Question

我是sass的新手，我写了一些sass代码，但未编译。

 $classes : primary secondary success warning danger;
    $colors : (primary:#007bff,secondary : #6c757d,success: #28a745,warning: #ffc107,dangaer: #dc3545);
    @each $class in $classes{
      .btn-#{$class}{
        $currentColor: map-get($colors,#{$class});
        background:linear-gradient(to right,$currentColor,lighten($currentColor,10%));
      }
    }

错误是：

$color: null is not a color.
stdin 14:55  root stylesheet on line 14 at column 55

但是当我用变量替换线性梯度时，它可以正常工作，即

$classes : primary secondary success warning danger;
$colors : (primary:#007bff,secondary : #6c757d,success: #28a745,warning: #ffc107,dangaer: #dc3545);

    @each $class in $classes{
      .btn-#{$class}{
        $currentColor: map-get($colors,#{$class});
        background:$currentColor;
        //background:linear-gradient(to right,$currentColor,lighten($currentColor,10%));
      }
    }

这是代码已成功编译。

linear-gradient（）函数中$ currentColor变量的作用或作用是什么

Answer 1

实际上，有些东西可以将变量从map-get传递给其他sass函数。

但是您可以稍微修改一下代码，它将起作用：

$classes: primary secondary success warning danger;
$colors: (
    primary: ( normal: #007bff, light: lighten(#007bff,10%) ),
    secondary: ( normal: #6c757d, light: lighten(#6c757d,10%) ),
    success: ( normal: #28a745, light: lighten(#28a745,10%) ),
    warning: ( normal: #ffc107, light: lighten(#ffc107,10%) ),
    danger: ( normal: #dc3545, light: lighten(#dc3545,10%) )
);
@each $class in $classes{
  .btn-#{$class}{
    $currentColor: map-get(map-get($colors,#{$class}), normal);
    $currentColorLighten: map-get(map-get($colors,#{$class}), light);

    background: linear-gradient(to right, $currentColor, $currentColorLighten);
  }
}

您为每个类定义了两种颜色（普通版和浅色版），只需通过double map-get即可使用。

sass complie错误，提示$ color：即使我声明了color，null也不是颜色

1 个答案: