Sass变量默认范围

Question

我在跨范围的Sass中使用变量默认值时遇到问题.我的测试例子是:

@mixin foo { 
        $val: 'red' !default; 
        .bar { 
                color: $val; 
        } 

} 

@include foo; 
.class1 { 
        $val: 'green'; 
        @include foo; 
        .class11 { 
                @include foo; 
        } 
} 

$val: 'black'; 
.class2 { 
        @include foo; 
} 

.class3 { 
        $val: 'blue'; 
        @include foo; 
} 

.class4 { 
        @include foo; 

}

它编译为:

.bar { 
  color: "red"; 

} 

.class1 .bar { 
  color: "red"; 
} 

.class1 .class11 .bar { 
  color: "red"; 
} 

.class2 .bar { 
  color: "black"; 
} 

.class3 .bar { 
  color: "blue"; 
} 

.class4 .bar { 
  color: "blue"; 

}

如您所见,变量$ val在mixin foo中被定义为'red'!default.我希望导入mixin会将$ val设置为'red',除非它已经定义.但是,在class1中,$ val在本地定义为'green',导入mixin foo会用'red'覆盖它.在其他类中,在$ val的全局定义为"black"之后,导入mixin按预期工作,$ val保留其已定义的值.

我究竟做错了什么？

Answer 1

$val: 'green'在class1中本地定义不会$val: 'red' !default在mixin中改变,因为它寻找全局$ val.此时,尚未定义全局$ val.

然后全局$ val定义为"黑色".在mixin中这个$ val之后寻找全球$ val.此时,全局$ val已定义为"黑色".

在本地再次定义$ val将改变已定义的全局$ val.

@mixin foo 
  $val: 'red' !default // defined locally
  .bar
    color: $val

@include foo // $val in mixin foo look for global $val. no global $val found, then 'red'

.class1
  $val: 'green'
  @include foo // $val in mixin foo look for global $val. no global $val found, then 'red'
  color: $val // local $val 'green'
  .class11 
    @include foo // $val in mixin foo look for global $val. no global $val found, then 'red'

$val: 'black' // defined globally at the first time

.class2 
  @include foo // $val in mixin foo look for global $val. $val found, 'black'

.class3
  $val: 'blue' // change the gobal $val
  @include foo // $val in mixin foo look for global $val. $val found, 'blue'

.class4
  @include foo // $val in mixin foo look for global $val. $val found, 'blue'