Stylus Iteration + Interpolation with nth-of-type

问题

I'm attempting to use the counter provided when looping thru a list of items like so:

colors = red blue orange green yellow

li
    for color, i in colors
        &:nth-of-type({i}n)
            background-color: color

This example does not work, but the intended output I'm looking for is:

li:nth-of-type(1n) {
    background-color: red;
}
li:nth-of-type(2n) {
    background-color: blue;
}
li:nth-of-type(3n) {
    background-color: orange;
}
...

Is this possible?

回答1:

Actually your example's output is almost correct. It starts with 0 and you need 1, so this should work:

colors = red blue orange green yellow

li
    for color, i in colors
        &:nth-of-type({i + 1}n)
            background-color: color

来源：https://stackoverflow.com/questions/34410889/stylus-iteration-interpolation-with-nth-of-type