javascript – 与MutationObserver等效的DOMNodeInserted DOMNodeRemoved?
作者:互联网
我目前有代码:
$('.example').bind('DOMNodeInserted DOMNodeRemoved', function(event) {
....
});
哪个工作完美,但效率不高,因此已被弃用.有什么更好的方法呢?
我一直在调查MutationObserver,但这段代码确实有用吗?
它给出错误“mutation.addedNodes不是一个函数”我还需要我知道的removedNodes.
var observer = new MutationObserver(function(mutations) {
mutations.forEach(function(mutation) {
mutation.addedNodes.forEach(function(node) {
if (node.className == 'example') {
....
}
});
});
});
observer.observe(document, {
childList: true,
subtree: true,
attributes: false,
characterData: false,
});
解决方法:
.addedNodes返回NodeList而不是Array,它没有.forEach()方法.尝试使用Array.prototype.slice()将.addedNodes转换为具有方法.forEach()的Array
var observer = new MutationObserver(function(mutations) {
mutations.forEach(function(mutation) {
var nodes = Array.prototype.slice.call(mutation.addedNodes);
nodes.forEach(function(node) {
if (node.parentElement.className == "example") {
alert(node.parentElement.className)
}
});
});
});
observer.observe(document.querySelector(".example"), {
childList: true,
subtree: true,
attributes: false,
characterData: false,
});
var el = document.createElement("div");
el.className = "example-child";
document.querySelector(".example").appendChild(el)
<div class="example"></div>
标签:jquery,javascript,dom,dom-node,mutation-observers 来源: https://codeday.me/bug/20190702/1355930.html