在尝试将文件上传到我们的服务器时，我收到以下异常
com.jcraft.jsch.JSchException: Auth fail
at com.jcraft.jsch.Session.connect(Session.java:464)
at com.jcraft.jsch.Session.connect(Session.java:158)
at FtpService.transferFileToReciever(FtpService.java:80)
at FtpService.transferFileToReciever(FtpService.java:54)
at FtpService.transferFileToRecievers(FtpService.java:44)
at FtpService.transferSingeFile(FtpService.java:241)
at FtpService.main(FtpService.java:26)
Auth fail
源文件中函数transferFileToReciever的部分是
JSch jsch = new JSch();
jsch.addIdentity("/root/.ssh/id_dsa");
Session session = jsch.getSession(username, host, 22);
session.setUserInfo(serverinfo);
session.connect(); //geting exception here
boolean ptimestamp = true;
密码正常，因为我可以使用ssh登录，但是使用JSCh它甚至不提供密钥，用户名和密码。 使用带有java版本“1.6.0_25”的id_dsa密钥。 可能是什么错误？
发现其他类似的问题，但不是答案。 提前谢谢。
答案 0 :(得分：21)
跟踪根本原因，我终于发现dsa类型的公钥未添加到远程服务器上的授权密钥。附加同样适用于我。
ssh正在使用rsa密钥，导致我回顾我的代码。
谢谢大家。
答案 1 :(得分：3)
示例情况，当我从远程服务器获取文件并将其保存在本地机器中时 包连接器;
import java.io.File;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import com.jcraft.jsch.ChannelSftp;
import com.jcraft.jsch.JSch;
import com.jcraft.jsch.JSchException;
import com.jcraft.jsch.Session;
import com.jcraft.jsch.SftpException;
public class Main {
public static void main(String[] args) throws JSchException, SftpException, IOException {
// TODO Auto-generated method stub
String username = "XXXXXX";
String host = "XXXXXX";
String passwd = "XXXXXX";
JSch conn = new JSch();
Session session = null;
session = conn.getSession(username, host, 22);
session.setPassword(passwd);
session.setConfig("StrictHostKeyChecking", "no");
session.connect();
ChannelSftp channel = null;
channel = (ChannelSftp)session.openChannel("sftp");
channel.connect();
channel.cd("/tmp/qtmp");
InputStream in = channel.get("testScp");
String lf = "OBJECT_FILE";
FileOutputStream tergetFile = new FileOutputStream(lf);
int c;
while ( (c= in.read()) != -1 ) {
tergetFile.write(c);
}
in.close();
tergetFile.close();
channel.disconnect();
session.disconnect();
}
}
答案 2 :(得分：3)
如果用户名/密码包含任何特殊字符，则在camel配置中使用RAW配置
等值 RAW(se+re?t&23)其中se+re?t&23是实际密码
RAW({abc.ftp.password})其中{abc.ftp.password}值来自spring属性文件。
通过使用RAW，解决了我的问题。
答案 3 :(得分：2)
发现其他类似的问题，但不是答案。
知道你在哪里找到这个问题会很有趣。
据我所知，根据com.jcraft.jsch.JSchException: Auth cancel
尝试向方法.addIdentity()添加密码。如果生成没有密钥文件的密钥文件，则可以使用""。
另一个错误来源是指纹字符串。如果它不匹配，您将获得身份验证失败（取决于目标服务器）。
最后我的工作源代码 - 在我解决丑陋的管理任务之后：
public void connect(String host, int port,
String user, String pwd,
String privateKey, String fingerPrint,
String passPhrase
) throws JSchException{
JSch jsch = new JSch();
String absoluteFilePathPrivatekey = "./";
File tmpFileObject = new File(privateKey);
if (tmpFileObject.exists() && tmpFileObject.isFile())
{
absoluteFilePathPrivatekey = tmpFileObject.getAbsolutePath();
}
jsch.addIdentity(absoluteFilePathPrivatekey, passPhrase);
session = jsch.getSession(user, host, port);
//Password and fingerprint will be given via UserInfo interface.
