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


Javascript Problem Overview

Is it possible to toggle the visibility of an element, using the functions .hide(), .show() or .toggle()?

How would you test if an element is visible or hidden?

Javascript Solutions

Solution 1 - Javascript

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]

// The same works with 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.

Solution 2 - Javascript

You can use the hidden selector:

// Matches all elements that are hidden

And the visible selector:

// Matches all elements that are visible

Solution 3 - Javascript

if ( $(element).css('display') == 'none' || $(element).css("visibility") == "hidden"){
    // 'element' is hidden

The above method does not consider the visibility of the parent. To consider the parent as well, you should use .is(":hidden") or .is(":visible").

For example,

<div id="div1" style="display:none">
  <div id="div2" style="display:block">Div2</div>

> The above method will consider div2 visible while :visible not. But the above might be useful in many cases, especially when you need to find if there is any error divs visible in the hidden parent because in such conditions :visible will not work.

Solution 4 - Javascript

None of these answers address what I understand to be the question, which is what I was searching for, "How do I handle items that have visibility: hidden?". Neither :visible nor :hidden will handle this, as they are both looking for display per the documentation. As far as I could determine, there is no selector to handle CSS visibility. Here is how I resolved it (standard jQuery selectors, there may be a more condensed syntax):

$(".item").each(function() {
    if ($(this).css("visibility") == "hidden") {
        // handle non visible state
    } else {
        // handle visible state

Solution 5 - Javascript

From How do I determine the state of a toggled element?

You can determine whether an element is collapsed or not by using the :visible and :hidden selectors.

var isVisible = $('#myDiv').is(':visible');
var isHidden = $('#myDiv').is(':hidden');

If you're simply acting on an element based on its visibility, you can just include :visible or :hidden in the selector expression. For example:

 $('#myDiv:visible').animate({left: '+=200px'}, 'slow');

Solution 6 - Javascript

Often when checking if something is visible or not, you are going to go right ahead immediately and do something else with it. jQuery chaining makes this easy.

So if you have a selector and you want to perform some action on it only if is visible or hidden, you can use filter(":visible") or filter(":hidden") followed by chaining it with the action you want to take.

So instead of an if statement, like this:

if ($('#btnUpdate').is(":visible"))
     $('#btnUpdate').animate({ width: "toggle" });   // Hide button

Or more efficient, but even uglier:

var button = $('#btnUpdate');
if (":visible"))
     button.animate({ width: "toggle" });   // Hide button

You can do it all in one line:

$('#btnUpdate').filter(":visible").animate({ width: "toggle" });

Solution 7 - Javascript

The :visible selector according to the jQuery documentation:

> * They have a CSS display value of none. > * They are form elements with type="hidden". > * Their width and height are explicitly set to 0. > * An ancestor element is hidden, so the element is not shown on the page. > > Elements with visibility: hidden or opacity: 0 are considered to be visible, since they still consume space in the layout.

This is useful in some cases and useless in others, because if you want to check if the element is visible (display != none), ignoring the parents visibility, you will find that doing .css("display") == 'none' is not only faster, but will also return the visibility check correctly.

If you want to check visibility instead of display, you should use: .css("visibility") == "hidden".

Also take into consideration the additional jQuery notes:

> Because :visible is a jQuery extension and not part of the CSS specification, queries using :visible cannot take advantage of the performance boost provided by the native DOM querySelectorAll() method. To achieve the best performance when using :visible to select elements, first select the elements using a pure CSS selector, then use .filter(":visible").

Also, if you are concerned about performance, you should check Now you see me… show/hide performance (2010-05-04). And use other methods to show and hide elements.

