nth-child with mod (or modulo) operator

Is it possible to use the nth-child with modulo? I know you can specify a formula, such as

nth-child(4n+2)

But I can't seem to find if there's a modulo operator. I've tried the following examples below, and none seem to work:

nth-child(n%7)
nth-chil开发者_运维技巧d(n % 7)
nth-child(n mod 7)

If you want to use nth-child with modulo k, just specify:

nth-child(kn)

For example, if you want to specify style for 3, 6, 9, .. elements, just specify (k = 3), and use:

nth-child(3n)

You can also specify an offset:

nth-child(kn+offset)

No, :nth-child() only supports addition, subtraction and coefficient multiplication.

I gather you're trying to pick up the first 6 elements (as n mod 7 for any positive integer n only gives you 0 to 6). For that, you can use this formula instead:

:nth-child(-n+6)

By negating n, element counting is done backwards starting from zero, so these elements will be selected:

 0 + 6 = 6
-1 + 6 = 5
-2 + 6 = 4
-3 + 6 = 3
-4 + 6 = 2
-5 + 6 = 1
...

jsFiddle demo

I know this topic is very old, but here's the answer that's close enough to modulo operator:

:not(:nth-child(7n))

This will select elements 1-6, 8-13 and so on...

:nth-child(an)

The code above will select elements divisible by "a" so adding :not() selector forces CSS to select elements not divisible by "a".

I know this topic is very old, but here's the answer:

    .parent :nth-child(4n):nth-last-child(2) ~ *,
    .parent :nth-child(4n):nth-last-child(3) ~ *,
    .parent :nth-child(4n):nth-last-child(4) ~ * {
        color: gold;
    }

This selector targets all children who are after the last child in the last full set of grouped children denoted by your nth-child formula.

If you want modulous with n=2 than you can use even or odd.

tr:nth-child(even) {
  background-color: #dddddd;
}