语法错误"transparent"不是'ie-hex-str'的颜色

Question

我正在使用指南针,在In下面的代码我想#fff用透明度代替@include filter-gradient.但是没有透明度的十六进制代码所以我使用transparent但是它给出了错误syntax error "transparent" is not a color for 'ie-hex-str'

@include filter-gradient(#f3f2f3, transparent, vertical);
$experimental-support-for-svg: true;
@include background-image(linear-gradient(top, #f3f2f3 0%,#eaeae9 68%,#cfcece 70%,transparent 73%,transparent 100%));

Answer 1

看起来filter-gradientmixin期待特定的颜色格式:3或6位十六进制或rgb.所以你想传递一种透明的颜色:

@include filter-gradient(#f3f2f3, transparentize(white, 1), vertical);