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
380 views
in Technique[技术] by (71.8m points)

javascript - How to handle ETIMEDOUT error?

How to handle etimedout error on this call ?

 var remotePath = "myremoteurltocopy"
 var localStream = fs.createWriteStream("myfil");;
        var out = request({ uri: remotePath });
        out.on('response', function (resp) {
            if (resp.statusCode === 200) {
                out.pipe(localStream);
                localStream.on('close', function () {
                    copyconcurenceacces--;
                    console.log('aftercopy');
                    callback(null, localFile);
                });
            }
            else
                callback(new Error("No file found at given url."), null);
        })

There are a way to wait for longer? or to request the remote file again?

What exactly can cause this error? Timeout only?

See Question&Answers more detail: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)

This is caused when your request response is not received in given time(by timeout request module option).

Basically to catch that error first, you need to register a handler on error, so the unhandled error won't be thrown anymore: out.on('error', function (err) { /* handle errors here */ }). Some more explanation here.

In the handler you can check if the error is ETIMEDOUT and apply your own logic: if (err.message.code === 'ETIMEDOUT') { /* apply logic */ }.

If you want to request for the file again, I suggest using node-retry or node-backoff modules. It makes things much simpler.

If you want to wait longer, you can set timeout option of request yourself. You can set it to 0 for no timeout.


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

...