javascript - 如何在条件中使用 style.display 值

Question

我正在尝试制作一个简单的 Accordion 。我正在使用 style.display 检查元素的 display 属性是 block 还是 none 但它没有返回任何值

var mybtn = document.querySelectorAll("button");

for (var i = 0; i < mybtn.length; i++) {
  mybtn[i].onclick = function() {
    var mySibling = this.nextElementSibling;

    console.log(this.style.display);

    if (mySibling.style.display == "block") {
      mySibling.style.display = "none";
    } else {
      mySibling.style.display = "block";
    }
  }
}

Answer 1

var mybtn = document.getElementById("click_me");

mybtn.onclick = function() {
  var mySibling = this.nextElementSibling;

  //console.log(this.style.display);

  if (mySibling.style.display === "none") {
    mySibling.style.display = "block";
  } else {
    mySibling.style.display = "none";
  }
}

 <button type="button" id="click_me">Click Me!</button>
 <button type="button">Show/Hide</button>