1
为了更加熟悉递归DOM &,我决定从头开始重新创建getElementsByClassName(仅适用于vanilla JS,不适用于jQuery)。目前,我可以在DOM中找到所有具有我想要的类的元素,但是遇到了只能获得具有两个特定类(或更多)的元素的方法的问题。从零开始重新实现getElementsByClassName(仅限于Vanilla JS)
<div class="one two">
<h1 class="one">
<span class="one two">
</h1>
</div>
我当前实现返回我的期望,返回包含类 '一' 的每一个元素:
我试图去:
getElementsByClassName('one two');
[<div class="one two"></div>, <span class="one two"</span>]
我遇到的问题之一是classList.contains:
element.classList;
// ['one, 'two'];
element.classList.contain("one");
//returns true since the value actually exists
//PROBLEM:
element.classList.contains("one two");
//this is looking for "one two" in the array and not 'one' and 'two'.
//this returns false & my code breaks
//How would I be able to do something like this, even if it
//means recreating my own helper contains function?
contains('one','two');
我的功能:
var getElementsByClassName = function(className){
var results = [];
function getClass(nodeList){
var childList = nodeList.children;
_forEach(childList, function(node) {
//1st level body check
if(node.classList && node.classList.contains(className)){
results.push(node);
}
//has children, recurse
if(node.children) {
getClass(node);
}
else {
getClass(node);
}
});
}
getClass(document.body);
return results;
}
//Helper forEach function to iterate over array like DOM objects
var _forEach = function(collection, func) {
for(var i = 0; i < collection.length; i++) {
func(collection[i], i, collection);
}
}
'[ '一',“二'] .every(function(className){return node.classList.contains(className); })' –
谢谢,几个星期前我刚开始学习几个月的编程和函数式编程。时间解剖这=)。 –
你很好,你有关键要素:真正努力学习。好一个。 (这就是为什么我没有这样做的原因,只是指出*你会怎么做。) –