Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
1.0k views
in Technique[技术] by (71.8m points)

javascript - How to check whether dynamically attached event listener exists or not?

Here is my problem: is it somehow possible to check for the existence of a dynamically attached event listener? Or how can I check the status of the "onclick" (?) property in the DOM? I have searched the internet just like Stack Overflow for a solution, but no luck. Here is my html:

<a id="link1" onclick="linkclick(event)"> link 1 </a>
<a id="link2"> link 2 </a> <!-- without inline onclick handler -->

Then in Javascript I attach a dynamically created event listener to the 2nd link:

document.getElementById('link2').addEventListener('click', linkclick, false);

The code runs well, but all my attempts to detect that attached listener fail:

// test for #link2 - dynamically created eventlistener
alert(elem.onclick); // null
alert(elem.hasAttribute('onclick')); // false
alert(elem.click); // function click(){[native code]} // btw, what's this?

jsFiddle is here. If you click "Add onclick for 2" and then "[link 2]", the event fires well, but the "Test link 2" always reports false. Can somebody help?

Question&Answers:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I did something like that:

const element = document.getElementById('div');

if (element.getAttribute('listener') !== 'true') {
     element.addEventListener('click', function (e) {
         const elementClicked = e.target;
         elementClicked.setAttribute('listener', 'true');
         console.log('event has been attached');
    });
}

Creating a special attribute for an element when the listener is attached and then checking if it exists.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...