Retrieve the position (X,Y) of an HTML element relative to the browser window

Question

I want to know how to get the X and Y position of HTML elements such as img and div in JavaScript relative to the browser window.

Answer 1

Difference between small and little

function getPosition( el ) {
    var x = 0;
    var y = 0;
    while( el && !isNaN( el.offsetLeft ) && !isNaN( el.offsetTop ) ) {
    x += el.offsetLeft - el.scrollLeft;
    y += el.offsetTop - el.scrollTop;
    el = el.offsetParent;
    }
    return { top: y, left: x };
}

Look a example coordinates: http://javascript.info/tutorial/coordinates

Answer 2

Just thought I'd throw this out there as well.
I haven't been able to test it in older browsers, but it works in the latest of the top 3. :)

Element.prototype.getOffsetTop = function() {
    return ( this.parentElement )? this.offsetTop + this.parentElement.getOffsetTop(): this.offsetTop;
};
Element.prototype.getOffsetLeft = function() {
    return ( this.parentElement )? this.offsetLeft + this.parentElement.getOffsetLeft(): this.offsetLeft;
};
Element.prototype.getOffset = function() {
    return {'left':this.getOffsetLeft(),'top':this.getOffsetTop()};
};

Answer 3

To retrieve the position relative to the page efficiently, and without using a recursive function: (includes IE also)

var element = document.getElementById('elementId'); //replace elementId with your element's Id.
var rect = element.getBoundingClientRect();
var elementLeft,elementTop; //x and y
var scrollTop = document.documentElement.scrollTop?
                document.documentElement.scrollTop:document.body.scrollTop;
var scrollLeft = document.documentElement.scrollLeft?                   
                 document.documentElement.scrollLeft:document.body.scrollLeft;
elementTop = rect.top+scrollTop;
elementLeft = rect.left+scrollLeft;