如何检查元素是否存在于可见的DOM中？ [英] How to check if element exists in the visible DOM?
问题描述
在不使用 getElementById 方法的情况下，如何测试元素的存在？我已经设置了一个现场演示供参考。我也会打印这里的代码：
How do you test an element for existence without the use of the getElementById method? I have setup a live demo for reference. I will also print the code on here as well:
<!DOCTYPE html>
<html>
<head>
<script>
var getRandomID = function (size) {
var str = "",
i = 0,
chars = "0123456789abcdefghijklmnopqurstuvwxyzABCDEFGHIJKLMNOPQURSTUVWXYZ";
while (i < size) {
str += chars.substr(Math.floor(Math.random() * 62), 1);
i++;
}
return str;
},
isNull = function (element) {
var randomID = getRandomID(12),
savedID = (element.id)? element.id : null;
element.id = randomID;
var foundElm = document.getElementById(randomID);
element.removeAttribute('id');
if (savedID !== null) {
element.id = savedID;
}
return (foundElm) ? false : true;
};
window.onload = function () {
var image = document.getElementById("demo");
console.log('undefined', (typeof image === 'undefined') ? true : false); // false
console.log('null', (image === null) ? true : false); // false
console.log('find-by-id', isNull(image)); // false
image.parentNode.removeChild(image);
console.log('undefined', (typeof image === 'undefined') ? true : false); // false ~ should be true?
console.log('null', (image === null) ? true : false); // false ~ should be true?
console.log('find-by-id', isNull(image)); // true ~ correct but there must be a better way than this?
};
</script>
</head>
<body>
<div id="demo"></div>
</body>
</html>
上面的代码演示的是一个元素存储到变量中，然后从dom中删除。即使元素已经从dom中删除，变量保留元素，就像首次声明一样。换句话说，它不是对元素本身的实时引用，而是复制品。因此，检查变量的值（元素）是否存在会提供意想不到的结果。
Basically what the above code demonstrates is an element being stored into a variable and then removed from dom. Even though the element has been removed from the dom, the variable retains the element as it was when first declared. In other words, it is not a live reference to the element itself, but rather a replica. As a result, checking the variable's value (the element) for existence will provide an unexpected result.
isNull 函数是我尝试从变量中检查元素的存在，并且它有效，但是我想知道是否有更简单的方法来实现相同的结果。
The isNull function is my attempt to check for an elements existence from a variable, and it works, but I would like to know if there is an easier way to accomplish the same result.
PS：我也有兴趣为什么JavaScript变量的行为如果有人知道一些与这个主题有关的好文章。
PS: I'm also interested in why JavaScript variables behave like this if anyone knows of some good articles related to the subject.
推荐答案
p>似乎有些人正在登陆，只想知道一个元素是否存在（与原始问题有点不同）。
It seems some people are landing here, and simply want to know if an element exists (a little bit different to the original question).
这就像使用任何浏览器的选择方法一样简单，并检查它是否为真实值（一般）。
That's as simple as using any of the browser's selecting method, and checking it for a truthy value (generally).
例如，如果我的元素具有 id find-me，我可以简单地使用...
For example, if my element had an id of "find-me", I could simply use...
var elementExists = document.getElementById("find-me");
这是指定返回对元素的引用或 null 。如果你必须有一个布尔值，只需在方法调用之前抛出一个 !! 。
This is spec'd to either return a reference to the element or null. If you must have a Boolean value, simply toss a !! before the method call.
另外，你可以使用一些存在的其他方法来查找元素，例如（所有生活从$ code>文档）：
In addition, you can use some of the many other methods that exist for finding elements, such as (all living off document):
-
querySelector（）/querySelectorAll（） -
getElementsByClassName（） -
getElementsByName（）
querySelector()/querySelectorAll()getElementsByClassName()getElementsByName()
其中一些方法返回一个 NodeList ，所以一定要检查它的长度属性，因为 NodeList 是一个对象，因此真实。
Some of these methods return a NodeList, so be sure to check its length property, because a NodeList is an object, and therefore truthy.
为了确定某个元素是否作为可见DOM的一部分存在（像原始问题一样）， Csuwldcat提供比滚动自己的更好的解决方案（如此回答包含）。也就是说，要使用 contains（） DOM元素的方法。
For actually determining if an element exists as part of the visible DOM (like the question originally asked), Csuwldcat provides a better solution than rolling your own (as this answer used to contain). That is, to use the contains() method on DOM elements.
您可以这样使用...
You could use it like so...
document.body.contains(someReferenceToADomElement);
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