防止:悬停冒泡

Question

如果我们将鼠标悬停在其子项上,如何阻止父元素悬停？

这是代码

const parent = document.getElementById('parent')
parent.onmouseover = function testAlert(e) {
  /* alert('parent') */
}
const childRight = document.getElementById('child-right')
childRight.onmouseover = function f(e) {
  e.stopPropagation()
  /* alert('child-right') */
}
const childLeft = document.getElementById('child-left')
childLeft.onmouseenter = function f(e) {
  e.stopPropagation()
  /* alert('child-right') */
}

#parent {
  background: green;
  width: 100px;
  height: 100px;
  position: relative;
  margin: 0 auto;
}

#parent:hover {
  background: rgba(0, 0, 0, 0.8);
}

#child-left {
  background: red;
  width: 50px;
  height: 50px;
  position: absolute;
  top: 0;
  left: -50px;
}

#child-right {
  background: red;
  width: 50px;
  height: 50px;
  position: absolute;
  top: 50px;
  left: 100px;
}

<div id="parent">
  <div id="child-left"></div>
  <div id="child-right"></div>
</div>

https://jsfiddle.net/3tjcsyov/48/

您可以看到,如果您将鼠标悬停在红色区域上,如果考虑CSS行为,绿色区域也会悬停.是的,我们可以使用,stopPropagation但它只会阻止js处理程序执行父元素,而CSS行为保持不变.

Answer 1

如果你将孩子设置pointer-events为none,你可以毫无任意地实现这一目标JavaScript.

#parent {
  background: green;
  width: 100px;
  height: 100px;
  position: relative;
  margin: 0 auto;
}

#parent:hover {
  background: rgba(0, 0, 0, 0.8);
}

#child-left {
  background: red;
  width: 50px;
  height: 50px;
  position: absolute;
  top: 0;
  left: -50px;
}

#child-right {
  background: red;
  width: 50px;
  height: 50px;
  position: absolute;
  top: 50px;
  left: 100px;
}

#child-left,
#child-right {
  pointer-events: none;
}

<div id="parent">
  <div id="child-left"></div>
  <div id="child-right"></div>
</div>

https://jsfiddle.net/bepLktoj/