scrollLeft = $(window).scrollLeft() ;
var scrollTop = $(window).scrollTop() ;

What you need is this:

document.getElementById('enlargedImgWrapper').style.top = (scrollTop+30) + 'px';