Solution 8 - Javascript

How element visibility and jQuery works;

An element could be hidden with display:none, visibility:hidden or opacity:0. The difference between those methods:

  • display:none hides the element, and it does not take up any space;

  • visibility:hidden hides the element, but it still takes up space in the layout;

  • opacity:0 hides the element as "visibility:hidden", and it still takes up space in the layout; the only difference is that opacity lets one to make an element partly transparent;

     if ($('.target').is(':hidden')) {
     } else {
     if ($('.target').is(':visible')) {
     } else {
     if ($('.target-visibility').css('visibility') == 'hidden') {
         visibility: "visible",
         display: ""
     } else {
         visibility: "hidden",
         display: ""
     if ($('.target-visibility').css('opacity') == "0") {
         opacity: "1",
         display: ""
     } else {
         opacity: "0",
         display: ""

    Useful jQuery toggle methods:

     $('.click').click(function() {
     $('.click').click(function() {
     $('.click').click(function() {

Solution 9 - Javascript

This works for me, and I am using show() and hide() to make my div hidden/visible:

if( $(this).css('display') == 'none' ){
	/* your code goes here */
} else {
	/* alternate logic   */

Solution 10 - Javascript

You can also do this using plain JavaScript:

function isRendered(domObj) {
    if ((domObj.nodeType != 1) || (domObj == document.body)) {
        return true;
    if (domObj.currentStyle && domObj.currentStyle["display"] != "none" && domObj.currentStyle["visibility"] != "hidden") {
        return isRendered(domObj.parentNode);
    } else if (window.getComputedStyle) {
		var cs = document.defaultView.getComputedStyle(domObj, null);
		if (cs.getPropertyValue("display") != "none" && cs.getPropertyValue("visibility") != "hidden") {
			return isRendered(domObj.parentNode);
    return false;


  1. Works everywhere

  2. Works for nested elements

  3. Works for CSS and inline styles

  4. Doesn't require a framework

Solution 11 - Javascript

I would use CSS class .hide { display: none!important; }.

For hiding/showing, I call .addClass("hide")/.removeClass("hide"). For checking visibility, I use .hasClass("hide").

It's a simple and clear way to check/hide/show elements, if you don't plan to use .toggle() or .animate() methods.

Solution 12 - Javascript

Demo Link

$('#clickme').click(function() {
  $('#book').toggle('slow', function() {
    // Animation complete.
    alert($('#book').is(":visible")); //<--- TRUE if Visible False if Hidden

<script src=""></script>
<div id="clickme">
  Click here
<img id="book" src="" alt="" width="300"/>

Source (from my blog):

Blogger Plug n Play - jQuery Tools and Widgets: How to See if Element is hidden or Visible Using jQuery

Solution 13 - Javascript

One can simply use the hidden or visible attribute, like:


Or you can simplify the same with is as follows.


Solution 14 - Javascript

ebdiv should be set to style="display:none;". It works for both show and hide:


Solution 15 - Javascript

Another answer you should put into consideration is if you are hiding an element, you should use jQuery, but instead of actually hiding it, you remove the whole element, but you copy its HTML content and the tag itself into a jQuery variable, and then all you need to do is test if there is such a tag on the screen, using the normal if (!$('#thetagname').length).

Solution 16 - Javascript

When testing an element against :hidden selector in jQuery it should be considered that an absolute positioned element may be recognized as hidden although their child elements are visible.

This seems somewhat counter-intuitive in the first place – though having a closer look at the jQuery documentation gives the relevant information:

> Elements can be considered hidden for several reasons: [...] Their width and height are explicitly set to 0. [...]

So this actually makes sense in regards to the box-model and the computed style for the element. Even if width and height are not set explicitly to 0 they may be set implicitly.

Have a look at the following example:

console.log($('.foo').is(':hidden')); // true
console.log($('.bar').is(':hidden')); // false

.foo {
  position: absolute;
  left: 10px;
  top: 10px;
  background: #ff0000;

.bar {
  position: absolute;
  left: 10px;
  top: 10px;
  width: 20px;
  height: 20px;
  background: #0000ff;

<script src=""></script>
<div class="foo">
  <div class="bar"></div>

Update for jQuery 3.x:

With jQuery 3 the described behavior will change! Elements will be considered visible if they have any layout boxes, including those of zero width and/or height.

