html - 如何删除链接 和它位于 中的列表项之间的所有填充？或者...点击 打开里面的链接？

Question

我的代码:

<ul class='sub-menu'>
  <li><a href='something'>something</a></li>
  <li><a href='else'>else</a></li>
  <li><a href='another'>another</a></li>
</ul>

我可以单击任何列表项 (li) 的外部，但链接不会打开(就像链接本身和它所在的列表项之间有填充)。我希望能够单击列表项的任何部分并打开其中的链接。

我一直在摆弄内联 CSS 以尝试强制执行我想要的行为，但仍然没有成功。帮忙？

Answer 1

在 anchor 上创建一个伪元素，覆盖它们的列表项，包括元素符号:

.sub-menu li {
  position: relative;  /* to contain the absolute-positioned pseudo-element */
}

.sub-menu a:before {
  content: '';         /* required for most pseudo-elements */
  position: absolute;  /* absolutely positioned */
  top: 0;              /* ... at top of its list item */
  left: -50px;         /* ... to the left of its list item, including the bullet */
  right: 0;            /* to its list item's right */
  height: 100%;        /* the height of its list item */
}

片段:

.sub-menu li {
  position: relative;  /* to contain the absolute-positioned pseudo-element */
}

.sub-menu a:before {
  content: '';         /* required for most pseudo-elements */
  position: absolute;  /* absolutely positioned */
  top: 0;              /* ... at top of its list item */
  left: -50px;         /* ... to the left of its list item, including the bullet */
  right: 0;            /* to its list item's right */
  height: 100%;        /* the height of its list item */
}

<ul class='sub-menu'>
  <li><a href='something'>something</a></li>
  <li><a href='else'>else</a></li>
  <li><a href='another'>another</a></li>
</ul>