Select last child when odd, 2 last childs when even

Question

I'm in a situation where the number of elements showed is variable, and I need a strange solution which I'm not able to achieve, I even doubt if it's achievable only with css.

I need to select the last-child if my number of elements is odd, and the last 2 child if the number of elements is even.

I've been trying with nth-last-child, :not(:nth-last-child()), odd and even, but never got a good solution.

Anyone has any idea/advice about this issue a part of adding a class "odd" like on html tables?

Thanks a lot in advance!

Answer 1

You can use CSS like so:

li:last-child:nth-child(odd) {
    /* Last child AND odd */
    background: red;
}

li:nth-last-child(2):nth-child(odd),
li:last-child:nth-child(even) {
    /* Before last child AND odd */
    /* Last child AND even */
    background: green;
}

Demo: https://jsfiddle.net/hw0ehrhy/

Answer 2

Here is one way...

.wrap div:last-child,
.wrap div:nth-last-of-type(-n+2):not(:nth-child(even)) {
    color: red;
}

<div class="wrap">
    <div>Odd</div>
    <div>Even</div>
    <div>Odd</div>
    <div>Even</div>
    <div>Odd</div>
    <div>Even</div>
</div>

<hr>

<div class="wrap">
    <div>Odd</div>
    <div>Even</div>
    <div>Odd</div>
    <div>Even</div>
    <div>Odd</div>
</div>