JSFiddle with jQuery 3.0.0-alpha1:

The same JavaScript code will then have this output:

console.log($('.foo').is(':hidden')); // false
console.log($('.bar').is(':hidden')); // false

Solution 17 - Javascript


Solution 18 - Javascript

$(document).ready(function() {
  if ($("#checkme:hidden").length) {

<script src=""></script>
<div id="checkme" class="product" style="display:none">
  <span class="itemlist"><!-- Shows Results for Fish --></span> Category:Fish
  <br>Product: Salmon Atlantic
  <br>Specie: Salmo salar
  <br>Form: Steaks

Solution 19 - Javascript

To check if it is not visible I use !:

if ( !$('#book').is(':visible')) {
    alert('#book is not visible')

Or the following is also the sam, saving the jQuery selector in a variable to have better performance when you need it multiple times:

var $book = $('#book')

if(!$':visible')) {
    alert('#book is not visible')

Solution 20 - Javascript

Use class toggling, not style editing . . .

Using classes designated for "hiding" elements is easy and also one of the most efficient methods. Toggling a class 'hidden' with a Display style of 'none' will perform faster than editing that style directly. I explained some of this pretty thoroughly in Stack Overflow question Turning two elements visible/hidden in the same div.

JavaScript Best Practices and Optimization

Here is a truly enlightening video of a Google Tech Talk by Google front-end engineer Nicholas Zakas:

Solution 21 - Javascript

After all, none of examples suits me, so I wrote my own.

Tests (no support of Internet Explorer filter:alpha):

a) Check if the document is not hidden

b) Check if an element has zero width / height / opacity or display:none / visibility:hidden in inline styles

c) Check if the center (also because it is faster than testing every pixel / corner) of element is not hidden by other element (and all ancestors, example: overflow:hidden / scroll / one element over another) or screen edges

d) Check if an element has zero width / height / opacity or display:none / visibility:hidden in computed styles (among all ancestors)

Tested on

Android 4.4 (Native browser/Chrome/Firefox), Firefox (Windows/Mac), Chrome (Windows/Mac), Opera (Windows Presto/Mac WebKit), Internet Explorer (Internet Explorer 5-11 document modes + Internet Explorer 8 on a virtual machine), and Safari (Windows/Mac/iOS).

var is_visible = (function () {
    var x = window.pageXOffset ? window.pageXOffset + window.innerWidth - 1 : 0,
        y = window.pageYOffset ? window.pageYOffset + window.innerHeight - 1 : 0,
        relative = !!((!x && !y) || !document.elementFromPoint(x, y));
        function inside(child, parent) {
                if (child === parent) return true;
                child = child.parentNode;
        return false;
    return function (elem) {
        if (
            document.hidden ||
            elem.offsetWidth==0 ||
            elem.offsetHeight==0 ||
  'hidden' ||
  'none' ||
        ) return false;
        var rect = elem.getBoundingClientRect();
        if (relative) {
            if (!inside(document.elementFromPoint(rect.left + elem.offsetWidth/2, + elem.offsetHeight/2),elem)) return false;
        } else if (
            !inside(document.elementFromPoint(rect.left + elem.offsetWidth/2 + window.pageXOffset, + elem.offsetHeight/2 + window.pageYOffset), elem) ||
       + elem.offsetHeight/2 < 0 ||
                rect.left + elem.offsetWidth/2 < 0 ||
                rect.bottom - elem.offsetHeight/2 > (window.innerHeight || document.documentElement.clientHeight) ||
                rect.right - elem.offsetWidth/2 > (window.innerWidth || document.documentElement.clientWidth)
        ) return false;
        if (window.getComputedStyle || elem.currentStyle) {
            var el = elem,
                comp = null;
            while (el) {
                if (el === document) {break;} else if(!el.parentNode) return false;
                comp = window.getComputedStyle ? window.getComputedStyle(el, null) : el.currentStyle;
                if (comp && (comp.visibility=='hidden' || comp.display == 'none' || (typeof comp.opacity !=='undefined' && comp.opacity != 1))) return false;
                el = el.parentNode;
        return true;

How to use:

is_visible(elem) // boolean

Solution 22 - Javascript

Example of using the visible check for adblocker is activated:

    $("#ablockermsg").text("Please disable adblocker.").show();

<script src=""></script>
<div class="ad-placement" id="ablockercheck"></div>
<div id="ablockermsg" style="display: none"></div>

"ablockercheck" is a ID which adblocker blocks. So checking it if it is visible you are able to detect if adblocker is turned On.

