如何制作一个钝角的按钮？

Question

如何制作一个钝角的按钮？

我想这样发生

我来到这里

我的代码 - 小提琴

*{
  box-sizing: border-box;
}

.btn{  
  display: inline-block;
  padding: 16px 30px;
  color: #fff; 
  border: 1px solid #4A803C;
  position: relative;
  border-radius: 3px;
  background: rgb(74,168,28); /* Old browsers */
background: -moz-linear-gradient(top, rgba(74,168,28,1) 0%, rgba(63,155,19,1) 100%, rgba(56,146,12,1) 100%); /* FF3.6-15 */
background: -webkit-linear-gradient(top, rgba(74,168,28,1) 0%,rgba(63,155,19,1) 100%,rgba(56,146,12,1) 100%); /* Chrome10-25,Safari5.1-6 */
background: linear-gradient(to bottom, rgba(74,168,28,1) 0%,rgba(63,155,19,1) 100%,rgba(56,146,12,1) 100%); /* W3C, IE10+, FF16+, Chrome26+, Opera12+, Safari7+ */
filter: progid:DXImageTransform.Microsoft.gradient( startColorstr='#4aa81c', endColorstr='#38920c',GradientType=0 );
}
.btn > span{  
 position:relative;
 z-index: 1; 
}
.btn:after {
    content: "";
    width: 35px;
    height: 35px;    
    display: block;
    position: absolute;
    top: 7px;
    right: -18px;
    border: 1px solid #4A803C;
    border-left: none;
    border-bottom: none;
    border-radius: 3px;
    -webkit-transform: rotate(47deg) skew(5deg);   
    transform: rotate(47deg) skew(5deg);
    background-image: -moz-linear-gradient( 143deg, rgb(74,168,28) 0%, rgb(63,155,19) 100%);
    background-image: -webkit-linear-gradient( 143deg, rgb(74,168,28) 0%, rgb(63,155,19) 100%);
    background-image: -ms-linear-gradient( 143deg, rgb(74,168,28) 0%, rgb(63,155,19) 100%);
    background-image: linear-gradient( 143deg, rgb(74,168,28) 0%, rgb(63,155,19) 100%);
}

.btn:hover{
  background: rgb(56,146,12); /* Old browsers */
background: -moz-linear-gradient(top, rgba(56,146,12,1) 0%, rgba(63,155,19,1) 0%, rgba(74,168,28,1) 100%); /* FF3.6-15 */
background: -webkit-linear-gradient(top, rgba(56,146,12,1) 0%,rgba(63,155,19,1) 0%,rgba(74,168,28,1) 100%); /* Chrome10-25,Safari5.1-6 */
background: linear-gradient(to bottom, rgba(56,146,12,1) 0%,rgba(63,155,19,1) 0%,rgba(74,168,28,1) 100%); /* W3C, IE10+, FF16+, Chrome26+, Opera12+, Safari7+ */
filter: progid:DXImageTransform.Microsoft.gradient( startColorstr='#38920c', endColorstr='#4aa81c',GradientType=0 );
}
.btn:hover:after{
  background-image: -moz-linear-gradient( -47deg, rgb(74,168,28) 0%, rgb(63,155,19) 100%);
  background-image: -webkit-linear-gradient( -47deg, rgb(74,168,28) 0%, rgb(63,155,19) 100%);
  background-image: -ms-linear-gradient( -47deg, rgb(74,168,28) 0%, rgb(63,155,19) 100%);
  background-image: linear-gradient( -47deg, rgb(74,168,28) 0%, rgb(63,155,19) 100%);
}

<a href="#" class="btn">
<span>?????????? ???????</span>
</a>

我会很高兴有任何帮助.谢谢

Answer 1

一个简单的解决方案是一个补充rotateY(Xdeg)的.btn:after元素.这将使元素的Y轴旋转,从而使其看起来比实际更窄.

可根据需要修改旋转角度.它可以是低于90度的任何值,具体取决于箭头的宽度或宽度.值越高,箭头越窄.

* {
  box-sizing: border-box;
}
.btn {
  display: inline-block;
  padding: 16px 30px;
  color: #fff;
  border: 1px solid #4A803C;
  position: relative;
  border-radius: 3px;
  background: rgb(74, 168, 28);
  background: linear-gradient(to bottom, rgba(74, 168, 28, 1) 0%, rgba(63, 155, 19, 1) 100%, rgba(56, 146, 12, 1) 100%);
}
.btn > span {
  position: relative;
  z-index: 1;
}
.btn:after {
  content: "";
  width: 35px;
  height: 35px;
  display: block;
  position: absolute;
  top: 7px;
  right: -18px;
  border: 1px solid #4A803C;
  border-left: none;
  border-bottom: none;
  border-radius: 3px;
  transform: rotateY(45deg) rotate(47deg) skew(5deg);
  background-image: linear-gradient(143deg, rgb(74, 168, 28) 0%, rgb(63, 155, 19) 100%);
  
}
.btn:hover {
  background: rgb(56, 146, 12);
  background: linear-gradient(to bottom, rgba(56, 146, 12, 1) 0%, rgba(63, 155, 19, 1) 0%, rgba(74, 168, 28, 1) 100%);
}
.btn:hover:after {
  background-image: linear-gradient(-47deg, rgb(74, 168, 28) 0%, rgb(63, 155, 19) 100%);
  
}

<a href="#" class="btn">
  <span>?????????? ???????</span>
</a>