适用于 Sass 3.3 及更高版本

从 Sass 3.3 开始，有一个 variable-exists() 函数.来自变更日志:

• 现在可以使用这些新函数确定不同 Sass 结构的存在:
  • variable-exists($name) 检查变量是否在当前范围内解析.
  • global-variable-exists($name) 检查给定名称的全局变量是否存在....

  • It is now possible to determine the existence of different Sass constructs using these new functions:
    • variable-exists($name) checks if a variable resolves in the current scope.
    • global-variable-exists($name) checks if a global variable of the given name exists. ...

  示例用法:

  $some_variable: 5;
  @if variable-exists(some_variable) {
      /* I get output to the CSS file */
  }
  @if variable-exists(nonexistent_variable) {
      /* But I don't */
  }
  

  对于 Sass 3.2.x 及更早版本(我的原始答案)

  我今天遇到了同样的问题:尝试检查是否设置了变量，如果设置了，则添加样式、使用 mixin 等.

  ---

  For Sass 3.2.x and earlier (my original answer)

  I ran into the same problem today: trying to check if a variable is set, and if so adding a style, using a mixin, etc.

  在阅读了 isset() 函数后 不会被添加到 sass，我找到了一个简单的解决方法，使用 !default 关键字:

  After reading that an isset() function isn't going to be added to sass, I found a simple workaround using the !default keyword:

  @mixin my_mixin() {
    // Set the variable to false only if it's not already set.
    $base-color: false !default;
  
    // Check the guaranteed-existing variable. If it didn't exist 
    // before calling this mixin/function/whatever, this will 
    // return false.
    @if $base-color {
       color: $base-color;
    }
  }
  

  如果 false 是变量的有效值，您可以使用:

  If false is a valid value for your variable, you can use:

  @mixin my_mixin() {
    $base-color: null !default;
  
    @if $base-color != null {
       color: $base-color;
    }
  }