Determine if an HTML element's content overflows

Question

Normally, you can compare the client[Height|Width] with scroll[Height|Width] in order to detect this... but the values will be the same when overflow is visible. So, a detection routine must account for this:

// Determines if the passed element is overflowing its bounds,
// either vertically or horizontally.
// Will temporarily modify the "overflow" style to detect this
// if necessary.
function checkOverflow(el)
{
   var curOverflow = el.style.overflow;

   if ( !curOverflow || curOverflow === "visible" )
      el.style.overflow = "hidden";

   var isOverflowing = el.clientWidth < el.scrollWidth 
      || el.clientHeight < el.scrollHeight;

   el.style.overflow = curOverflow;

   return isOverflowing;
}

Tested in FF3, FF40.0.2, IE6, Chrome 0.2.149.30.

Try comparing element.scrollHeight / element.scrollWidth to element.offsetHeight / element.offsetWidth

http://developer.mozilla.org/en/DOM/element.offsetWidth
http://developer.mozilla.org/en/DOM/element.offsetHeight
http://developer.mozilla.org/en/DOM/element.scrollWidth
http://developer.mozilla.org/en/DOM/element.scrollHeight

I don't think this answer is perfect. Sometimes the scrollWidth/clientWidth/offsetWidth are the same even though the text is overflow.

This works well in Chrome, but not in IE and Firefox.

At last, I tried this answer: HTML text-overflow ellipsis detection

It's perfect and works well anywhere. So I choose this, maybe you can try, you won't disappoint.

Another way is compare the element width with its parent's width:

function checkOverflow(elem) {
    const elemWidth = elem.getBoundingClientRect().width
    const parentWidth = elem.parentElement.getBoundingClientRect().width

    return elemWidth > parentWidth
}

With jQuery you could do:

if ( $(".inner-element").prop('scrollHeight') > $(".inner-element").height() ) {

    console.log("element is overflowing");

} else {

    console.log("element is not overflowing");

}

Change to .prop('scrollWidth') and .width() if needed.