自定义旋转框箭头

Question

我想知道如何自定义旋转框的箭头。

input[type="number"]::-webkit-inner-spin-button, 
input[type="number"]::-webkit-outer-spin-button {
    -webkit-appearance: inner-spin-button !important;
    opacity: 1 !important;
}

<input type="number" />

这就是我想要得到的。

我应该从哪里开始？（我必须使用 jQuery 小部件或 javascript 来自定义它吗？）

Answer 1

屏幕截图（Chrome、macOS）：

代码：

input[type="number"] {
  height: 32px;
  border-radius: 4px;
  border: 1px solid #d8d8d8;
  position: relative;
  text-align: center;
  font-size: 20px;
  width: 80px;
  outline: none;
  background-image: url('data:image/svg+xml;utf8,%3Csvg%20version%3D%221.1%22%20viewBox%3D%220%200%2050%2067%22%20xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F2000%2Fsvg%22%3E%3Cg%20fill%3D%22none%22%20stroke-width%3D%222%22%3E%3Cline%20x1%3D%221%22%20x2%3D%2250%22%20y1%3D%2233.5%22%20y2%3D%2233.5%22%20stroke%3D%22%23D8D8D8%22%2F%3E%3Cpolyline%20transform%3D%22translate(25%2020)%20rotate(45)%20translate(-25%20-20)%22%20points%3D%2219%2026%2019%2014%2032%2014%22%20stroke%3D%22%23000%22%2F%3E%3Cpolyline%20transform%3D%22translate(25%2045)%20rotate(225)%20translate(-25%20-45)%22%20points%3D%2219%2052%2019%2039%2032%2039%22%20stroke%3D%22%23000%22%2F%3E%3C%2Fg%3E%3C%2Fsvg%3E');
  background-position: center right;
  background-size: contain;
  background-repeat: no-repeat;
  caret-color: transparent;
}

input[type="number"]::-webkit-inner-spin-button {
  -webkit-appearance: none !important;
  opacity: 1 !important;
  background: transparent !important;
  border-width: 0px;
  margin: 0;
  border-left: 1px solid #d8d8d8;
  height: 34px;
  width: 23px;
  cursor: pointer;
}

<input type="number" value="1" />

所以，这是可以做到的。但像这样的自定义通常非常脆弱（即使您添加了必要的跨浏览器属性，我还没有这样做；这只适用于 webkit），并且可能会损害可用性。使用平台的本机控件可能会更好。