Hover and Active only when not disabled

Question

You can use :enabled pseudo-class, but notice IE<9 does not support it:

button:hover:enabled{
    /*your styles*/
}
button:active:enabled{
    /*your styles*/
}

.button:active:hover:not([disabled]) {
    /*your styles*/
}

You can try this..

Why not using attribute "disabled" in css. This must works on all browsers.

button[disabled]:hover {
    background: red;
}
button:hover {
    background: lime;
}

If you are using LESS or Sass, You can try this:

.btn {
  &[disabled] {
    opacity: 0.6;
  }
  &:hover, &:active {
    &:not([disabled]) {
      background-color: darken($orange, 15);
    }
  }
}

In sass (scss):

 button {
  color: white;
  cursor: pointer;
  border-radius: 4px;

  &:disabled{
    opacity: 0.4;

    &:hover{
      opacity: 0.4;  //this is what you want
    }
  }

  &:hover{
    opacity: 0.9;
  }
}

Use the lightest touch: overriding via rule order.

.btn {
  /* base styles */
}

.btn:hover {
  color: red;
}

.btn:active {
  color: green;
}

.btn:disabled {
  color: #aaa;
}

The trick is the order -- apply all the non-disabled states first, makes sure they all have the same specificity, and do disabled last, with the same specificity.

This won't work for a class added to links, or non-interactive elements, which don't have the disabled property.

(Edited to remove higher-specificity rules, and messing with pointer-events)

One way is to add a partcular class while disabling buttons and overriding the hover and active states for that class in css. Or removing a class when disabling and specifying the hover and active pseudo properties on that class only in css. Either way, it likely cannot be done purely with css, you'll need to use a bit of js.