How to make a mixin with the same name as the value of a variable?

Question

I have variable and need to construct mixin with its value:

$param = 'color';

{$param}() {
  color: green;
}

.foo {
  color: red;
}

It's renders as

color() {
  color: #008000;
}
.foo {
  color: #f00;
}

But I need

.foo {
  color: #008000;
}

Maybe, I don't know, how to do it right.
How can I get it?

@Ser-Gen, Okay that would be `#{}` interpolation syntax. And that doesn't work with mixins unfortunately. See updated answer. Also see.. http://stackoverflow.com/questions/15748937/how-to-define-a-dynamic-mixin-name-in-sass — khollenbeck, Jul 14 '14 at 19:27
@kariem, no, I need variable value in name of mixin. This example is just to demonstrate, not use case. — Ser-Gen, Jul 14 '14 at 19:32
@KrisHollenbeck, my question is about Stylus. But thanks, your answer is interesting. Maybe the same problem in Stylus too. — Ser-Gen, Jul 14 '14 at 19:35
Ah, my bad.. I thought you where I asking about SASS. Guess I should have looked at the tags. — khollenbeck, Jul 14 '14 at 19:36

score 2 · Accepted Answer · answered Jul 15 '14 at 16:58

2

I have got answer from maintainer of Stylus:

You can do this with define bif and anonymous functions:

$param = 'color'

define($param, @() {
  color green
})

.foo
  color red

answered Jul 15 '14 at 16:58

Ser-Gen

1 Answers1