jQuery 在jQuery中选择第N个前一个兄弟？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/1022595/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Select Nth previous sibling in jQuery?
提问by Paolo Bergantino
I'm trying to set up my jQuery selector but I'm not exactly sure how I need to write it.
我正在尝试设置我的 jQuery 选择器，但我不确定我需要如何编写它。
I've got a unordered list that looks something like this:
我有一个看起来像这样的无序列表：
<ul>
<li>something</li>
<li>something</li>
<li>something</li>
<li>something</li>
<li class="last">something</li>
</ul>
Now I know if I want to select the last child I can do that by using either "ul li.last" or "ul.li:last" but say if I wanted the second from last, or third, or 20th? Now would I do that?
现在我知道如果我想选择最后一个孩子，我可以使用“ul li.last”或“ul.li:last”来做到这一点，但是说我是否想要倒数第二个，第三个，还是第 20 个？现在我会这样做吗？
回答by Paolo Bergantino
You use eq:
你使用eq：
$('ul li').eq(X);
Where X is a 0-based index of which element you want. You could also use the selector form:
其中 X 是您想要的元素的基于 0 的索引。您还可以使用选择器形式：
$('ul li:eq(X)');
And you could also achieve a similar result by using nth-child:
你也可以通过使用来获得类似的结果nth-child：
$('ul li:nth-child(X)');
The documentation has this to say about the difference between nth-childand eq:
该文件有这样说的区别nth-child和eq：
While :eq(index) matches only a single element, this matches more than one: One for each parent with index. Multiple for each parent with even, odd, or equation. The specified index is one-indexed, in contrast to :eq() which starts at zero.
虽然 :eq(index) 仅匹配一个元素，但它匹配多个元素：每个具有索引的父元素一个。每个具有偶数、奇数或等式的父级的倍数。与从零开始的 :eq() 相比，指定的索引是一索引的。
So by doing $('ul li').eq(19);you would get the single 20th matched element of the query (so if there is more than one list in the document this won't get you "all 19th children", while $('ul li:nth-child(20)');would get you the individual 20th matched list element of each <ul>in the document.
因此，通过这样做，$('ul li').eq(19);您将获得查询的第 20 个匹配元素（因此，如果文档中有多个列表，则不会获得“所有第 19 个孩子”，而$('ul li:nth-child(20)');将获得每个第 20 个匹配的列表元素<ul>在文件中。
EDIT:
编辑：
To do something like "second to last", the sane way to do it would be like:
要执行诸如“倒数第二”之类的操作，明智的做法是：
var $ul = $('#mylist > li');
var $el = $ul.eq($ul.length-2);
But you should probably do a check to make sure length-2 is not less than 0.
但是您可能应该进行检查以确保长度 2 不小于 0。
回答by Jeff Meatball Yang
Do you want the 3rd from last?
你想要倒数第三个吗？
var $ul = $("ul li");
$ul.find(":eq(" + Math.max(0, $ul.length - 3) + ")");
Quick explanation: The length of $ul (which is a jQuery object) is how many li's are found. So you can use that along with the ":eq(n)" selector to get the li at the desired index.
快速解释：$ul（它是一个 jQuery 对象）的长度是找到了多少个 li。因此，您可以将它与 ":eq(n)" 选择器一起使用以获取所需索引处的 li。
I used Math.max in case the list was too short (less than 3 li's).
我使用 Math.max 以防列表太短（少于 3 里）。
This will also work, but it's obviously not the best way:
这也可以，但显然不是最好的方法：
var $ul = $("ul li:last").prev().prev();
回答by Magnus
The jQuery team added this feature in version 1.9
jQuery 团队在 1.9 版本中添加了此功能
You can read about it at http://api.jquery.com/nth-last-child-selector/
回答by chaos
回答by karim79
alert($('ul li').eq(3).text());
That will alert the text of the third list item.
这将提醒第三个列表项的文本。
