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csscss-selectors

CSS Last Odd Child?


I have an unordered list, which can contain either an even or odd number of items. I'm looking for a CSS-only way to remove the border from the last 2 li tags if the number of lis is even. The :last-child pseudo-selector removes the last one regardless.

li {
float: left;
border-bottom: 1px solid #000;
}

li:last-child{
border-bottom: none;
}

Works for Odd Numbers of lis

+============================================+
+          1          |          2           +
+--------------------------------------------+
+          3          |                      +
+============================================+

But for even numbers, I need to remove the bottom of cell #3

+============================================+
+          1          |          2           +
+--------------------------------------------+
+          3          |          4           +
+---------------------                       +
+============================================+

So I figured I could use li:nth-last-child() but I can't figure out what should be the equation to grab the last odd child.

It's not 2n+1, 2n-1, n-1, or anything I can think of. Please help.


Solution

  • nth-last-child counts backwards from the last child, so to grab the second to last, the expression is:

    li:nth-last-child(2)
    

    You can combine pseudo-selectors, so to select the 2nd to last child, but only when it's odd, use:

    li:nth-last-child(2):nth-child(odd) {border-bottom: none;}
    

    And so, the whole thing should be:

    li:last-child,
    li:nth-last-child(2):nth-child(odd) {border-bottom: none;}
    

    In answer to @ithil's question, here's how I'd write it in SASS:

    li
      &:last-child,
      &:nth-last-child(2):nth-child(odd)
        border-bottom: none
    

    It's not that much simpler, since the selection of the 'second-to-last odd child' is always going to require the 'two step' selector.

    In answer to @Caspert's question, you can do this for arbitrarily many last elements by grouping more selectors (there feels like there should be some xn+y pattern to do this without grouping, but AFAIU these patterns just work by counting backwards from the last element).

    For three last elements:

    li:last-child,
    li:nth-last-child(2):nth-child(odd),
    li:nth-last-child(3):nth-child(odd) {border-bottom: none;}
    

    This is a place where something like SASS can help, to generate the selectors for you. I would structure this as a placeholder class, and extend the element with it, and set the number of columns in a variable like this:

    $number-of-columns: 3
    
    %no-border-on-last-row
     @for $i from 1 through $number-of-columns
       &:nth-last-child($i):nth-child(odd)
         border-bottom: none
    
    //Then, to use it in your layout, just extend:
    
    .column-grid-list li
      @extend %no-border-on-last-row