Check element CSS display with JavaScript

Question

As sdleihssirhc says below, if the element's display is being inherited or being specified by a CSS rule, you'll need to get its computed style:

return window.getComputedStyle(element, null).display;

Elements have a style property that will tell you what you want, if the style was declared inline or with JavaScript:

console.log(document.getElementById('someIDThatExists').style.display);

will give you a string value.

If the style was declared inline or with JavaScript, you can just get at the style object:

return element.style.display === 'block';

Otherwise, you'll have to get the computed style, and there are browser inconsistencies. IE uses a simple currentStyle object, but everyone else uses a method:

return element.currentStyle ? element.currentStyle.display :
                              getComputedStyle(element, null).display;

The null was required in Firefox version 3 and below.

For jQuery, do you mean like this?

$('#object').css('display');

You can check it like this:

if($('#object').css('display') === 'block')
{
    //do something
}
else
{
    //something else
}

This answer is not exactly what you want, but it might be useful in some cases. If you know the element has some dimensions when displayed, you can also use this:

var hasDisplayNone = (element.offsetHeight === 0 && element.offsetWidth === 0);

EDIT: Why this might be better than direct check of CSS display property? Because you do not need to check all parent elements. If some parent element has display: none, its children are hidden too but still has element.style.display !== 'none'.

yes.

var displayValue = document.getElementById('yourid').style.display;

Basic JavaScript:

if (document.getElementById("elementId").style.display == 'block') { 
  alert('this Element is block'); 
}

To find out if it's visible with plain JavaScript, check whether the display property is 'none' (don't check for 'block', it could also be blank or 'inline' and still be visible):

var isVisible = (elt.style.display != "none");

If you are using jQuery, you can use this instead:

var isVisible = $elt.is(":visible");