How to get an element's top position relative to the browser's viewport?

Question

The existing answers are now outdated. The native getBoundingClientRect() method has been around for quite a while now, and does exactly what the question asks for. Plus it is supported across all browsers (including IE 5, it seems!)

From MDN page:

The returned value is a TextRectangle object, which contains read-only left, top, right and bottom properties describing the border-box, in pixels, with the top-left relative to the top-left of the viewport.

You use it like so:

var viewportOffset = el.getBoundingClientRect();
// these are relative to the viewport, i.e. the window
var top = viewportOffset.top;
var left = viewportOffset.left;

On my case, just to be safe regarding scrolling, I added the window.scroll to the equation:

var element = document.getElementById('myElement');
var topPos = element.getBoundingClientRect().top + window.scrollY;
var leftPos = element.getBoundingClientRect().left + window.scrollX;

That allows me to get the real relative position of element on document, even if it has been scrolled.

var element =  document.querySelector('selector');
var bodyRect = document.body.getBoundingClientRect(),
    elemRect = element.getBoundingClientRect(),
    offset   = elemRect.top - bodyRect.top;

Edit: Add some code to account for the page scrolling.

function findPos(id) {
    var node = document.getElementById(id);     
    var curtop = 0;
    var curtopscroll = 0;
    if (node.offsetParent) {
        do {
            curtop += node.offsetTop;
            curtopscroll += node.offsetParent ? node.offsetParent.scrollTop : 0;
        } while (node = node.offsetParent);

        alert(curtop - curtopscroll);
    }
}

The id argument is the id of the element whose offset you want. Adapted from a quirksmode post.

jQuery implements this quite elegantly. If you look at the source for jQuery's offset, you'll find this is basically how it's implemented:

var rect = elem.getBoundingClientRect();
var win = elem.ownerDocument.defaultView;

return {
    top: rect.top + win.pageYOffset,
    left: rect.left + win.pageXOffset
};