30

我想在左边有一个 4px 厚的粉红色和 1px 灰色的边框:

border: 1px solid #E5E5E5;
border-left: 4px solid #F24495;

问题是连接是对角的,所以我得到了一个可怕的覆盖。我试过:

.item:before{ 
  border-left: 4px solid #F24495;
}

但没有运气。

jsFiddle 示例

截屏

截屏

4

4 回答 4

35

.item::before是正确的方法,但它需要在单个border-left属性之外进行一些工作。您需要使伪元素可见 ( display: block; content: "";),将伪元素放置在 的左侧.item,然后将其拉伸以与顶部和底部边框正确对齐。

虽然这可以手动完成,但我强烈建议使用CSS 变量(或预处理器中的变量),因为它可以减少更新边框宽度的错误和痛苦。

.item {
  display: inline-block;
  padding: 0.2em 0.3em;
  background: #f8f8f8;
  color: #454545;

  /* Set border widths with variables */
  --top-border-width: 4px;
  --bottom-border-width: var(--top-border-width);
  --left-border-width: 16px;
  --right-border-width: var(--top-border-width);

  /* Set border styles for each side */
  border-top: var(--top-border-width) solid #e4e4e4;
  border-bottom: var(--bottom-border-width) solid #e4e4e4;
  border-right: var(--right-border-width) solid #e4e4e4;

  /* Remove the left border and add blank space where the border should be placed */
  border-left: 0;
  margin-left: var(--left-border-width);

  /* Contain the ::before */
  position: relative;
}

.item::before {
  /* Give the pseudo element substance */
  display: block;
  content: "";

  /* Add a left border with a straight edge */
  border-left: var(--left-border-width) solid #f84995;

  /* Position pseudo element's border where the normal border would have been placed */
  position: absolute;
  top: calc(0px - var(--top-border-width));
  bottom: calc(0px - var(--bottom-border-width));
  left: calc(0px - var(--left-border-width));
}
<h1 class="item">Gen.2</h1>

于 2012-06-15T14:15:42.820 回答
6

这应该可以,但需要额外的标记:

.outer {
    border: 1px solid #E5E5E5;
    border-left: 0;
}

.inner {
    border-left: 4px solid #F24495;
}

<div class="outer">
    <div class="inner">
        ...
    </div>
</div>
于 2012-06-15T14:07:26.777 回答
6

如果您希望使用:before伪选择器,您还需要设置一些内容。例如,请参阅带有以下示例代码的这个 jsfiddle :

<div>Container</div>

CSS:

div {
    border: 10px solid black;
    border-left-width: 0;
}
div::before {
    border: 10px solid orange;
    border-right-width: 0;
    content: '';
}

显示为:

工作代码的屏幕截图

编辑
嗯,虽然这应该严格回答问题,但在尝试将我的解决方案调整为问题的小提琴时,我发现这与填充效果不太好。打开可以处理该位的建议/编辑/其他答案:(...

于 2012-06-15T14:14:35.560 回答
1

背景

默认情况下,CSS 对所有边框关节使用斜接关节(45° 角)。因此,要为任何边框实现方形关节(90° 角),您可以使用 (1) 内部box-shadow,(2)伪元素或 (3) background-image和多个linear-gradients

假设您有以下要设置样式的元素:

<div></div>

选项1:使用方形接头box-shadow

div {
  /* downside of using box-shadow, you need to add the   */
  /* inner border size to the padding on top of any      */
  /* additional padding you might want                   */
  padding: 20px;
  /* by changing the order of your box-shadows, you      */
  /* can modify which borders overlap each other         */
  box-shadow:
    /* left "inner border" */
    inset 20px 0 0 0 red,
    /* right "inner border" */
    inset -20px 0 0 0 grey,
    /* top "inner border" */
    inset 0 20px 0 0 grey,
    /* bottom "inner border" */
    inset 0 -20px 0 0 grey;
}

选项2:方形接头pseudo-elements

div {
  border: 20px solid grey;
}

div::before {
  position: absolute;
  background-color: red;
  content: "";
  width: 20px;
  /* we need to add the height of the top and bottom    */
  /* border to the pseudo-elements' height as the       */
  /* inset border is not included in the height of the  */
  /* div even when "box-sizing: border-box" is set.     */
  height: calc(100% + 20px + 20px);
  top: -20px;
  left: -20px;
}

选项 3:方形接头使用background-image和多个linear-gradients

div {
  /* downside of using multiple linear-gradients, you   */
  /* need to add the inner border size to the padding   */
  /* on top of any additional padding you might want    */
  padding: calc(20px + 10px);
  background-image: 
    /* left "inner border" */
    linear-gradient(to right, red 20px, transparent 20px),
    /* right "inner border" */
    linear-gradient(to left, grey 20px, transparent 20px),
    /* top "inner border" */
    linear-gradient(grey 20px, transparent 20px),
    /* bottom "inner border" */
    linear-gradient(to top, grey 20px, transparent 20px);
}

于 2019-01-27T16:10:42.643 回答