How To Compute GetBoundingClientRect() Without Considering Transforms?
Solution 1:
Get element position without considering any transformation on the element and up the DOM tree :
var el = element,
offsetLeft = 0,
offsetTop = 0;
do{
offsetLeft += el.offsetLeft;
offsetTop += el.offsetTop;
el = el.offsetParent;
} while( el );
Get element position without considering transformation applied to it but keeping any transformation up the DOM tree.
To do so you could try to revert the transform.
You must first set transform-origin
to 0,0,0
and surround yourself your transformation (scale, rotate) width translate(50%,50%) ... translate(-50%, -50%)
.
Here is an example,
change that :
transform: scale(2) rotate(45deg) translate(20px);
transform-origin: 50% 50%; //default value
into
transform: translate(50%, 50%) scale(2) rotate(45deg) translate(-50%,-50%) translate(20px);
transform-origin: 0 0 0;
We need to do that because the matrix returned by getComputedStyle() does not include stuff done with transform-origin. Don't know really why.
Then you can use this code :
function parseTransform(transform){
//add sanity check
return transform.split(/\(|,|\)/).slice(1,-1).map( function(v){
return parseFloat(v);
});
}
function convertCoord(transformArr, x, y, z){
//add sanity checks and default values
if( transformArr.length == 6 ){
//2D matrix
//need some math to apply inverse of matrix
var t = transformArr,
det = t[0]*t[3] - t[1]*t[2];
return {
x: ( x*t[3] - y*t[2] + t[2]*t[5] - t[4]*t[3] )/det,
y: ( -x*t[1] + y*t[0] + t[4]*t[1] - t[0]*t[5] )/det
}
}
else /*if (transformArr.length > 6)*/{
//3D matrix
//haven't done the calculation to apply inverse of 4x4 matrix
}
}
var elRect = element.getBoundingClientRect(),
st = window.getComputedStyle(element),
topLeft_pos = convertCoord(
parseTransform( st.transform ),
elRect.left,
elRect.top,
st.perspective
);
I won't explain the math part because I think it's beyond the scope of this post. Could still explain it somewhere else (another question maybe ? ).
Solution 2:
The answer above that inverts the transform mathematically is a nice try, but not quite correct (and more complex than it needs to be). A more correct inversion is below.
This doesn't account for skew or rotate translations, but at least it produces correct edge positions when scale is used and does not impose much of a performance penalty when there is no transform.
It produces accurate results even with scale(0) (albeit losing subpixel precision on width/height).
Note that iOS with the software keyboard open produces different results between getBoundingClientRect()
and offsetTop
/offsetLeft
- and the latter do not support subpixel precision on any browser. This produces results consistent with getBoundingClientRect()
.
function adjustedBoundingRect(el) {
var rect = el.getBoundingClientRect();
var style = getComputedStyle(el);
var tx = style.transform;
if (tx) {
var sx, sy, dx, dy;
if (tx.startsWith('matrix3d(')) {
var ta = tx.slice(9,-1).split(/, /);
sx = +ta[0];
sy = +ta[5];
dx = +ta[12];
dy = +ta[13];
} else if (tx.startsWith('matrix(')) {
var ta = tx.slice(7,-1).split(/, /);
sx = +ta[0];
sy = +ta[3];
dx = +ta[4];
dy = +ta[5];
} else {
return rect;
}
var to = style.transformOrigin;
var x = rect.x - dx - (1 - sx) * parseFloat(to);
var y = rect.y - dy - (1 - sy) * parseFloat(to.slice(to.indexOf(' ') + 1));
var w = sx ? rect.width / sx : el.offsetWidth;
var h = sy ? rect.height / sy : el.offsetHeight;
return {
x: x, y: y, width: w, height: h, top: y, right: x + w, bottom: y + h, left: x
};
} else {
return rect;
}
}
var div = document.querySelector('div');
console.log(div.getBoundingClientRect(), adjustedBoundingRect(div));
div.classList.add('transformed');
console.log(div.getBoundingClientRect(), adjustedBoundingRect(div));
.transformed {
transform: translate(8px,8px) scale(0.5);
transform-origin: 16px 16px;
}
<div>Hello</div>
Solution 3:
I liked Ghetolay's answer. I used it but I made it a bit more performant by avoiding the loop.
I have a draggable tag cloud and I have to update the drag position using transforms, but keep track of the original position (without transform).
The previous answer suggested to loop thru the offsetParents. In my case, and I think in a lot of cases, the tags are transformed but the container isn't. So I only have to get the first offsetParent and use getBoundingClientRect()
there. No need to keep looping. I solved it doing this:
var el = element;
var parentRect = element.offsetParent.getBoundingClientRect();
var offsetLeft = parentRect.left + element.offsetLeft;
var offsetTop = parentRect.top + element.offsetTop;
Solution 4:
Improve Ghetolay's answer:
Get element position without considering any transformation on the element and up the DOM tree :
function getOffset(element)
{
var offsetLeft = 0;
var offsetTop = 0;
while (element)
{
offsetLeft += element.offsetLeft;
offsetTop += element.offsetTop;
element = element.offsetParent;
}
return [offsetLeft, offsetTop];
}
Improvements:
- Contained in a function
- loop will handle empty elements without error (
element
=null
/undefined
)
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