Selector for one tag directly followed by another tag

Question

This selects all tags directly preceded by tags:

A+B {
    /* styling */
}

What is t

Answer 1

Although it's not very handy, nowadays you could achieve this behavior by reversing the order of your elements both when you generate the HTML and by applying the CSS rules: display: flex and flex-direction: column-reverse

ul {
  display: flex;
  flex-direction: column-reverse;
}

.b ~ .a {
  color: red;
}

<ul>
    <li class="a">A 3</li>
    <li class="c">C 2</li>
    <li class="c">C 1</li>
    <li class="b">B 1</li>
    <li class="a">A 2</li>
    <li class="a">A 1</li>
</ul>

Also, if you have 2 or more inline elements, you could achieve it by applying float: right, as they will be displayed in reverse order:

ul {
  float: left;
  list-style-type: none;
}

li {
  float: right;
}

li:not(:first-child) {
  margin-right: 20px;
}

.b ~ .a {
  color: red;
}

<ul>
    <li class="a">A 3</li>
    <li class="c">C 2</li>
    <li class="c">C 1</li>
    <li class="b">B 1</li>
    <li class="a">A 2</li>
    <li class="a">A 1</li>
</ul>

Answer 2

You can ONLY do the converse: This selects all tags directly preceded by tags.

This is logically equivalent to your request.

I often use this to style a row of many checkboxes with labels

CSS:

label+input {
    margin-left: 4px;
}

DOM:

<input id="a" name="a" type="checkbox"/><label for="a">...</label>
<input id="b" name="b" type="checkbox"/><label for="b">...</label>
<input id="c" name="c" type="checkbox"/><label for="c">...</label>

Answer 3

You can’t in css.

Edit: To be a bit more helpful, if you use for example jQuery (a JavaScript library), you can use .prev().

Answer 4

Do you mean to style A given that it has a B element directly inside or followed? Like this:

<A>
    <B>
    </B>
</A>

// OR

<A>
</A>
<B>
</B>

You can't do such a thing in CSS (yet). Eric Meyer states that this kind of selector has been discussed quite a few times on the CSS mailing list, and isn’t doable. Dave Hyatt, one of the core WebKit developers, comments with a good explanation of why it can’t be done.

Check out: Shaun Inman's blog post and the comment by Eric Meyer.
David Hyatt weighs in, too.