带有后代的queryselectorAll未正确选择 [英] queryselectorAll with descendant not selecting correctly
问题描述
我具有以下DOM结构:
I have the following DOM structure:
var parent = document.querySelector(".test");
var navChild = parent.querySelectorAll("ul > li > a");
for (i = 0; i < navChild.length; i++) {
navChild[i].style.backgroundColor = "red";
}
console.log(navChild.lenght);
console.log(navChild);
<!-- begin snippet: js hide: false console: true babel: false -->
<ul>
<li class="test">
<a>ssss</a> <!--this will also be affected -->
<ul>
<li><a>fsfsf</a></li>
<li><a>sff</a></li>
<li><a>ggg</a></li>
</ul>
</li>
</ul>
我想使用香草javascript并从 li
开始使用类test选择3个 a
标记.我将其放入 parent
变量中,然后使用querySelectorAll尝试仅获取下一个列表中的 a
标记.但是,返回的节点列表也包括第一个 a
标记,即使它不是后代.这是我的代码:
I want to select the 3 a
tags using vanilla javascript and starting from the li
with the class test. I put this in a parent
variable and then use querySelectorAll to try to get only the a
tags that are inside the next list. However, the nodelist that is returned also includes the very first a
tag even though it's not a descendant. This is my code:
var navChild = parent.querySelectorAll("ul > li > a");
真正奇怪的是,如果我仅查询 li
后代,我只会得到3.所以我不明白为什么当我查询那个 li 的子代后
我还获得了第一个 a
标记,该标记在DOM上位于两个级别.
What's really weird is that if I query for just the li
descendants I get only the 3. So I don't understand why when I query for the child descendant of that li
I also get that first a
tag that is two levels up on the DOM.
我知道我可以使用jQuery来做到这一点,但我不想使用它.
I know I could do this using jQuery but I don't want to use it.
已编辑以显示如何设置父变量
我尝试在第一个 a
a 标记添加到 childs
变量中>我不想包含在childs节点列表中的标签.
I'm trying to add those a
tags to the childs
variable after I've hit the enter keyboard button while on the first a
tag that I don't want to include in the childs node list.
var alinks = document.querySelectorAll('.test > a');
[].forEach.call(alinks, function(link){
link.addEventListener('keyup', function(e){
e.preventDefault();
if(e.keyCode === 13) {
var linkParent = e.target.parentElement;
var childs = menuParent.querySelectorAll('ul > li > a');
}
})
});
当我登录 linkParent
时,它已被正确设置为 li
类别为 test
的目录.我不明白为什么然后从那里运行查询仍然会包含第一个 a
标记.如前所述,如果我只查询 ul>li
,它为我提供了正确的 li
标记.
When I log linkParent
it is correctly being set as the li
with class test
. I don't understand why then if I run the query from there it still includes the very first a
tag. As I previously stated, if I query just ul > li
it gives me the correct li
tags.
推荐答案
这是因为您要忽略的第一个< a>
标记仍属于选择器规则 ul>&一个
.因此,即使您以< li class ="test">
作为根来开始查询(顺便说一句也可以,但是我不知道为什么其他答案会说document仍然是根目录),它找到的第一个子元素是< a>
标记,并且实际上,它是&li; li
的子元素.< ul>
的子级.结束这一事实的事实超过"了您指定的根.
This is because the first <a>
tag you're trying to ignore still falls into the selector rule ul > li > a
. So, even though you start the query with the <li class="test">
as the root (which does work by the way, I don't know why the other answers say that the document is still the root), the first child element it finds is the <a>
tag and, indeed, it is the child of an <li>
which is the child of a <ul>
. The fact that this winds up going "above" your specified root is ignored.
修改:如果您希望选择器规则也作用于根目录,则可以改用它:
If you want the selector rule to be scoped to the root as well, you can use this instead:
parent.querySelectorAll(":scope ul > li > a");
为进一步说明,浏览器CSS引擎进行评估CSS规则从右到左.在您看来,您希望浏览器从根目录(父< li class ="test">
)启动,然后找到一个< ul>
标记然后找到< li>
标签,然后找到< a>
标签.
And to further clarify, browser CSS engines evaluate CSS rules right-to-left. In your mind, you want the browser to start at the root (the parent <li class="test">
) and then find a <ul>
tag and then find a <li>
tag and then find an <a>
tag.
但是,浏览器从根目录开始,然后在根目录下查找所有< a>
标记.然后,它检查< a>
标记是否在< li>
内,然后检查< li>
code>< ul> .因此< li class ="parent">
标记确实是根,但此后它不遵循选择器规则的从左到右的层次结构,因此从右到右左.
However, the browser starts with the root, but then looks for all of the <a>
tags below the root. Then it checks if the <a>
tag is within an <li>
and then if the <li>
is within a <ul>
. So the <li class="parent">
tag really is the root, but it does not follow the left-to-right hierarchy of your selector rule after that, it goes right-to-left.
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