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android之如何以编程方式在android中运行SSH命令

third_qq_23965379c3878727 2025年12月25日 编程语言 17 0

我想运行一个简单的 SSH 命令,例如 ssh -R 80:localhost:1337 serveo.net ,我知道有图书馆jsch要做到这一点,但没有用户名和密码它将无法工作。在我的情况下,不需要身份验证。我怎样才能做到这一点 ?
更新
当我运行命令 ssh -R 80:localhost:1337 serveo.net -v ,我得到以下输出

debug1: Server host key: ssh-rsa SHA256:07jcXlJ4SkBnyTmaVnmTpXuBiRx2+Q2adxbttO9gt0M 
The authenticity of host 'serveo.net (159.89.214.31)' can't be established. 
RSA key fingerprint is SHA256:07jcXlJ4SkBnyTmaVnmTpXuBiRx2+Q2adxbttO9gt0M. 
Are you sure you want to continue connecting (yes/no/[fingerprint])? yes 
Warning: Permanently added 'serveo.net,159.89.214.31' (RSA) to the list of known hosts. 
debug1: rekey out after 134217728 blocks 
debug1: SSH2_MSG_NEWKEYS sent 
debug1: expecting SSH2_MSG_NEWKEYS 
debug1: SSH2_MSG_NEWKEYS received 
debug1: rekey in after 134217728 blocks 
debug1: Will attempt key: /home/paranoid/.ssh/id_rsa  
debug1: Will attempt key: /home/paranoid/.ssh/id_dsa  
debug1: Will attempt key: /home/paranoid/.ssh/id_ecdsa  
debug1: Will attempt key: /home/paranoid/.ssh/id_ecdsa_sk  
debug1: Will attempt key: /home/paranoid/.ssh/id_ed25519  
debug1: Will attempt key: /home/paranoid/.ssh/id_ed25519_sk  
debug1: Will attempt key: /home/paranoid/.ssh/id_xmss  
debug1: SSH2_MSG_SERVICE_ACCEPT received 
debug1: Authentications that can continue: publickey,keyboard-interactive 
debug1: Next authentication method: publickey 
debug1: Trying private key: /home/paranoid/.ssh/id_rsa 
debug1: Trying private key: /home/paranoid/.ssh/id_dsa 
debug1: Trying private key: /home/paranoid/.ssh/id_ecdsa 
debug1: Trying private key: /home/paranoid/.ssh/id_ecdsa_sk 
debug1: Trying private key: /home/paranoid/.ssh/id_ed25519 
debug1: Trying private key: /home/paranoid/.ssh/id_ed25519_sk 
debug1: Trying private key: /home/paranoid/.ssh/id_xmss 
debug1: Next authentication method: keyboard-interactive 
debug1: Authentication succeeded (keyboard-interactive). 
Authenticated to serveo.net ([159.89.214.31]:22). 
debug1: Remote connections from LOCALHOST:80 forwarded to local address localhost:1337 
debug1: channel 0: new [client-session] 
debug1: Entering interactive session. 
debug1: pledge: network 
debug1: Sending environment. 
debug1: Sending env LC_ADDRESS = en_US.UTF-8 
debug1: Sending env LC_NAME = en_US.UTF-8 
debug1: Sending env LC_MONETARY = en_US.UTF-8 
debug1: Sending env LC_PAPER = en_US.UTF-8 
debug1: Sending env LANG = en_US.UTF-8 
debug1: Sending env LC_IDENTIFICATION = en_US.UTF-8 
debug1: Sending env LC_TELEPHONE = en_US.UTF-8 
debug1: Sending env LC_MEASUREMENT = en_US.UTF-8 
debug1: Sending env LC_TIME = en_US.UTF-8 
debug1: Sending env LC_NUMERIC = en_US.UTF-8 
debug1: remote forward success for: listen 80, connect localhost:1337 
debug1: All remote forwarding requests processed 
Forwarding HTTP traffic from https://vacuus.serveousercontent.com 
 
