如果没有点击链接,则显示所有div

Question

我有一个页面名称url.com/yourfirstpage/当我转到页面时默认隐藏所有div(显示:无)如果我们将#sec1定位为url.com/yourfirstpage/#sec1它只显示sec1并隐藏其他.我想知道我们是否在没有像url.com/yourfirstpage/这样的锚ID的情况下去了网址,它需要显示所有的div.

#sec1, #sec2, #sec3{
	display:none;
}
#sec1:target{
	display:block;
}
#sec2:target{
	display:block;
}
#sec3:target{
	display:block;
}

<a href="#sec1">sec1</a>
<a href="#sec2">sec2</a>
<a href="#sec3">sec3</a>

<div id="sec1" class="page"> this is sec1</div>
<div id="sec2" class="page"> this is sec2</div>
<div id="sec3" class="page"> this is sec3</div>

Answer 1

如果您能够修改 HTML 结构，这里有一个技巧。我们的想法是让元素可见，然后我们使用 隐藏它们:target。由于我们没有先前的同级选择器或父选择器，因此我在父元素中使用 id 以便能够选择任何元素：

#sec1:target .page:nth-child(n+2){
  display: none;
}

#sec2:target .page:nth-child(2n+1){
  display: none;
}

#sec3:target .page:nth-last-child(n+2){
  display: none;
}

<a href="#sec1">sec1</a>
<a href="#sec2">sec2</a>
<a href="#sec3">sec3</a>

<div id="sec1">
  <div id="sec2">
    <div id="sec3">
      <div class="page"> this is sec1</div>
      <div class="page"> this is sec2</div>
      <div class="page"> this is sec3</div>
    </div>
  </div>
</div>

它可以与任意数量的部分一起使用，我们可以如下改进 CSS 代码：

#sec1:target .page:not(:nth-child(1)),
#sec2:target .page:not(:nth-child(2)),
#sec3:target .page:not(:nth-child(3)),
#sec4:target .page:not(:nth-child(4)),
#sec5:target .page:not(:nth-child(5)) {
  display: none;
}

<a href="#sec1">sec1</a>
<a href="#sec2">sec2</a>
<a href="#sec3">sec3</a>
<a href="#sec4">sec4</a>
<a href="#sec5">sec5</a>

<div id="sec1">
  <div id="sec2">
    <div id="sec3">
      <div id="sec4">
        <div id="sec5">
          <div class="page"> this is sec1</div>
          <div class="page"> this is sec2</div>
          <div class="page"> this is sec3</div>
          <div class="page"> this is sec4</div>
          <div class="page"> this is sec5</div>
        </div>
      </div>
    </div>
  </div>
</div>