你能在SASS/SCSS CSS中使用多个条件语句吗？

Question

我正在使用SCSS代码为我的ruby应用程序进行样式设置，并尝试编写自己的“rounded” mixin以帮助多浏览器边框圆角化。

目前我有以下代码：

@mixin rounded($corner: all , $radius: 8px) {
  @if $corner==all || $corner==bottom || $corner == right || $corner==bottom-right{webkit-border-bottom-right-radius: $radius;}
  @if $corner==all || $corner==bottom || $corner == left || $corner==bottom-left{-webkit-border-bottom-left-radius: $radius;}
  @if $corner==all || $corner==top || $corner == right || $corner==top-right{-webkit-border-top-right-radius: $radius;}
  @if $corner==all || $corner==bottom || $corner == left || $corner==top-left{-webkit-border-top-left-radius: $radius;}

  @if $corner==all || $corner==bottom || $corner == right || $corner==bottom-right{-khtml-border-radius-bottomright: $radius;}
  @if $corner==all || $corner==bottom || $corner == left || $corner==bottom-left{-khtml-border-radius-bottomleft: $radius;}
  @if $corner==all || $corner==top || $corner == right || $corner==top-right{-khtml-border-radius-topright: $radius;}
  @if $corner==all || $corner==bottom || $corner == left || $corner==top-left{-khtml-border-radius-topleft: $radius;}

  @if $corner==all || $corner==bottom || $corner == right || $corner==bottom-right{-moz-border-radius-bottomright: $radius;}
  @if $corner==all || $corner==bottom || $corner == left || $corner==bottom-left{-moz-border-radius-bottomleft: $radius;}
  @if $corner==all || $corner==top || $corner == right || $corner==top-right{-moz-border-radius-topright: $radius;}
  @if $corner==all || $corner==bottom || $corner == left || $corner==top-left{-moz-border-radius-topleft: $radius;}

  @if $corner==all || $corner==bottom || $corner == right || $corner==bottom-right{border-bottom-right-radius: $radius;}
  @if $corner==all || $corner==bottom || $corner == left || $corner==bottom-left{border-bottom-left-radius: $radius;}
  @if $corner==all || $corner==top || $corner == right || $corner==top-right{border-top-right-radius: $radius;}
  @if $corner==all || $corner==bottom || $corner == left || $corner==top-left{border-top-left-radius: $radius;}
}

然而，似乎 SASS 只能处理 if 语句中的一个条件？有方法可以解决这个问题吗，还是我必须为每个圆角做四个 if 语句？

- Pete Hamilton

Answer 1

你需要使用or，而不是||。请参阅Sass文档。

另外，在每个块的最后一个@if语句中似乎有一个拼写错误：

$corner==bottom should be $corner==top

- Bob Nadler

Answer 2

我这样写: 希望你会觉得它有用。

@mixin rounded($amount: "10px",$position: null) {
  @if $position != null {
    @if $position == "top" or $position == "bottom" {
      // top or bottom 
      -webkit-border#{$position}-left-radius: $amount;
      -moz-border#{$position}-left-radius: $amount;
      border#{$position}-left-radius: $amount;

      -webkit-border#{$position}-right-radius: $amount;
      -moz-border#{$position}-right-radius: $amount;
      border#{$position}-right-radius: $amount;

    } @else {
      // top-left or top-right or bottom-left or bottom-right
      -webkit-border#{$position}-radius: $amount;
      -moz-border#{$position}-radius: $amount;
      border#{$position}-radius: $amount;      
    }
  } @else {
    // ALL IF EMPTY
    -webkit-border-radius: $amount;
    -moz-border-radius: $amount;
    border-radius: $amount; 
  }

}

你可以这样使用它：

  @include rounded(); /*as default 10px on all corners*/
  @include rounded(15px); /*all corners*/ 

  @include rounded(15px, top); /*all top corners*/ 
  @include rounded(15px, bottom); /* all bottom corners*/ 

  @include rounded(15px, top-right); /*this corner only*/ 
  @include rounded(15px, top-left); /*this corner only*/ 
  @include rounded(15px, bottom-right); /*this corner only*/ 
  @include rounded(15px, bottom-left); /*this corner only*/ 

- DevWL

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一个更好的解决方案 - Dan Osborne