它还创建 known_hosts .ssh 内的文件文件夹。 known_hosts的内容id 关注
|1|2K3SBFWPCPqI3poBW2X99LiuP8c=|cqCa7m1e23x1P9UpcUuPac+KKI8= ssh-rsa AAAAB3NzaC1yc2EAAAADAQABAAABAQDxYGqSKVwJpQD1F0YIhz+bd5lpl7YesKjtrn1QD1RjQcSj724lJdCwlv4J8PcLuFFtlAA8AbGQju7qWdMN9ihdHvRcWf0tSjZ+bzwYkxaCydq4JnCrbvLJPwLFaqV1NdcOzY2NVLuX5CfY8VTHrps49LnO0QpGaavqrbk+wTWDD9MHklNfJ1zSFpQAkSQnSNSYi/M2J3hX7P0G2R7dsUvNov+UgNKpc4n9+Lq5Vmcqjqo2KhFyHP0NseDLpgjaqGJq2Kvit3QowhqZkK4K77AA65CxZjdDfpjwZSuX075F9vNi0IFpFkGJW9KlrXzI4lIzSAjPZBURhUb8nZSiPuzj 
|1|F1SIE4/IIEjZPJfHBIx90xnSjSU=|NKCdGqv3SFcGcAqPLvVfuRXI4Ok= ssh-rsa AAAAB3NzaC1yc2EAAAADAQABAAABAQDxYGqSKVwJpQD1F0YIhz+bd5lpl7YesKjtrn1QD1RjQcSj724lJdCwlv4J8PcLuFFtlAA8AbGQju7qWdMN9ihdHvRcWf0tSjZ+bzwYkxaCydq4JnCrbvLJPwLFaqV1NdcOzY2NVLuX5CfY8VTHrps49LnO0QpGaavqrbk+wTWDD9MHklNfJ1zSFpQAkSQnSNSYi/M2J3hX7P0G2R7dsUvNov+UgNKpc4n9+Lq5Vmcqjqo2KhFyHP0NseDLpgjaqGJq2Kvit3QowhqZkK4K77AA65CxZjdDfpjwZSuX075F9vNi0IFpFkGJW9KlrXzI4lIzSAjPZBURhUb8nZSiPuzj 

请您参考如下方法:

下面是一个通过 Jsch 使用 serveo 的例子。这里唯一真正的“技巧”似乎是特定于serveo的。在 session 打开“shell”或“exec” channel 之前,Serveo 不会接受隧道请求。 Serveo 使用 shell/exec channel 向客户端发送状态消息。所以需要先打开一个shell或者exec channel ,然后再请求远程转发。
SSH 协议(protocol)期望客户端提供一个用户名来登录。 ssh如果您不提供,命令行实用程序默认使用本地用户名。 Jsch 同样默认使用 system property "user.name" 的值。 . Serveo 有点不寻常,因为它会接受来自任何人的 SSH 连接,因此您可以使用任何您喜欢的名称。你可以让 jsch 使用它的默认值——假设它适用于 android——或者你可以使用像“x”或“foo”这样的硬编码名称。

import com.jcraft.jsch.*; 
 
public class App { 
    public static void main(String[] arg) { 
        JSch.setLogger(new MyLogger()); 
        JSch jsch = new JSch(); 
        try { 
            final Session session = jsch.getSession("serveo.net"); 
            session.setHostKeyRepository(new NoSecurityRepo()); 
            session.connect(); 
 
            // Establish a shell channel to receive messages from serveo 
            ChannelShell cc = (ChannelShell)session.openChannel("shell"); 
            cc.setPty(false); 
            cc.setInputStream(System.in); 
            cc.setOutputStream(System.out); 
            cc.setExtOutputStream(System.err); 
            cc.connect(); 
 
            // Now open a remote forward 
            session.setPortForwardingR(null, 80, "localhost", 3000); 
        } catch (Exception e) { 
            e.printStackTrace(); 
        } 
    } 
 
    private static class MyLogger implements Logger { 
        @Override 
        public boolean isEnabled(int level) { return true; } 
        @Override 
        public void log(int level, String message) { 
            System.err.printf("%s\t%s%n", level, message); 
        } 
    } 
 
    /* 
     * This implementation of HostKeyRepository is just for demonstration. 
     * Every time you use it in production, a kitten dies. Please think of 
     * the kittens. 
     */ 
    private static class NoSecurityRepo implements HostKeyRepository { 
        @Override 
        public int check(String host, byte[] key) { return HostKeyRepository.OK; } 
        @Override 
        public void add(HostKey hostkey, UserInfo ui) { } 
        @Override 
        public void remove(String host, String type) { } 
        @Override 
        public void remove(String host, String type, byte[] key) { } 
        @Override 
        public String getKnownHostsRepositoryID() { return "NoSecurityRepo"; } 
        @Override 
        public HostKey[] getHostKey() { return new HostKey[0]; } 
        @Override 
        public HostKey[] getHostKey(String host, String type) { return new HostKey[0]; } 
    } 
} 


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