element.style.display不是浏览器中呈现的内容

Question

可能是一个重复的问题但无法找到答案.

element.style.display不是浏览器中呈现的内容.

它不返回实际值(即块或内联等),而是返回空.在Chrome 56.0.2924.87(64位)中测试过.

如何获得实际渲染值？

function displayStyle(aEvent) {
aEvent.target.textContent=aEvent.target.style.display;
}
window.onload = function() { 
var  top_array = document.getElementsByClassName("top");
for(var i = 0; i <  top_array.length; i++)
{
   top_array[i].addEventListener("click", displayStyle, false);
}
}

.top{
background-color:#FFF8DC;
}

<div class="top">top (click it and it will disappear because its style.display is empty)</div>

Answer 1

CSS样式不可用于JavaScript,除非它们以前是在JavaScript中设置的,或者它们已被硬编码为内联样式.

改为使用getComputedStyle():

function displayStyle(aEvent) {
  var target = aEvent.target;
  target.textContent = window.getComputedStyle(target).getPropertyValue('display');
}

window.onload = function() {
  var top_array = document.getElementsByClassName("top");
  for (var i = 0; i < top_array.length; i++) {
    top_array[i].addEventListener("click", displayStyle, false);
  }
}

.top {
  background-color: #FFF8DC;
}

<div class="top">top (click it and it will now show "block" since we're getting its computed style.)</div>