Sass 错误“函数 X 在没有@return 的情况下完成" [英] Sass error "Function X finished without @return"

问题描述

我在尝试将相当复杂的渐变转换为 SASS 的 mixin 时遇到了一些麻烦，大多数渐变都很好，而且我相信我的色标也正确，我相信这是我的 MIXIN 不允许的定位我这样做(左上，右下).

I'm having some trouble trying to convert a fairly complex gradient, into a mixin for SASS, most gradients are fine, and I believe my colour stops are also correct, I believe it's the positioning that my MIXIN doesn't let me do this (left top, right bottom).

我需要这个:

.progress.stripes.animate > .bar > span {
  background-image: -webkit-gradient(
  linear, left top, right bottom, 
  color-stop(.25, rgba(255, 255, 255, .2)), 
  color-stop(.25, transparent), 
  color-stop(.5, transparent), 
  color-stop(.5, rgba(255, 255, 255, .2)), 
  color-stop(.75, rgba(255, 255, 255, .2)), 
  color-stop(.75, transparent), to(transparent));
  background-image: -moz-linear-gradient(-45deg, rgba(255, 255, 255, .2) 25%, transparent 25%, transparent 50%, rgba(255, 255, 255, .2) 50%, rgba(255, 255, 255, .2) 75%, transparent 75%, transparent);
}

要使用此 mixin:

@mixin linear-gradient($gradientLine, $colorStops...) {
  background-color: nth($colorStops,1);
  background-image: -webkit-gradient(linear, $gradientLine, $colorStops);
  background-image: -webkit-linear-gradient($gradientLine, $colorStops);
  background-image:    -moz-linear-gradient($gradientLine, $colorStops);
  background-image:      -o-linear-gradient($gradientLine, $colorStops);
  background:             -ms-linear-gradient($gradientLine, $colorStops);
  @if length($colorStops) == 2 {
    $colour1:nth($colorStops,1);
    $colour2:nth($colorStops,2);
    filter: progid:DXImageTransform.Microsoft.gradient( startColorstr='#{$colour1}', endColorstr='#{$colour2}',GradientType=0 );
  }  
  @if length($gradientLine) == 2 {
  background-image:         linear-gradient(to #{inverse-side(nth($gradientLine, 1))} #{inverse-side(nth($gradientLine, 2))}, $colorStops);
  } @else {
  background-image:         linear-gradient(to #{inverse-side($gradientLine)}, $colorStops);
  }
}

这是我尝试过的，但不起作用...

$grad: rgba(255, 255, 255, .2);
.progress.stripes.animate > .bar > span {
  @include linear-gradient((left top, right bottom),$grad 25%, $transparent 25%, $transparent 5%, $grad 5%, $grad 75%, $transparent 75%);
}

推荐答案

Sass 在错误消息中告诉您这不起作用的原因:Function inverse-side finished without @return.

Sass is telling you the reason this isn't working in the error message: Function inverse-side finished without @return.

@function inverse-side($side) {
    @if $side == top {
        @return bottom;
    } 
    @else if $side == bottom {
        @return top;
    }
    @else if $side == left {
        @return right;
    }
    @else if $side == right {
        @return left;
    }
}

您在这里只考虑了 4 个条件:上、右、下、左.有第五个选项被忽略了:以上都不是.如果应该只有 4 个选项，那么您需要 3 个 if/else 块和最后一个返回值，这就是您的全部.

You're only accounting for 4 conditions here: top, right, bottom, left. There's a 5th option that's being overlooked: none of the above. If there should only be 4 options, then you want 3 if/else blocks and a return value at the very end that's your catch all.

@function inverse-side($side) {
    @if $side == top {
        @return bottom;
    } 
    @else if $side == bottom {
        @return top;
    }
    @else if $side == left {
        @return right;
    }
    @return left;
}

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