【发布时间】:2017-05-10 11:38:03
【问题描述】:
如何断开我的变异观察者与其回调函数的连接?正在按照应有的方式观察更改,但我想在第一次更改后断开观察者的连接。由于观察者变量超出了范围,因此它没有按应有的方式断开连接。如何将观察者变量传递给回调函数以便代码正常工作?
function mutate(mutations) {
mutations.forEach(function(mutation) {
if ( mutation.type === 'characterData' ) {
console.log('1st change.');
observer.disconnect(); // Should disconnect here but observer variable is not defined.
}
else if ( mutation.type === 'childList' ) {
console.log('2nd change. This should not trigger after being disconnected.');
}
});
}
jQuery(document).ready(function() {
setTimeout(function() {
document.querySelector('div#mainContainer p').innerHTML = 'Some other text.';
}, 2000);
setTimeout(function() {
jQuery('div#mainContainer').append('<div class="insertedDiv">New div!<//div>');
}, 4000);
var targetOne = document.querySelector('div#mainContainer');
var observer = new MutationObserver( mutate );
var config = { attributes: true, characterData: true, childList: true, subtree: true };
observer.observe(targetOne, config);
});
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<body>
<div id="mainContainer">
<h1>Heading</h1>
<p>Paragraph.</p>
</div>
</body>
【问题讨论】:
标签: javascript disconnect mutation-observers