Get the position of a div/span tag

JavascriptHtml

Javascript Problem Overview


Can someone show me how to get the top & left position of a div or span element when one is not specified?

ie:

<span id='11a' style='top:55px;' onmouseover="GetPos(this);">stuff</span>
<span id='12a' onmouseover="GetPos(this);">stuff</span>

In the above, if I do:

document.getElementById('11a').style.top

The value of 55px is returned. However, if I try that for span '12a', then nothing gets returned.

I have a bunch of div/spans on a page that I cannot specify the top/left properties for, but I need to display a div directly under that element.

Javascript Solutions


Solution 1 - Javascript

You can call the method getBoundingClientRect() on a reference to the element. Then you can examine the top, left, right and/or bottom properties...

var offsets = document.getElementById('11a').getBoundingClientRect();
var top = offsets.top;
var left = offsets.left;

If using jQuery, you can use the more succinct code...

var offsets = $('#11a').offset();
var top = offsets.top;
var left = offsets.left;

Solution 2 - Javascript

This function will tell you the x,y position of the element relative to the page. Basically you have to loop up through all the element's parents and add their offsets together.

function getPos(el) {
    // yay readability
    for (var lx=0, ly=0;
         el != null;
         lx += el.offsetLeft, ly += el.offsetTop, el = el.offsetParent);
    return {x: lx,y: ly};
}

However, if you just wanted the x,y position of the element relative to its container, then all you need is:

var x = el.offsetLeft, y = el.offsetTop;

To put an element directly below this one, you'll also need to know its height. This is stored in the offsetHeight/offsetWidth property.

var yPositionOfNewElement = el.offsetTop + el.offsetHeight + someMargin;

Solution 3 - Javascript

While @nickf's answer works. If you don't care for older browsers, you can use this pure Javascript version. Works in IE9+, and others

var rect = el.getBoundingClientRect();

var position = {
  top: rect.top + window.pageYOffset,
  left: rect.left + window.pageXOffset
};

Solution 4 - Javascript

As Alex noted you can use jQuery offset() to get the position relative to the document flow. Use position() for its x,y coordinates relative to the parent.

EDIT: Switched document.ready for window.load because load waits for all of the elements so you get their size instead of simply preparing the DOM. In my experience, load results in fewer incorrectly Javascript positioned elements.

$(window).load(function(){ 
  // Log the position with jQuery
  var position = $('#myDivInQuestion').position();
  console.log('X: ' + position.left + ", Y: " + position.top );
});

Solution 5 - Javascript

For anyone needing just top or left position, slight modifications to @Nickf's readable code does the trick.

function getTopPos(el) {
    for (var topPos = 0;
        el != null;
        topPos += el.offsetTop, el = el.offsetParent);
    return topPos;
}

and

function getLeftPos(el) {
    for (var leftPos = 0;
        el != null;
        leftPos += el.offsetLeft, el = el.offsetParent);
    return leftPos;
}

Solution 6 - Javascript

I realize this is an old thread, but @alex 's answer needs to be marked as the correct answer

element.getBoundingClientRect() is an exact match to jQuery's $(element).offset()

And it's compatible with IE4+ ... https://developer.mozilla.org/en-US/docs/Web/API/Element.getBoundingClientRect

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionschmoopyView Question on Stackoverflow
Solution 1 - JavascriptalexView Answer on Stackoverflow
Solution 2 - JavascriptnickfView Answer on Stackoverflow
Solution 3 - JavascriptBenjamin IntalView Answer on Stackoverflow
Solution 4 - JavascriptDylan ValadeView Answer on Stackoverflow
Solution 5 - JavascriptJens FrandsenView Answer on Stackoverflow
Solution 6 - JavascriptlukeedView Answer on Stackoverflow