Ways to retrieve the precise value of a CSS attribute even when the property is overridden in a CSS class

If I have an li element like this:

<li id="1" class="show" style="display: none;">

and the CSS class contains:

.show {
  display: list-item !important;
}

I attempted to retrieve the value of the display attribute with:

document.getElementById("1").style.display

which returned "none".

Despite the CSS class, the real display value is list-item. How can I use JavaScript to get the correct and applied display value?

Answer №1

When it comes to determining the style properties of an element, Window.getComputedStyle() is a useful tool. You can see an example of it in action at this link.

var target = document.getElementById("list-item-1");
var displayProperty = window.getComputedStyle(target, null).getPropertyValue("display");

Answer №3

JavaScript Code: utilizing the getComputedStyle method

var element = document.getElementById("1");
 alert(window.getComputedStyle(element, null).getPropertyValue("display"));
.show {
  display: list-item !important;
}
<li id="1" class="show" style="display: none;">

jQuery Example: utilizing the .css method

alert($('#1').css('display'));
.show {
  display: list-item !important;
}
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<li id="1" class="show" style="display: none;">

Answer №4

Searching for the particular style being used to display the element? You'll want to utilize window.getComputedStyle:

var displayProperty = window.getComputedStyle(element).display

On a side note, please remember that an id cannot start with a number as it violates the rules of HTML.

Answer №5

It is simply retrieving the value of the display property from the style attribute that is present in your markup.

console.log(document.getElementById("1").style.display)
.show {
  display: list-item !important;
}
<li id="1" class="show" style="display: none;">adsfds</li>

However, when you inspect the element in the browser, you can see that the display: list-item !important; style is being applied.

Refer to the image linked below for a visual representation.

https://i.sstatic.net/kjiUr.jpg

To obtain the actual applied style, you can use the getComputedStyle function as shown below.

var elem = document.getElementById("1");

console.log(window.getComputedStyle(elem).display)
.show {
  display: list-item !important;
}
<li id="1" class="show" style="display: none;">adsfds</li>

Note : It is unnecessary to explicitly set the display property to 'list-item' for a list item, as it is the default behavior.

Answer №6

Avoid using inline CSS styles by eliminating style="display: none;" from your HTML code. Instead, establish a .hidden { display: none; } class within your CSS file and utilize it to hide or show elements as needed. This way, you can easily determine the visibility of an element by examining its assigned CSS class.

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