Question

我正在尝试制作一些特殊的菜单，但我在选择最嵌套的元素（div）时遇到了问题。菜单将是动态的，因此它可以更改将div嵌套在一个div中的数量。（父母将使用新的孩子创建）所以我需要选择最后一个（最嵌套的）而不使用更多的类od ID。这是我直到现在写的代码：

<div id="strategy">
  <div class="selected">
  0
    <div class="selected">
    some text
      <div class="selected"> this is the last div, but it can be anytime     changed and more childs of this element can be created</div>
    </div>
  </div>
<div class="selected">
1
</div>
<div>
2
</div>
</div>

我试过的一些css：

div.selected:only-of-type {background: #F00;} 还试过nth：last-child，only-child ..我认为一切都必须有一些方法如何去做。

Answer 1

如果您对jQuery开放......

＆＃13;

$(document).ready(function() {
  var $target = $('#strategy').children();

  while( $target.length ) {
    $target = $target.children();
  }

  var last = $target.end(); // You need .end() to get to the last matched set
  var lastHtml = last.html();
  $('body').append('<strong>deepest child is: ' + lastHtml + '</strong>');
  last.css('color', 'blue');
});

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="strategy">
  <div class="selected">
    0
    <div class="selected">
      some text
      <div class="selected"> this is the last div, but it can be anytime     changed and more childs of this element can be created</div>
    </div>
  </div>
  <div class="selected">
    1
  </div>
  <div>
    2
  </div>
</div>

＆＃13;

选择具有相同类的最嵌套元素

1 个答案: