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

java - com.jcraft.jsch.JSchException: UnknownHostKey

I'm trying to use Jsch to establish an SSH connection in Java. My code produces the following exception:

com.jcraft.jsch.JSchException: UnknownHostKey: mywebsite.com. 
RSA key fingerprint is 22:fb:ee:fe:18:cd:aa:9a:9c:78:89:9f:b4:78:75:b4

I cannot find how to verify the host key in the Jsch documentation. I have included my code below.

import com.jcraft.jsch.JSch;
import com.jcraft.jsch.Session;

public class ssh {
    public static void main(String[] arg) {

        try {
            JSch jsch = new JSch();

            //create SSH connection
            String host = "mywebsite.com";
            String user = "username";
            String password = "123456";

            Session session = jsch.getSession(user, host, 22);
            session.setPassword(password);
            session.connect();

        } catch(Exception e) {
            System.out.println(e);
        } 
    }
}
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 would either:

  1. Try to ssh from the command line and accept the public key (the host will be added to ~/.ssh/known_hosts and everything should then work fine from Jsch) -OR-
  2. Configure JSch to not use "StrictHostKeyChecking" (this introduces insecurities and should only be used for testing purposes), using the following code:

    java.util.Properties config = new java.util.Properties(); 
    config.put("StrictHostKeyChecking", "no");
    session.setConfig(config);
    

Option #1 (adding the host to the ~/.ssh/known_hosts file) has my preference.


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

...