Solution 23 - Javascript

You need to check both... Display as well as visibility:

if ($(this).css("display") == "none" || $(this).css("visibility") == "hidden") {
    // The element is not visible
} else {
    // The element is visible

If we check for $(this).is(":visible"), jQuery checks for both the things automatically.

Solution 24 - Javascript

$(document).ready(function() {
   var visible = $('#tElement').is(':visible');

   if(visible) {
                    // Code

<script src=""></script>

<input type="text" id="tElement" style="display:block;">Firstname</input>

Solution 25 - Javascript

Simply check visibility by checking for a boolean value, like:

if (this.hidden === false) {
    // Your code

I used this code for each function. Otherwise you can use is(':visible') for checking the visibility of an element.

Solution 26 - Javascript

Because Elements with visibility: hidden or opacity: 0 are considered visible, since they still consume space in the layout (as described for jQuery :visible Selector) - we can check if element is really visible in this way:

function isElementReallyHidden (el) {
	return $(el).is(":hidden") || $(el).css("visibility") == "hidden" || $(el).css('opacity') == 0;

var booElementReallyShowed = !isElementReallyHidden(someEl);
$(someEl).parents().each(function () {
    if (isElementReallyHidden(this)) {
        booElementReallyShowed = false;

Solution 27 - Javascript

But what if the element's CSS is like the following?

    position: absolute;left:-9999;    

So this answer to Stack Overflow question How to check if an element is off-screen should also be considered.

Solution 28 - Javascript

A function can be created in order to check for visibility/display attributes in order to gauge whether the element is shown in the UI or not.

function checkUIElementVisible(element) {
    return ((element.css('display') !== 'none') && (element.css('visibility') !== 'hidden'));

Working Fiddle

Solution 29 - Javascript

Also here's a ternary conditional expression to check the state of the element and then to toggle it:

$('someElement').on('click', function(){ $('elementToToggle').is(':visible') ? $('elementToToggle').hide('slow') : $('elementToToggle').show('slow'); });

Solution 30 - Javascript

if($('#postcode_div').is(':visible')) {
    if($('#postcode_text').val()=='') {
    } else {

Solution 31 - Javascript

.is(":not(':hidden')") /*if shown*/

Solution 32 - Javascript


Solution 33 - Javascript

I searched for this, and none of the answers are correct for my case, so I've created a function that will return false if one's eyes can't see the element

  isvisible: function() {
	//	This function call this: $("div").isvisible()
	//	Return true if the element is visible
	//	Return false if the element is not visible for our eyes
	if ( $(this).css('display') == 'none' ){
		console.log("this = " + "display:none");
		return false;
	else if( $(this).css('visibility') == 'hidden' ){
		console.log("this = " + "visibility:hidden");	
		return false;
	else if( $(this).css('opacity') == '0' ){
		console.log("this = " + "opacity:0");
		return false;
		console.log("this = " + "Is Visible");
		return true;

Solution 34 - Javascript

As hide(), show() and toggle() attaches inline css (display:none or display:block) to element. Similarly, we can easily use the ternary operator to check whether the element is hidden or visible by checking display CSS.


  • You also need to check if element CSS set to visibility: "visible" or visibility: "hidden"
  • The element will be also visible if display property set to inline-block, block, flex.

So we can check for the property of an element that makes it invisible. So they are display: none and visibility: "hidden";

We can create an object for checking property responsible for hiding element:

var hiddenCssProps = {
display: "none",
visibility: "hidden"

We can check by looping through each key value in object matching if element property for key matches with hidden property value.

var isHidden = false;
for(key in hiddenCssProps) {
  if($('#element').css(key) == hiddenCssProps[key]) {
     isHidden = true;

If you want to check property like element height: 0 or width: 0 or more, you can extend this object and add more property to it and can check.

Solution 35 - Javascript

There are quite a few ways to check if an element is visible or hidden in jQuery.

Demo HTML for example reference

<div id="content">Content</div>
<div id="content2" style="display:none">Content2</div>

Use Visibility Filter Selector $('element:hidden') or $('element:visible')

  • $('element:hidden'): Selects all elements that are hidden.

