Performant way to find out if an element or any of its parent element has display: none

I need to find a very performant way to find out if a custom element or any of its parent elements has display: none;

First approach:

checkVisible() {
  let parentNodes = [];
  let el = this;
  while (!!(el = el.parentNode)) {
    parentNodes.push(el);
  }
  return [this, ...parentNodes].some(el => el.style.display === 'none') 
}

Is there anything that runs faster than this? Is this even a safe method?

The reason I need this: We have a <data-table> custom element (native webcomponent) which does very heavy lifting in its connectedCallback(). We have an application that has like 20-30 of those custom elements in a single page, which leads to IE 11 taking like 15 seconds until the page is rendered.

I need to delay initialisation of those <data-table> components which are initially not even visible, so I need a way to test inside the connectedCallback() if the element is visible (which it is not if it is in one of the 18 tabs initially not shown).

Answers:

Answer

The easiest way to see if an element or its parent has display:none is to use el.offsetParent.

const p1 = document.getElementById('parent1');
const p2 = document.getElementById('parent2');
const c1 = document.getElementById('child1');
const c2 = document.getElementById('child2');
const btn = document.getElementById('btn');
const output = document.getElementById('output');

function renderVisibility() {
  const p1state = isElementVisible(p1) ? 'is visible' : 'is not visible';
  const p2state = isElementVisible(p2) ? 'is visible' : 'is not visible';
  const c1state = isElementVisible(c1) ? 'is visible' : 'is not visible';
  const c2state = isElementVisible(c2) ? 'is visible' : 'is not visible';
  
  output.innerHTML = `Parent 1 ${p1state}<br>Parent 2 ${p2state}<br/>Child 1 ${c1state}<br/>Child 2 ${c2state}`;
}

function isElementVisible(el) {
  return !!el.offsetParent;
}

function toggle() {
  p1.style.display = (p1.style.display ? '' : 'none');
  p2.style.display = (p2.style.display ? '' : 'none');
  renderVisibility();
}

btn.addEventListener('click', toggle),
renderVisibility();
<div id="parent1" style="display:none">
  <div id="child1">child 1</div>
</div>
<div id="parent2">
  <div id="child2">second child</div>
</div>
<button id="btn">Toggle</button>
<hr>
<div id="output"></div>

This code converts el.offsetParent into a boolean that indicates if the element is showing or not.

This only works for display:none

Answer

Not sure about performance, but it should be faster than your approach at least:

HTMLElement.prototype.isInvisible = function() {
  if (this.style.display == 'none') return true;
  if (getComputedStyle(this).display === 'none') return true;
  if (this.parentNode.isInvisible) return this.parentNode.isInvisible();
  return false;
};

Tags

Recent Questions

Top Questions

Home Tags Terms of Service Privacy Policy DMCA Contact Us

©2020 All rights reserved.