如何在css中显示块的前N个元素并隐藏其他元素?
我试图隐藏块 .container
内具有类 .row
的前 3 个元素.
I am trying to hide the first 3 elements having the class .row
inside the block .container
.
我正在做的是首先隐藏所有 .row
,然后我尝试使用 .row 显示前 3 个
.row
:nth-child(-n+3)
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 在这里:http://jsfiddle.net/z8fMr/1/
jsfiddle here: http://jsfiddle.net/z8fMr/1/
.row {
display: none;
}
.row:nth-child(-n+3) {
display: block;
}
<div class="content">
<div class="notarow">I'm not a row and I must remain visible</div>
<div class="row">Row 1</div>
<div class="row">Row 2</div>
<div class="row">Row 3</div>
<div class="row">Row 4</div>
<div class="row">Row 5</div>
<div class="row">Row 6</div>
</div>
我这里有两个问题:
- 第 3 行不显示,是不是我用错了 nth-child?
- 有没有比隐藏所有内容然后创建特定规则来显示我想要的前 n 个元素更好的做法?css 中有没有办法只显示前 3 个
.row
然后隐藏所有其他.row
?
- Row 3 is not displayed, am I using nth-child in the wrong way?
- 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
?
谢谢.
推荐答案
您有一个
.notarow
作为第一个孩子,因此您必须在:nth-child()
公式中考虑这一点.由于那个.notarow
,你的第一个.row
成为了父级的第二个孩子,所以你必须从第二个到第四个开始计数:
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;
}
更新小提琴
.row {
display: none;
}
.row:nth-child(-n+4) {
display: block;
}
<div class="content">
<div class="notarow">I'm not a row and I must remain visible</div>
<div class="row">Row 1</div>
<div class="row">Row 2</div>
<div class="row">Row 3</div>
<div class="row">Row 4</div>
<div class="row">Row 5</div>
<div class="row">Row 6</div>
</div>
你做的很好.
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