Is there anyway to the positioning of any element without using e.pageX and e.pageY.
Check this fiddle
The fiddle is actually a poor attempt at what im tryi
You may use
document.getElementById("elementID").offsetTop;
document.getElementById("elementID").offsetLeft;
Refer to MDN. They return the offset from the parent element. If you need the offset of the element in respect to the whole body it may get more tricky, as you will have to sum the offsets of each element in the chain.
Respectively for .getElementsByTagName
, as each object in the DOM has these attributes.
.getBoundingClientRect
is also worth a look.
var clientRectangle = document.getElementById("elementID").getBoundingClientRect();
console.log(clientRectangle.top); //or left, right, bottom
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<script>
$("#something").click(function(e){
var parentOffset = $(this).parent().offset();
//or $(this).offset(); if you really just want the current element's offset
var relX = e.pageX - parentOffset.left;
var relY = e.pageY - parentOffset.top;
});
</script>
If Im understanding right:
UIEvent.PageX/PageY: relative to the document (https://developer.mozilla.org/en-US/docs/Web/API/UIEvent/pageX)
Elem.getBoundingClientRect: relative to viewport.(https://developer.mozilla.org/en-US/docs/Web/API/Element/getBoundingClientRect)
Window.pageXOffset/pageYOffset:number of pixels that the document is currently scrolled. (https://developer.mozilla.org/en-US/docs/Web/API/Window/scrollX)
Based on the following description:
const PageX = (elem) => window.pageXOffset + elem.getBoundingClientRect().left
const PageY = (elem) => window.pageYOffset + elem.getBoundingClientRect().top
I think offset() should work.
$(element).offset() //will get {top:.., left:....}
You can find almost all jquery stuff here: http://youmightnotneedjquery.com/
$(el).offset();
the same as:
var rect = el.getBoundingClientRect();
{
top: rect.top + document.body.scrollTop,
left: rect.left + document.body.scrollLeft
}