SASS #if statement using variable name - if-statement

Question

I'm having an issue generating the correct code with the #if statement based on a variable.
SASS
=band($name, $color)
.band-#{$name}
background-color: #{$color}
#if $name == white
font-size: 100px
#else
font-size: 20px
+band('white', $white)
+band('black', $black)
to generate
.band-white{background-color:white; font-size:100px;}
.band-black{background-color:black; font-size:50px;}

Answer 1

The issue is that your mixin is comparing the variable that is coming in with a color.
You are passing in the string 'white' which is not the same as the HTML color white.
Have a look at the SassScript DataTypes.
So, to fix your Sass, you will have to either change your code to
=band($name, $color)
.band-#{$name}
background-color: #{$color}
#if $name == 'white'
font-size: 100px
#else
font-size: 20px
+band('white', $white)
+band('black', $black)
or
=band($name, $color)
.band-#{$name}
background-color: #{$color}
#if $name == white
font-size: 100px
#else
font-size: 20px
+band(white, $white)
+band(black, $black)
The latter will not give you anything useful if you pass in anything else than a color as the first argument.
By the way, you could also simplify the condition by using a single-line if in case you are using Sass 3.3 or higher:
font-size: if($name == white, 100px, 20px)

Answer 2

I suspect it's a white space issue. Do you get anything useful in the logs?
I was able to get a version of your code working:
#mixin band($name)
background-color: $name
#if $name == white
font-size: 50px
#else
font-size: 20px
.band-white
#include band($name: white)
.band-black
#include band($name: black)
And you can see it working here https://codepen.io/sitrobotsit/pen/aWQxNE