CSS3 :unchecked pseudo-class

CssCss SelectorsPseudo Class

Css Problem Overview


I know there is an official CSS3 :checked pseudo-class, but is there an :unchecked pseudo-class, and do they have the same browser support?

Sitepoint's reference doesn't mention one, however this whatwg spec (whatever that is) does.

I know the same result can be achieved when the :checked and :not() pseudo-classes are combined, but i'm still curious:

input[type="checkbox"]:not(:checked) {
    /* styles */
}

Edit:

The w3c recommends the same technique

> An unchecked checkbox can be selected by using the negation pseudo-class: > > :not(:checked)

Css Solutions


Solution 1 - Css

:unchecked is not defined in the Selectors or CSS UI level 3 specs, nor has it appeared in level 4 of Selectors.

In fact, the quote from W3C is taken from the Selectors 4 spec. Since Selectors 4 recommends using :not(:checked), it's safe to assume that there is no corresponding :unchecked pseudo. Browser support for :not() and :checked is identical, so that shouldn't be a problem.

This may seem inconsistent with the :enabled and :disabled states, especially since an element can be neither enabled nor disabled (i.e. the semantics completely do not apply), however there does not appear to be any explanation for this inconsistency.

(:indeterminate does not count, because an element can similarly be neither unchecked, checked nor indeterminate because the semantics don't apply.)

Solution 2 - Css

I think you are trying to over complicate things. A simple solution is to just style your checkbox by default with the unchecked styles and then add the checked state styles.

input[type="checkbox"] {
  // Unchecked Styles
}
input[type="checkbox"]:checked {
  // Checked Styles
}

I apologize for bringing up an old thread but felt like it could have used a better answer.

EDIT (3/3/2016):

W3C Specs state that :not(:checked) as their example for selecting the unchecked state. However, this is explicitly the unchecked state and will only apply those styles to the unchecked state. This is useful for adding styling that is only needed on the unchecked state and would need removed from the checked state if used on the input[type="checkbox"] selector. See example below for clarification.

input[type="checkbox"] {
  /* Base Styles aka unchecked */
  font-weight: 300; // Will be overwritten by :checked
  font-size: 16px; // Base styling
}
input[type="checkbox"]:not(:checked) {
  /* Explicit Unchecked Styles */
  border: 1px solid #FF0000; // Only apply border to unchecked state
}
input[type="checkbox"]:checked {
  /* Checked Styles */
  font-weight: 900; // Use a bold font when checked
}

Without using :not(:checked) in the example above the :checked selector would have needed to use a border: none; to achieve the same effect.

Use the input[type="checkbox"] for base styling to reduce duplication.

Use the input[type="checkbox"]:not(:checked) for explicit unchecked styles that you do not want to apply to the checked state.

Solution 3 - Css

There is no :unchecked pseudo class however if you use the :checked pseudo class and the sibling selector you can differentiate between both states. I believe all of the latest browsers support the :checked pseudo class, you can find more info from this resource: http://www.whatstyle.net/articles/18/pretty_form_controls_with_css

Your going to get better browser support with jquery... you can use a click function to detect when the click happens and if its checked or not, then you can add a class or remove a class as necessary...

Solution 4 - Css

The way I handled this was switching the className of a label based on a condition. This way you only need one label and you can have different classes for different states... Hope that helps!

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionWeb_DesignerView Question on Stackoverflow
Solution 1 - CssBoltClockView Answer on Stackoverflow
Solution 2 - CssMichael StramelView Answer on Stackoverflow
Solution 3 - CssJeff WoodenView Answer on Stackoverflow
Solution 4 - CssDaniel RamView Answer on Stackoverflow