Is there a way to increment attribute values using n in nth-child(n)
to output the result of:
div:nth-child(1){
top: -5px;
}
div:nth-child(2){
top: -10px;
}
div:nth-child(3){
top: -15px;
}
div:nth-child(4){
top: -20px;
}
div:nth-child(5){
top: -25px;
}
Syntax. :nth-child() takes a single argument that describes a pattern for matching element indices in a list of siblings. Element indices are 1-based. :nth-child( <nth> [ of <complex-selector-list> ]? )
You can pass in a positive number as an argument to :nth-child() , which will select the one element whose index inside its parent matches the argument of :nth-child() . For example, li:nth-child(3) will select the list item with an index value 3; that is, it will select the third list item.
As a general rule, if you want to select an interval of a selector regardless of the type of element it is, use nth-child . However, if you want to select a specific type only and apply an interval selection from there, use nth-of-type .
You could if you use a preprocessor
like sass
.
An example:
div {
$var: 5;
@for $i from 1 through 5 {
&:nth-child(#{$i}) {
top: -#{$var}px;
$var: $var + 5;
}
}
}
A demo: http://sassmeister.com/gist/7d7a8ed86e857be02d1a
Another way to achieve this:
div {
$list: 1 2 3 4 5;
@each $i in $list {
&:nth-child(#{$i}) {
top: -5px * $i;
}
}
}
A demo: https://www.sassmeister.com/gist/fb5ee7aa774aafb896cd71a828882b99
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