How to Check If an Element Is Really Visible with JavaScript

Check if element is visible in DOM

According to this MDN documentation, an element's offsetParent property will return null whenever it, or any of its parents, is hidden via the display style property. Just make sure that the element isn't fixed. A script to check this, if you have no position: fixed; elements on your page, might look like:

// Where el is the DOM element you'd like to test for visibility
function isHidden(el) {
    return (el.offsetParent === null)
}

On the other hand, if you do have position fixed elements that might get caught in this search, you will sadly (and slowly) have to use window.getComputedStyle(). The function in that case might be:

// Where el is the DOM element you'd like to test for visibility
function isHidden(el) {
    var style = window.getComputedStyle(el);
    return (style.display === 'none')
}

Option #2 is probably a little more straightforward since it accounts for more edge cases, but I bet its a good deal slower, too, so if you have to repeat this operation many times, best to probably avoid it.

How to check if the element is visible

 "none" == document.getElementById("element").style.display //Check for hide

 "block" == document.getElementById("element").style.display //Check for show

you can use like also

  if ($('#element').css('display') == 'none') {
    alert('element is hidden');
 }

How can I tell if a DOM element is visible in the current viewport?

Update: Time marches on and so have our browsers. This technique is no longer recommended and you should use Dan's solution if you do not need to support version of Internet Explorer before 7.

Original solution (now outdated):

This will check if the element is entirely visible in the current viewport:

function elementInViewport(el) {
  var top = el.offsetTop;
  var left = el.offsetLeft;
  var width = el.offsetWidth;
  var height = el.offsetHeight;

  while(el.offsetParent) {
    el = el.offsetParent;
    top += el.offsetTop;
    left += el.offsetLeft;
  }

  return (
    top >= window.pageYOffset &&
    left >= window.pageXOffset &&
    (top + height) <= (window.pageYOffset + window.innerHeight) &&
    (left + width) <= (window.pageXOffset + window.innerWidth)
  );
}

You could modify this simply to determine if any part of the element is visible in the viewport:

function elementInViewport2(el) {
  var top = el.offsetTop;
  var left = el.offsetLeft;
  var width = el.offsetWidth;
  var height = el.offsetHeight;

  while(el.offsetParent) {
    el = el.offsetParent;
    top += el.offsetTop;
    left += el.offsetLeft;
  }

  return (
    top < (window.pageYOffset + window.innerHeight) &&
    left < (window.pageXOffset + window.innerWidth) &&
    (top + height) > window.pageYOffset &&
    (left + width) > window.pageXOffset
  );
}

Finding if element is visible (JavaScript )

Display is not an attribute, it's a CSS property inside the style attribute.

You may be looking for

var myvar = document.getElementById("mydivID").style.display;

var myvar = $("#mydivID").css('display');

How do I check if an element is really visible with JavaScript?

For the point 2.

I see that no one has suggested to use document.elementFromPoint(x,y), to me it is the fastest way to test if an element is nested or hidden by another. You can pass the offsets of the targetted element to the function.

Here's PPK test page on elementFromPoint.

From MDN's documentation:

The elementFromPoint() method—available on both the Document and ShadowRoot objects—returns the topmost Element at the specified coordinates (relative to the viewport).

Check if an element is visible on screen

Here is a script returning a promise using the new IntersectionObserver API for checking whether or not an element is actually visible in the viewport:

function isVisible(domElement) {
  return new Promise(resolve => {
    const o = new IntersectionObserver(([entry]) => {
      resolve(entry.intersectionRatio === 1);
      o.disconnect();
    });
    o.observe(domElement);
  });
}

Which you can use in your code:

const visible = await isVisible(document.querySelector('#myElement'));
console.log(visible);

How do I check if an element is hidden in jQuery?

Since the question refers to a single element, this code might be more suitable:

// Checks CSS content for display:[none|block], ignores visibility:[true|false]
$(element).is(":visible");

// The same works with hidden
$(element).is(":hidden");

It is the same as twernt's suggestion, but applied to a single element; and it matches the algorithm recommended in the jQuery FAQ.

We use jQuery's is() to check the selected element with another element, selector or any jQuery object. This method traverses along the DOM elements to find a match, which satisfies the passed parameter. It will return true if there is a match, otherwise return false.

Check if element is partially in viewport

Late answer, but about a month ago I wrote a function that does exactly that, it determines how much an element is visible measured in percent in the viewport. Ive tested it in chrome, firefox, ie11, ios on iphone/ipad. The function returns true when X percent (as a number from 0 to 100) of the element is visible. Only determines if the measurements of the element are visible and not if the element is hidden with opacity, visibility etc..

const isElementXPercentInViewport = function(el, percentVisible) {
  let
    rect = el.getBoundingClientRect(),
    windowHeight = (window.innerHeight || document.documentElement.clientHeight);

  return !(
    Math.floor(100 - (((rect.top >= 0 ? 0 : rect.top) / +-rect.height) * 100)) < percentVisible ||
    Math.floor(100 - ((rect.bottom - windowHeight) / rect.height) * 100) < percentVisible
  )
};