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

Question

I want to get the position of an element relative to the browser's viewport (the viewport in which the page is displayed, not the whole page). How can this be done in JavaScript?

Many thanks

Answer 1

The existing answers are now outdated.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;

Answer 2

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.

Answer 3

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

Answer 4

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 .

Answer 5

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
};

Answer 6

function inViewport(element) {
    let bounds = element.getBoundingClientRect();
    let viewWidth = document.documentElement.clientWidth;
    let viewHeight = document.documentElement.clientHeight;

    if (bounds['left'] < 0) return false;
    if (bounds['top'] < 0) return false;
    if (bounds['right'] > viewWidth) return false;
    if (bounds['bottom'] > viewHeight) return false;

    return true;
}

source

Answer 7

The function on this page will return a rectangle with the top, left, height and width co ordinates of a passed element relative to the browser view port.

    localToGlobal: function( _el ) {
       var target = _el,
       target_width = target.offsetWidth,
       target_height = target.offsetHeight,
       target_left = target.offsetLeft,
       target_top = target.offsetTop,
       gleft = 0,
       gtop = 0,
       rect = {};

       var moonwalk = function( _parent ) {
        if (!!_parent) {
            gleft += _parent.offsetLeft;
            gtop += _parent.offsetTop;
            moonwalk( _parent.offsetParent );
        } else {
            return rect = {
            top: target.offsetTop + gtop,
            left: target.offsetLeft + gleft,
            bottom: (target.offsetTop + gtop) + target_height,
            right: (target.offsetLeft + gleft) + target_width
            };
        }
    };
        moonwalk( target.offsetParent );
        return rect;
}

Answer 8

You can try:

node.offsetTop - window.scrollY

It works on Opera with viewport meta tag defined.

Answer 9

Thanks for all the answers. It seems Prototype already has a function that does this (the page() function). By viewing the source code of the function, I found that it first calculates the element offset position relative to the page (ie the document top), then subtracts the scrollTop from that. See the source code of prototype for more details.

Answer 10

I am assuming an element having an id of btn1 exists in the web page, and also that jQuery is included. This has worked across all modern browsers of Chrome, FireFox, IE >=9 and Edge. jQuery is only being used to determine the position relative to document.

var screenRelativeTop =  $("#btn1").offset().top - (window.scrollY || 
                                            window.pageYOffset || document.body.scrollTop);

var screenRelativeLeft =  $("#btn1").offset().left - (window.scrollX ||
                                           window.pageXOffset || document.body.scrollLeft);

Answer 11

Sometimes getBoundingClientRect() object's property value shows 0 for IE. In that case you have to set display = 'block' for the element. You can use below code for all browser to get offset.

Extend jQuery functionality :

(function($) {
    jQuery.fn.weOffset = function () {
        var de = document.documentElement;
        $(this).css("display", "block");
        var box = $(this).get(0).getBoundingClientRect();
        var top = box.top + window.pageYOffset - de.clientTop;
        var left = box.left + window.pageXOffset - de.clientLeft;
        return { top: top, left: left };
    };
}(jQuery));

Use :

var elementOffset = $("#" + elementId).weOffset();

Answer 12

Based on Derek's answer.

/**
 * Gets element's x position relative to the visible viewport.
 */
function getClientX(el) {
  let offset = 0;
  let currentElement = el;

  while (currentElement !== null) {
    offset += currentElement.offsetLeft;
    offset -= currentElement.scrollLeft;
    currentElement = currentElement.offsetParent;
  }

  return offset;
}

/**
 * Gets element's y position relative to the visible viewport.
 */
function getClientY(el) {
  let offset = 0;
  let currentElement = el;

  while (currentElement !== null) {
    offset += currentElement.offsetTop;
    offset -= currentElement.scrollTop;
    currentElement = currentElement.offsetParent;
  }

  return offset;
}

Answer 13

here is something for Angular2 +. Tested on version 13

event.srcElement.getBoundingClientRect().top;