html - 带虚线边框的 CSS3 六边形

Question

我可以实现如下所示的具有实心边框的六边形:

.hex {
  margin-top: 70px;
  width: 208px;
  height: 120px;
  background: red;
  position: relative;
}
.hex:before,
.hex:after {
  content: "";
  border-left: 104px solid transparent;
  border-right: 104px solid transparent;
  position: absolute;
}
.hex:before {
  top: -59px;
  border-bottom: 60px solid red;
}
.hex:after {
  bottom: -59px;
  border-top: 60px solid red;
}
.hex.inner {
  background-color: lightgreen;
  -webkit-transform: scale(.98, .98);
  -moz-transform: scale(.98, .98);
  transform: scale(.98, .98);
  z-index: 1;
}
.hex.inner:before {
  border-bottom: 60px solid lightgreen;
}
.hex.inner:after {
  border-top: 60px solid lightgreen;
}

<div class="hex">
  <div class="hex inner">
  </div>
</div>

但是，我想创建一个带虚线边框的六边形，如下图所示:

Answer 1

这是一个带有 inline svg 的方法使用:

• polygon element对于六边形
• stroke-dasharray对于虚线边框(控制点之间的大小和间距)
• stroke-linecap对于虚线边框

svg{width:30%;margin:0 auto;}

<svg viewbox="0 0 100 100" version="1.1" xmlns="http://www.w3.org/2000/svg">
  <polygon fill="#92D050" 
           stroke="red"
           stroke-width="1"
           stroke-linecap="round"
           stroke-dasharray="0.5,3"
           points="50 1 95 25 95 75 50 99 5 75 5 25"/>
</svg>