css选择器排除某些后代及其jquery对象的后代？

Question

给定一个具有多个级别后代的jquery对象，如何排除某些后代及其后代？假设在要排除的元素的顶部节点上有一个类（在这种情况下为div1），并且您的jquery对象是下面的.foo。

编辑：澄清：我想排除$('#button1').click(function() { var selector = '#test ' + $('#input1').val(); var numberElements = $(selector).length; $('#numberElements').text(numberElements); var elementsAsCommaSeparatedList = _.pluck($(selector), 'id'); $('#elementList').text(elementsAsCommaSeparatedList.join(', ')); });的所有后代，而不仅仅是直接的孩子。

https://sandipanweb.wordpress.com/2017/03/24/solving-4-puzzles-with-reinforcement-learning-q-learning-in-python/?frame-nonce=8531adb87d

#input1 {
  width: 230px;
  font-size: 16px;
  font-weight: bold;
}

#numberElements,
#elementList {
  color: green;
  font-weight: bold;
  font-size: 20px;
}

<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
<label for="input1">Please supply a selector... $('<input id="input1" value="div:not(.foo, .foo *)"></label>')
<button id="button1">Apply Selector</button>
<div>(default value is from @DaniP's answer)</div>
<div id="result"><span>Number of elements selected:  </span><span id="numberElements">0</span></div>
<div>
  <span>Elements selected (by id):</span>
  <span id="elementList"></span>
</div>
<div id="test">
  <div id="div1">
    <div id="div2">
      <div id="div4">
        <div id="div8"></div>
        <div id="div9"></div>
      </div>
      <div id="div5" class="foo">
        <div id="div10"></div>
        <div id="div11"></div>
      </div>
    </div>
    <div id="div3">
      <div id="div6">
        <div id="div12"></div>
        <div id="div13"></div>
      </div>
      <div id="div7" class="foo">
        <div id="div14"></div>
        <div id="div15"></div>
      </div>
    </div>
  </div>
</div>

build-essential

Answer 1

JQuery的

使用Jquery，您可以在:not选择器上使用多个参数，这样很容易排除.foo和子元素：

$('div:not(.foo, .foo *)')

_{示例代码段}

$('div:not(.foo, .foo *)').css('border-color','blue')

div {
  padding-left:20px;
  margin:5px;
  border:thin red solid;
}
.foo {
  background:rgba(0,0,0,.1)
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div>
DIV1
  <div>
  DIV2
    <div>
    DIV4
      <div>DIV8</div>
      <div>DIV9</div>
    </div>
    <div class="foo">
    DIV5
      <div>DIV10</div>
      <div>DIV11</div>
    </div>
  </div>
  <div>
  DIV3
    <div>
    DIV6
      <div>DIV12</div>
      <div>DIV13</div>
    </div>
    <div class="foo">
    DIV7
      <div>DIV14</div>
      <div>DIV15</div>
    </div>
  </div>
</div>

  

但CSS不允许这样做，refer to this answer。

CSS

在CSS上，你需要定位所有不是.foo的div，但是还要再次覆盖.foo的子元素：

div:not(.foo) {
  border-color:blue;
}
div.foo * {
  border-color:red;
}

_{示例代码段}

div {
  padding-left:20px;
  margin:5px;
  border:thin red solid;
}
.foo {
  background:rgba(0,0,0,.1)
}
div:not(.foo) {
  border-color:blue;
}
div.foo * {
  border-color:red;
}

<div>
DIV1
  <div>
  DIV2
    <div>
    DIV4
      <div>DIV8</div>
      <div>DIV9</div>
    </div>
    <div class="foo">
    DIV5
      <div>DIV10</div>
      <div>DIV11</div>
    </div>
  </div>
  <div>
  DIV3
    <div>
    DIV6
      <div>DIV12</div>
      <div>DIV13</div>
    </div>
    <div class="foo">
    DIV7
      <div>DIV14</div>
      <div>DIV15</div>
    </div>
  </div>
</div>

Answer 2

也许使用：not selector？

像

这样的东西

div:not(.foo)