  • $('element:visible'): Selects all elements that are visible.

        $('#content:visible').css('color', '#EEE');

>

Use is() Filtering

       $('#content').is(":visible").css('color', '#EEE');

    Or checking condition
    if ($('#content').is(":visible")) {
         // Perform action

>

Solution 36 - Javascript

Just simply check if that element is visible and it will return a boolean. jQuery hides the elements by adding display none to the element, so if you want to use pure JavaScript, you can still do that, for example:

if (document.getElementById("element").style.display === 'block') {
  // Your element is visible; do whatever you'd like

Also, you can use jQuery as it seems the rest of your code is using that and you have smaller block of code. Something like the below in jQuery does the same trick for you:

if ($(element).is(":visible")) {
    // Your element is visible, do whatever you'd like

Also using the css method in jQuery can result in the same thing:

if ($(element).css('display') === 'block') {
    // Your element is visible, do whatever you'd like

Also in case of checking for visibility and display, you can do the below:

if ($(this).css("display") === "block" || $(this).css("visibility") === "visible") {
   // Your element is visible, do whatever you'd like

Solution 37 - Javascript

This is how jQuery internally solves this problem:

jQuery.expr.pseudos.visible = function( elem ) {
	return !!( elem.offsetWidth || elem.offsetHeight || elem.getClientRects().length );

If you don't use jQuery, you can just leverage this code and turn it into your own function:

function isVisible(elem) {
	return !!( elem.offsetWidth || elem.offsetHeight || elem.getClientRects().length );

Which isVisible will return true as long as the element is visible.

Solution 38 - Javascript

You can use this:


Example code


       if( $('#content').is(':visible') )
          alert("visible"); // Put your code for visibility

<script src=""></script>

<p id="content">This is a Content</p>

<button id="toggle">Toggle Content Visibility</button>
<button id="visibility">Check Visibility</button>

Solution 39 - Javascript

I just want to clarify that, in jQuery,

> Elements can be considered hidden for several reasons: > > - They have a CSS display value of none. > - They are form elements with type="hidden". > - Their width and height are explicitly set to 0. > - An ancestor element is hidden, so the element is not shown on the page. > > Elements with visibility: hidden or opacity: 0 are considered to be visible, since they still consume space in the layout. During animations that hide an element, the element is considered to be visible until the end of the animation. > > Source: :hidden Selector | jQuery API Documentation

if($('.element').is(':hidden')) {
  // Do something

Solution 40 - Javascript

This is some option to check that tag is visible or not

 // using a pure CSS selector  
   if ($('p:visible')) {  
      alert('Paragraphs are visible (checked using a CSS selector) !');  
   // using jQuery's is() method  
   if ($('p').is(':visible')) {  
      alert('Paragraphs are visible (checked using is() method)!');  
   // using jQuery's filter() method  
   if ($('p').filter(':visible')) {  
      alert('Paragraphs are visible (checked using filter() method)!');  
   // you can use :hidden instead of :visible to reverse the logic and check if an element is hidden  
   // if ($('p:hidden')) {  
   //    do something  
   // };  

Solution 41 - Javascript

You can just add a class when it is visible. Add a class, show. Then check for it have a class:


It returns true if you have the show class.

Add CSS like this:

.show{ display: block; }

Solution 42 - Javascript

$( "div:visible" ).click(function() {
  $( this ).css( "background", "yellow" );
$( "button" ).click(function() {
  $( "div:hidden" ).show( "fast" );

API Documentation: visible Selector

Solution 43 - Javascript

1 • jQuery solution

Methods to determine if an element is visible in jQuery
if ($("#myelement").is(":visible")){alert ("#myelement is visible");}
if ($("#myelement").is(":hidden")){alert ("#myelement is hidden"); }
Loop on all visible div children of the element of id 'myelement':
$("#myelement div:visible").each( function() {
 //Do something
Peeked at source of jQuery

This is how jQuery implements this feature:

jQuery.expr.filters.visible = function( elem ) {
    return !!( elem.offsetWidth || elem.offsetHeight || elem.getClientRects().length );

