使用Javascript从具有属性选择器的CSS类更改属性的状态

Question

I have this HTML line:

<div id="content" class="slide" data-position="left" ></div>

and this CSS class with this [attribute] selector:

.slide[data-position="left"] {
    display: none;
}

What I'm trying to do is after I click a button the css class will change like this using Javascript

.slide[data-position="left"] {
    display: block;
}

I tried different methods like adding a new class in the HTML line or adding a inline style="display: block;" using Javascript:

$(".slide").attr('data-position', 'left').css({'display':'block'});

that generates

<div id="content" class="slide" data-position="left" style="display: block;"></div>

but the .css() method doesn't help me at all because I want to change/override .slide[data-position="left"]. The inline doesn't change anything in the CSS stylesheet.

I want to know how to change the state of a property for a specific css class that has a certain [attribute] selector.

I tried to google this issue but it seems that nobody has an answer for this.

Any ideas?

Thanks!

Answer 1

Instead of altering the CSS rule directly, you can give the element another class corresponding to another rule:

.slide.active { display: block; }

or, if the attribute has importance still:

.slide.active[data-position="left"] { display: block; }

That rule should go after the existing rule to ensure that a "slide" element that also has the "active" class will be visible.

You'd add the class in JavaScript with

$("#content").addClass("active");

Answer 2

You can achieve this by changing your jQuery selector. At the moment you're using jQuerys attr method to target the data-position attribute and set its value to left. Instead you need to use the same attribute selector you previously used in your CSS declaration: .slide[data-position=left]

What you need to do is change this:

$(".slide").attr('data-position', 'left').css({'display':'block'});

to this:

$(".slide[data-position=left]").css({'display':'block'});

The complete jQuery would look like this:

$("#clickMe").on("click",function(){

    $(".slide[data-position=left]").css({'display':'block'});

});

Demo