为什么SASS不能像`&:: not(:first-child)那样编译pseudo_expr？

Question

我通过css2sass获得了以下片段(转换为SCSS)

.floating-label-form-group {
  position: relative;
  margin-bottom: 0;
  padding-bottom: .5em;
  border-bottom: 1px solid #e1e1e1;
  input, textarea {
    z-index: 1;
    position: relative;
    padding-right: 0;
    padding-left: 0;
    border: 0;
    border-radius: 0;
    font-size: 1.5em;
    background: 0 0;
    box-shadow: none!important;
    resize: none;
  }
  label {
    display: block;
    z-index: 0;
    position: relative;
    top: 2em;
    margin: 0;
    font-size: .85em;
    line-height: 1.764705882em;
    vertical-align: middle;
    vertical-align: baseline;
    opacity: 0;
    -webkit-transition: top .5s ease,opacity .5s ease;
    -moz-transition: top .5s ease,opacity .5s ease;
    -ms-transition: top .5s ease,opacity .5s ease;
    transition: top .5s ease,opacity .5s ease;
  }
  &::not(:first-child) {
    padding-left: 14px;
    border-left: 1px solid #e1e1e1;
  }

当我使用SASS 3.4.9编译它时,它会抱怨:

Error: Invalid CSS after "&::not(": expected pseudo_expr, was ":first-child)"

预期的CSS代码应如下所示:

.floating-label-form-group::not(:first-child){
     padding-left:14px;border-left:1px solid #e1e1e1}

但是,似乎SASS不知道如何编译&::not(成CSS.有没有人有关于如何解决这个问题的想法？

Answer 1

实际上,Sass是对的:那确实是无效的CSS.:not()是一个伪类,而不是伪元素,所以它应该只有一个冒号:

  &:not(:first-child) {
    padding-left: 14px;
    border-left: 1px solid #e1e1e1;
  }