2 • How to check if an element is off-screen - CSS

Using Element.getBoundingClientRect() you can easily detect whether or not your element is within the boundaries of your viewport (i.e. onscreen or offscreen):

jQuery.expr.filters.offscreen = function(el) {
  var rect = el.getBoundingClientRect();
  return (
           (rect.x + rect.width) < 0 
             || (rect.y + rect.height) < 0
             || (rect.x > window.innerWidth || rect.y > window.innerHeight)

You could then use that in several ways:

// Returns all elements that are offscreen

// Boolean returned if element is offscreen

If you use Angular, check: Don’t use hidden attribute with Angular

Solution 44 - Javascript

There are too many methods to check for hidden elements. This is the best choice (I just recommended you):

> Using jQuery, make an element, "display:none", in CSS for hidden.

The point is:


And an example for use:


Solution 45 - Javascript

Simply check for the display attribute (or visibility depending on what kind of invisibility you prefer). Example:

if ($('#invisible').css('display') == 'none') {
    // This means the HTML element with ID 'invisible' has its 'display' attribute set to 'none'

Solution 46 - Javascript

Using hidden selection you can match all hidden elements


Using Visible selection you can match all visible elements


Solution 47 - Javascript

If you want to check if an element is visible on the page, depending on the visibility of its parent, you can check if width and height of the element are both equal to 0.


$element.width() === 0 && $element.height() === 0


element.clientWidth === 0 && element.clientHeight === 0


element.offsetWidth === 0 && element.offsetHeight === 0

Solution 48 - Javascript

A jQuery solution, but it is still a bit better for those who want to change the button text as well:

    var btn = $(this);
    $("#content").toggle(function () {
      btn.text($(this).css("display") === 'none' ? "Show" : "Hide");

<script src=""></script>

<button id="showHide">Hide</button>
<div id="content">
  <h2>Some content</h2>
  What is Lorem Ipsum? Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. It has survived not only five centuries, but also the leap into electronic typesetting, remaining essentially unchanged.

Solution 49 - Javascript

Extended function for checking if element is visible, display none, or even the opacity level

It returns false if the element is not visible.

function checkVisible(e) {
    if (!(e instanceof Element)) throw Error('not an Element');
    const elementStyle = getComputedStyle(e);
    if (elementStyle.display === 'none' || elementStyle.visibility !== 'visible' || elementStyle.opacity < 0.1) return false;
    if (e.offsetWidth + e.offsetHeight + e.getBoundingClientRect().height +
        e.getBoundingClientRect().width === 0) {
        return false;
    const elemCenter   = {
        x: e.getBoundingClientRect().left + e.offsetWidth / 2,
        y: e.getBoundingClientRect().top + e.offsetHeight / 2
    if (elemCenter.x < 0 || elemCenter.y < 0) return false;
    if (elemCenter.x > (document.documentElement.clientWidth || window.innerWidth)) return false;
    if (elemCenter.y > (document.documentElement.clientHeight || window.innerHeight)) return false;
    let pointContainer = document.elementFromPoint(elemCenter.x, elemCenter.y);
    do {
        if (pointContainer === e) return true;
    } while (pointContainer = pointContainer.parentNode);
    return false;

Solution 50 - Javascript

To be fair the question pre-dates this answer.

I add it not to criticise the OP, but to help anyone still asking this question.

The correct way to determine whether something is visible is to consult your view-model;

If you don't know what that means then you are about to embark on a journey of discovery that will make your work a great deal less difficult.

Here's an overview of the model-view-view-model architecture (MVVM).

KnockoutJS is a binding library that will let you try this stuff out without learning an entire framework.

And here's some JavaScript code and a DIV that may or may not be visible.

<script src=""></script>
    var vm = {
        IsDivVisible: ko.observable(true);
    vm.toggle = function(data, event) {
      // Get current visibility state for the div
      var x = IsDivVisible();
      // Set it to the opposite
<div data-bind="visible: IsDivVisible">Peekaboo!</div>
<button data-bind="click: toggle">Toggle the div's visibility</button>

Notice that the toggle function does not consult the DOM to determine the visibility of the div; it consults the view-model.

Solution 51 - Javascript

There are two ways to check visibility of element.

Solution #1

    // element is visible
    // element is hidden

Solution #2

    // element is visible
    // element is hidden

Solution 52 - Javascript

    // your code..

Solution 53 - Javascript

You can use a CSS class when it visible or hidden by toggling the class:

.show{ display :block; }

Set your jQuery toggleClass() or addClass() or removeClass();.

As an example,


The above code will add show css class when the element don't have show and will remove when it has show class.

And when you are checking if it visible or not, You can follow this jQuery code,


Above code will return a boolean (true) when #myID element has our class (show) and false when it don't have the (show) class.

Solution 54 - Javascript

isHidden = function(element){
    return ( === "none");

if(isHidden($("element")) == true){
    // Something

Solution 55 - Javascript

You can use jQuery's is() function to check the selected element visible or hidden. 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 returns false.

	($("#myelement").is(":visible"))? alert("#myelement is visible") : alert("#myelement is hidden");

Solution 56 - Javascript != 'none'

function toggle() {
  let visible= != 'none'
  console.log('visible:', visible);

<script src=""></script>

<button onclick="toggle()">Show/hide</button>
<div id="content">ABC</div>

Solution 57 - Javascript

Instead of writing an event for every single element, do this:

  if($(this).css('display') === 'none'){

Also you can use it on the inputs:

  if($(this).attr('type') === 'hidden'){
    $(this).attr('type', 'text');

Solution 58 - Javascript

The below code checks if an element is hidden in jQuery or visible

// You can also do this...

            // show hide paragraph on button click
            $("p").toggle("slow", function(){
                // check paragraph once toggle effect is completed
                    alert("The paragraph  is visible.");
                } else{
                    alert("The paragraph  is hidden.");

Solution 59 - Javascript

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

  else if ($("#accordionZiarat").is(":hidden")) {



Solution 60 - Javascript

The easiest answer to this question is this:

function checkUIElementVisible(element) {
    return ((element.css('display') !== 'none') && (element.css('visibility') !== 'hidden'));

Solution 61 - Javascript

$('someElement').on('click', function(){ $('elementToToggle').is(':visible')

Solution 62 - Javascript

if($(element).is(":visible")) {
  console.log('element is visible');
} else {
  console.log('element is not visible');

Solution 63 - Javascript

Use any of visible Selector or hidden Selector to check visiblity:

  1. Use :visible Selector - jQuery( ":visible" )
  2. Use :hidden Selector - jQuery( ":hidden" )

use .toggle() - Display and hide element

function checkVisibility() {
    // check if element is hidden or not and return true false

    // check if element is visible or not and return true false

    if ( $('#element').css('display') == 'none' || $('#element').css("visibility") == "hidden"){
        console.log('element is hidden');
    } else {
        console.log('element is visibile');

$('#toggle').click(function() {

<script src=""></script>
<button id='toggle'>Toggle</button>
<div style='display:none' id='element'>
    <p>it's visible</p>

Solution 64 - Javascript

if you hide with class - d-none

if (!$('#ele').hasClass('d-none')) {
        $('#ele').addClass('d-none'); //hide 




