SASS ignores variables, defined in if-statement

问题

I have one file named style.scss with the following code:

@import 'variables';

body {
    color: $text-color;
    background: $background-color;
}

And one partial named _variables.scss:

$colorscheme: white;

@if $colorscheme == white {
    $text-color: #333;
    $background-color: #fff;
}
@else {
    $text-color: #ccc;
    $background-color: #333;
}

The if-statement works properly, but the variables defined inside, do not work. When I try to compile it, I keep getting:

Syntax error: Undefined variable: “$text-color”.

回答1:

That's completely expected. Variables have a scope to them. If you define them inside of a control block (like an if statement), then they won't be available outside. So, what you need to do is initialize it outside like so:

$text-color: null;
$background-color: null;
@if $colorscheme == white {
    $text-color: #333;
    $background-color: #fff;
}
@else {
    $text-color: #ccc;
    $background-color: #333;
}

Or...

$text-color: #ccc;
$background-color: #333;
@if $colorscheme == white {
    $text-color: #333;
    $background-color: #fff;
}

Though it would be less verbose to use the if() function like this:

$text-color: if($colorscheme == white, #333, #ccc);
$background-color: if($colorscheme == white, #fff, #333);

回答2:

While this example is even more verbose, it eliminates the need to set two empty variables:

@if $colorscheme == white {
  $text-color: #333 !global;
  $background-color: #fff !global;
} @else {
  $text-color: #ccc !global;
  $background-color: #333 !global;
}

@cimmanon's 2nd and 3rd examples are much better, though.

来源：https://stackoverflow.com/questions/15371332/sass-ignores-variables-defined-in-if-statement

标签

variables

if-statement

Sass