GTS*_*Joe 7 javascript disconnect mutation-observers
如何断开我的变异观察者与其回调函数的连接？正如他们应该观察到的变化,但我想在第一次变化后断开观察者.由于观察者变量超出范围,因此它不会断开连接.如何将observer变量传递给回调函数,以便代码可以工作？
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>log*_*yth 11
最简单的方法是调整回调
function mutate(mutations) {
至
function mutate(mutations, observer) {
因为与突变关联的观察者实例会自动作为第二个参数传递给突变处理函数.
然后你可以observer.disconnect()在任何需要的时候打电话.