CODEWITHSUNDEEP
javascriptcssweb-componentcustom-elementnative-web-component
connexo
connexo

Reputation: 56754

Performant way to find out if an element or any of its ancestor elements 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 => getComputedStyle(el).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).

Upvotes: 4

Views: 3829

Answers (4)

farinspace
farinspace

Reputation: 8801

Similar to the answer above, without HTMLElement.prototype and the addition of style.visibility check

function isHidden(elem) {
    if (document === elem) return false;
    if ('none' == elem.style.display || 'hidden' == elem.style.visibility) return true;
    var cs = getComputedStyle(elem);
    if ('none' === cs.display || 'hidden' === cs.visibility) return true;
    if (elem.parentNode) return isHidden(elem.parentNode);
    return false;
}

Upvotes: 0

KingRanTheMan
KingRanTheMan

Reputation: 740

For a pure function that:

  • returns TRUE if and only if neither the element itself nor any of its parents up to the document itself have style display === 'none'
  • returns FALSE if and only if either the element itself or any parent element up to the document itself have style display === 'none'

You can define the below function and then call it on the element you wish to validate:

function isVisible(element) {
    // Start with the element itself and move up the DOM tree
    for (let el = element; el && el !== document; el = el.parentNode) {
        // If current element has display property 'none', return false
        if (getComputedStyle(el).display === "none") {
            return false;
        }
    }
    // Neither element itself nor any parents have display 'none', so return true
    return true;
}

Upvotes: 0

user5734311
user5734311

Reputation:

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;
};

Upvotes: 4

Intervalia
Intervalia

Reputation: 10945

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

Upvotes: 10

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