如何显示一个块的前N个元素,并隐藏其他在css? [英] How to show the first N elements of a block and hide the others in css?

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问题描述

我试图隐藏块 .container 内有 .row / p>

我所做的是先隐藏所有 .row ,然后我试图显示前3个 .row 使用 .row:nth-​​child(-n + 3)



jsfiddle这里: http://jsfiddle.net/z8fMr/1/



这里有两个问题:


  1. 第3行不显示,我使用nth-child

  2. 有没有比隐藏所有内容更好的做法,然后创建一个特定的规则来显示我想要的n个第一个元素?在css中有一种方法只显示第一个3 .row ,然后隐藏所有其他 .row

感谢。

解决方案


  1. 你有一个 .notarow 作为第一个孩子,所以你必须在你的:nth- child()公式。因为 .notarow ,你的第一个 .row 成为了父的第二个孩子,所以你必须计数从第二个到第四个:

      .row:nth-​​child(-n + 4){
    display :块;
    }

    >更新小提琴


  2. ol>

    I am trying to hide the first 3 elements having the class .row inside the block .container.

    What I'm doing is hiding all the .row first, and then I am trying to display the first 3 .row by using .row:nth-child(-n+3)

    jsfiddle here: http://jsfiddle.net/z8fMr/1/

    I have two problems here:

    1. Row 3 is not displayed, am I using nth-child in the wrong way?
    2. Is there a better practice than hiding everything and then creating a specific rule to display the n first elements that I want? Is there a way in css to just display the first 3 .row and then hide all the other .row ?

    Thanks.

    解决方案

    1. You have a .notarow as the first child, so you have to account for that in your :nth-child() formula. Because of that .notarow, your first .row becomes the second child overall of the parent, so you have to count starting from the second to the fourth:

      .row:nth-child(-n+4){
          display:block;
      }
      

      Updated fiddle

    2. What you're doing is fine.

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