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 

文章来源: Stylus Iteration + Interpolation with nth-of-type