Java JSch:JSch tty代码用于执行交互命令，我做错了什么？

我正在尝试使用JSch连接到服务器，然后执行交互式

su

命令

你能告诉我我错过了什么吗？这种东西每次都挂着，我看不出有什么理由这样做。 特别是在发送密码之前，我特别等待密码字符串（实际上是

assword

而不是

：

）

这就是我正在做的：

以user1/pass1身份登录

然后执行

su-user2-c命令行

在通道输入流中传递密码

无论如何，这里是主要的执行函数，它接受user1/pass1、user2/pass2和命令行

public String execute (String host, String nuser, String npass,
                       String puser, String ppass, String commandLine)
    {
      try{
        synchronized(this) 
        {
          session = jsch.getSession(nuser,host,22);
          session.setConfig("StrictHostKeyChecking", "no");
          pstr=npass;
          //session.setPassword(npass);
          String authmethods=  session.getConfig("PreferredAuthentications");
          System.out.println(authmethods);
          UserInfo ui=new SUSessionExecution.UInfo();
          Thread.sleep(150);
          if(authmethods.contains("keyboard-interactive"))
          { System.out.print("keyboard-interactive"); session.setUserInfo(ui); } 
          else if ( authmethods.contains("password") )
          { System.out.print("password"); session.setPassword(pstr);  }

          session.connect();
          channel = session.openChannel("exec");
          nuser=null;
          npass=null;


          ((ChannelExec)channel).setPty(true);
          ((ChannelExec)channel).setPtyType("vt100");
          String command= "su - " + puser + " -c " + commandLine  + "\n";
          //((ChannelExec)channel).setCommand(commandLine);

          byte[] cmdBuffer=command.getBytes();
          ByteArrayInputStream bi = new ByteArrayInputStream(cmdBuffer);
          channel.setInputStream(bi);

          ByteArrayOutputStream bo = new ByteArrayOutputStream();
          ((ChannelExec)channel).setErrStream(bo);

          sessionOutput = channel.getInputStream();
          //sessionError = channel.getExtInputStream();


          channel.connect();

          session_open=true;
          // it is only here our session is fully functional.

          boolean sustatus;//=establishSU(commandIO, channel,puser,ppass);
          // NEEDS REPLACE

          //commandIO.write(command.getBytes());
          //commandIO.flush();



          String standardOutBuffer="";
          String standardErrBuffer="";

          int counter;
          byte[] byteBuffer = new byte[1024];


          while(sessionOutput.available() > 0)
            { counter=0; //byteBuffer=null;
              counter=sessionOutput.read(byteBuffer, 0, byteBuffer.length);
              if(counter < 0) { throw new java.io.IOException(); }
              standardOutBuffer += new String(byteBuffer,0,counter);
              if(standardOutBuffer.contains("assword")){break;}
            }

            /*if(sessionError.available() > 0)
            { counter=0; //byteBuffer=null;
              counter=sessionError.read(byteBuffer, 0, byteBuffer.length);
              if(counter < 0) { throw new java.io.IOException(); }
              standardErrBuffer += new String(byteBuffer,0,counter);
              if(standardErrBuffer.contains("assword")){break;}}*/


        commandIO = new PipedOutputStream();
        sessionInput = new PipedInputStream(commandIO);
        channel.setInputStream(sessionInput);
        commandIO.write(new String(ppass+"\n").getBytes());
        commandIO.flush();

        counter=0; standardOutBuffer="";
        while((counter = sessionOutput.read(byteBuffer,0,byteBuffer.length)) != -1)
        { standardOutBuffer += new String(byteBuffer,0,counter); }

        closeComponents();
        return standardOutBuffer;
        }
      }
      catch(com.jcraft.jsch.JSchException jse) 
      { session_open=false;su_space_open=false;jse.printStackTrace();
        closeComponents(); return null; }
      catch(java.io.IOException ioe)
      { session_open=false;su_space_open=false;ioe.printStackTrace();
        closeComponents(); return null; }
      catch(InterruptedException ie)
      { ie.printStackTrace(); }

        return null;
    }

---

看来你实施了

  UIKeyboardInteractive#promptKeyboardInteractive()

返回包含“null”的字符串数组

至于sudo，我建议参考

---

在伙计们之间，这里有一条错误消息：com.jcraft.jsch.Buffer.putString（Buffer.java:59）com.jcraft.jsch.UserAuthKeyboardInteractive.start（UserAuthKeyboardInteractive.java:183）com.jcraft.jsch.Session.connect上的线程“main”java.lang.NullPointerException中出现异常（Session.java:442）在com.jcraft.jsch.Session.connect（Session.java:162）在susessionexecution.SUExecute.execute（SUExecute.java:53）在Tester.main（Tester.java:17）处（Tester.java:12）\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuues到1，但是，我尝试了可变长度数组并发送它们。结果是一样的。有趣的是它只运行了一次，没有给出错误。我真的看不到与计时有关的任何东西，但现在看到这种行为，我觉得似乎不得不与计时有关。谢谢，ChandanThanks，我将研究promptKeyboardInteractive黑客再次出现。从代码中可以看出，使其非交互是一种令人不快的黑客行为。在几乎所有情况下，session.setPassword（pstr）都适用于“键盘交互”auth方法。你试过了吗？@paulo，嘿，非常感谢你将问题重新编排得井井有条。看起来更清晰了。我刚刚检查了promptKeyboardInteractive（），使它只传递一个字符串（因此无法发送null），但是，这个错误仍然出现。这是一个糟糕的测试技巧。

public string[]promptKeyboardInteractive（字符串目标、字符串名称、字符串指令、字符串[]提示符、布尔[]回显）{System.out.println（“\n”+prompt.length+”\n\n”）；字符串[]响应=新字符串[prompt.length]；响应[0]=passwd；返回响应；}

我正在尝试其他一些事情，将保持此线程的发布。我猜响应[0]为null，因为“passwd”为null。

public String[] promptKeyboardInteractive (String destination, String name,
                                           String instruction, String[] prompt,
                                           boolean[] echo)
{
    System.out.println("\n"+prompt.length+"\n\n");
    String[] response=new String[prompt.length];
    response[0] = passwd;
    return response;
}

  UIKeyboardInteractive#promptKeyboardInteractive()