javascript jQuery递归查找子项，但忽略某些元素
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jQuery finding children recursively, but ignore certain elements
提问by Arun
Lets assume the following HTML
让我们假设以下 HTML
<div id=main>
<div id=a></div>
<div id=b></div>
<div id=c></div>
<div id=d>
<div id=d1 class="no">
<div id=d11></div>
<div id=d12></div>
</div>
</div>
<div id=e>
<div id=e1 class="no">
<div id=e11></div>
<div id=e12></div>
<div id=e13></div>
</div>
</div>
</div>
I want to select all div tags that are children of main, but want to ignore children of divs that have a class as "no".
我想选择作为 main 子级的所有 div 标签，但想忽略类为“no”的 div 子级。
I have currently written a recursive function to do the job. But was wondering if there is a jQuery selector to get what I want.
我目前已经编写了一个递归函数来完成这项工作。但想知道是否有一个 jQuery 选择器来获得我想要的。
I want the DIVs with ids a,b,c,d,d1,e,e1
我想要 ID 为 a,b,c,d,d1,e,e1 的 DIV
Thanks
谢谢
EDIT: Created a test page here - http://jsfiddle.net/mRENV/
编辑：在这里创建了一个测试页面 - http://jsfiddle.net/mRENV/
回答by Felix Kling
It should be:
它应该是：
$('#main div').not('.no div')
Btw. the term childrenonly refers to the directdescendants of an element. You want to get all descendants (not only children) of #main.
顺便提一句。术语children仅指元素的直接后代。您想获得 . 的所有后代（不仅仅是孩子）#main。
Reference:.not()
参考：.not()
Edit:Updated your demo so that it works properly: http://jsfiddle.net/fkling/mRENV/1/
编辑：更新您的演示，使其正常工作：http: //jsfiddle.net/fkling/mRENV/1/
回答by jAndy
Several syntactical ways to achieve that using jQuery, my suggestion is:
使用 jQuery 实现的几种语法方法，我的建议是：
var $result = $('#main').find('div:not(.no > div)');
Invoking .find()helpalong with the pseudo :not()helpselector. This line is equivalent to:
与伪帮助选择器一起调用.find()帮助。这一行相当于：:not()
var $result = $('#main').find('div').not('.no > div');
while the latter might be slightly faster.
而后者可能会稍微快一点。
回答by Useless Code
In a single selector with no further traversal you could do this:
$('#main div:not(div.no > div)');
在没有进一步遍历的单个选择器中，您可以这样做：
$('#main div:not(div.no > div)');
