Is it possible to transfer the class name to the sass through the mixin variable? like this

@mixin colors ($qq){ $qq{ color:red; } } @include colors(myClass);

in the end I want to get this result

 .myClass{ color:red; }

    2 answers 2

    figured out, suddenly someone needs an answer

     @mixin colors ($qq){ .#{$qq}{ color:red; } } @include colors(myClass); 
       $color: ( $blue: blue, $red: red, $yellow: yellow, ); $border: map-get($color, $blue);

      Would such an example suit, if you make a start from this? :)