[英]I've added a variable to CSSStyleDeclaration prototype, but can't modify it
I don't really understand prototypes, so it might be my fault, but theorically if I add a variable to a prototype, I will be able to change it in its instances, right? 我不太了解原型,所以这可能是我的错,但是从理论上讲,如果我向原型添加变量,则可以在其实例中进行更改,对吗?
Here the code: 这里的代码:
<head>
<script>
CSSStyleDeclaration.prototype["foo"] = "something";
</script>
</head>
<body>
<div style="foo:'maybe'" id ="myId"></div>
<script>
var el = document.getElementById("myId");
console.log(el.style.foo);
</script>
</body>
The console returns "something", why? 控制台返回“某物”,为什么?
That's because foo
is not a standard property, so 那是因为
foo
不是标准属性,所以
cssText
, the declaration is ignored cssText
,声明将被忽略 getPropertyValue
would not have been able to retrieve the value. getPropertyValue
将无法检索该值。 However, on browsers that support CSS variables , you could use them: 但是,在支持CSS变量的浏览器上,可以使用它们:
function getFoo(el) { return getComputedStyle(el).getPropertyValue('--foo'); } snippet.log("body: " + getFoo(document.body)); snippet.log("#myId: " + getFoo(document.getElementById('myId')));
* { --foo: 'something'; }
<div style="--foo: 'maybe'" id="myId"></div> <!-- Provides the `snippet` object, see http://meta.stackexchange.com/a/242144/134069 --><script src="http://tjcrowder.github.io/simple-snippets-console/snippet.js"></script>
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