All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionPhilip MortonView Question on Stackoverflow
Solution 1 - JavascriptTsvetomir TsonevView Answer on Stackoverflow
Solution 2 - JavascripttwerntView Answer on Stackoverflow
Solution 3 - JavascriptMoteView Answer on Stackoverflow
Solution 4 - JavascriptaaronLileView Answer on Stackoverflow
Solution 5 - Javascriptuser574889View Answer on Stackoverflow
Solution 6 - JavascriptSimon_WeaverView Answer on Stackoverflow
Solution 7 - JavascriptPedro RainhoView Answer on Stackoverflow
Solution 8 - JavascriptwebvitalyView Answer on Stackoverflow
Solution 9 - JavascriptAbiyView Answer on Stackoverflow
Solution 10 - JavascriptMatt BrockView Answer on Stackoverflow
Solution 11 - JavascriptDavid LevinView Answer on Stackoverflow
Solution 12 - JavascriptCode SpyView Answer on Stackoverflow
Solution 13 - JavascriptScoRpionView Answer on Stackoverflow
Solution 14 - JavascriptVaishuView Answer on Stackoverflow
Solution 15 - JavascriptLucasView Answer on Stackoverflow
Solution 16 - JavascriptconceptdeluxeView Answer on Stackoverflow
Solution 17 - JavascriptManeesh KumarView Answer on Stackoverflow
Solution 18 - JavascriptIrfan DANISHView Answer on Stackoverflow
Solution 19 - JavascriptMatthias WegtunView Answer on Stackoverflow
Solution 20 - JavascriptRoss BrasseauxView Answer on Stackoverflow
Solution 21 - JavascriptAlekoView Answer on Stackoverflow
Solution 22 - JavascriptRoman LosevView Answer on Stackoverflow
Solution 23 - JavascriptPremshankar TiwariView Answer on Stackoverflow
Solution 24 - JavascriptMathias StavrouView Answer on Stackoverflow
Solution 25 - JavascriptpixellabmeView Answer on Stackoverflow
Solution 26 - JavascriptAndronView Answer on Stackoverflow
Solution 27 - JavascriptRN KushwahaView Answer on Stackoverflow
Solution 28 - JavascriptV31View Answer on Stackoverflow
Solution 29 - JavascriptcssimsekView Answer on Stackoverflow
Solution 30 - JavascriptGauravView Answer on Stackoverflow
Solution 31 - JavascriptKareemView Answer on Stackoverflow
Solution 32 - JavascriptPrabhagaranView Answer on Stackoverflow
Solution 33 - JavascriptlmcDevloperView Answer on Stackoverflow
Solution 34 - JavascriptNo oneView Answer on Stackoverflow
Solution 35 - JavascriptArun KarnawatView Answer on Stackoverflow
Solution 36 - JavascriptAlirezaView Answer on Stackoverflow
Solution 37 - JavascriptOriolView Answer on Stackoverflow
Solution 38 - JavascriptAbrar JahinView Answer on Stackoverflow
Solution 39 - JavascriptSky YipView Answer on Stackoverflow
Solution 40 - JavascriptSangeet ShahView Answer on Stackoverflow
Solution 41 - JavascriptDisapamokView Answer on Stackoverflow
Solution 42 - JavascriptWolfackView Answer on Stackoverflow
Solution 43 - JavascriptL Y E S - C H I O U K HView Answer on Stackoverflow
Solution 44 - JavascriptAbdul Aziz Al BasyirView Answer on Stackoverflow
Solution 45 - JavascriptAntoine AuffrayView Answer on Stackoverflow
Solution 46 - JavascriptHasee AmarathungaView Answer on Stackoverflow
Solution 47 - JavascriptProfesor08View Answer on Stackoverflow
Solution 48 - JavascriptMuhammadView Answer on Stackoverflow
Solution 49 - JavascriptBraneView Answer on Stackoverflow
Solution 50 - JavascriptPeter WoneView Answer on Stackoverflow
Solution 51 - JavascriptVicky PView Answer on Stackoverflow
Solution 52 - JavascriptcbertelegniView Answer on Stackoverflow
Solution 53 - JavascriptDisapamokView Answer on Stackoverflow
Solution 54 - Javascriptuser8903269View Answer on Stackoverflow
Solution 55 - JavascriptAravinda MeewalaarachchiView Answer on Stackoverflow
Solution 56 - JavascriptKamil KiełczewskiView Answer on Stackoverflow
Solution 57 - Javascriptuser10145552View Answer on Stackoverflow
Solution 58 - JavascriptAnkush KumarView Answer on Stackoverflow
Solution 59 - JavascriptMojtaba NavaView Answer on Stackoverflow
Solution 60 - JavascriptEnriproView Answer on Stackoverflow
Solution 61 - Javascriptuser6119825View Answer on Stackoverflow
Solution 62 - Javascriptcentralhubb.comView Answer on Stackoverflow
Solution 63 - JavascriptSahil ThummarView Answer on Stackoverflow
Solution 64 - Javascriptudorb bView Answer on Stackoverflow