UserInfo ui = new UserInfoImpl(pwd, fingerPrint);
session.setUserInfo(ui);
session.connect();
Channel channel = session.openChannel("sftp");
channel.connect();
c = (ChannelSftp) channel;
}
答案 4 :(得分：2)
我也面临Auth Fail问题，我的代码问题是我有
＆＃xA;＆＃xA; channelSftp.cd（“”）;＆＃xA ; ＆＃xA;＆＃xA; 将其更改为
＆＃xA;＆＃xA; channelSftp.cd（“。”） ;＆＃xA; ＆＃xA;＆＃xA; 然后就可以了。
＆＃xA;答案 5 :(得分：1)
尝试明确添加auth方法，如下所示，因为有时它是必需的：
session.setConfig("PreferredAuthentications", "password");
答案 6 :(得分：0)
在我的情况下，我正在使用以下依赖项
$ valgrind ./bin/person_struct
==2078== Memcheck, a memory error detector
==2078== Copyright (C) 2002-2015, and GNU GPL'd, by Julian Seward et al.
==2078== Using Valgrind-3.12.0 and LibVEX; rerun with -h for copyright info
==2078== Command: ./bin/person_struct
==2078==
***PROGRAM TO SORT PERSONS***
Firstname: John
Lastname: Wayne
Age: 91
***NEXT PERSON***
Firstname: Jane
Lastname: Doe
Age: 101
***NEXT PERSON***
Firstname: Mickey
Lastname: Mouse
Age: 99
***NEXT PERSON***
***CHOOSE HOW TO SORT***
By firstname: 1
By lastname: 2
By age: 3
Exit Program: 4
1
***SORTING BY FIRSTNAME...***
***DONE***
Firstname: Jane | Lastname: Doe | Age: 101
Firstname: John | Lastname: Wayne | Age: 91
Firstname: Mickey | Lastname: Mouse | Age: 99
***CHOOSE HOW TO SORT***
By firstname: 1
By lastname: 2
By age: 3
Exit Program: 4
2
***SORTING BY LASTNAME...***
***DONE***
Firstname: Jane | Lastname: Doe | Age: 101
Firstname: Mickey | Lastname: Mouse | Age: 99
Firstname: John | Lastname: Wayne | Age: 91
***CHOOSE HOW TO SORT***
By firstname: 1
By lastname: 2
By age: 3
Exit Program: 4
3
***SORTING BY AGE...***
***DONE***
Firstname: John | Lastname: Wayne | Age: 91
Firstname: Mickey | Lastname: Mouse | Age: 99
Firstname: Jane | Lastname: Doe | Age: 101
***CHOOSE HOW TO SORT***
By firstname: 1
By lastname: 2
By age: 3
Exit Program: 4
4
***EXITING PROGRAM***
==2078==
==2078== HEAP SUMMARY:
==2078== in use at exit: 0 bytes in 0 blocks
==2078== total heap usage: 3 allocs, 3 frees, 396 bytes allocated
==2078==
==2078== All heap blocks were freed -- no leaks are possible
==2078==
==2078== For counts of detected and suppressed errors, rerun with: -v
==2078== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)
并获得相同的Auth异常失败，但是更新了对以下版本的依赖关系并解决了问题。
#include <iostream>
#include <stdio.h>
using namespace std;
int main(void)
{
int testCases, i, j, k, n, num, digits, carry = 0, temp;
scanf("%d", &testCases);
int testArr[160];
for (i = 0; i < testCases; i++)
{
scanf("%d", &n);
num = n;
if (n == 0 || n == 1)
{
testArr[0] = 1;
digits = 1;
}
else
{
k = 0;
for (j = 10; n != 0; j = j * 10)
{
testArr[k] = n % j;
n = n / j;
k++;
}
digits = k;
for (j = 1; j < num; j++)
{
for (k = 0; k < digits; k++)
{
temp = testArr[k] * j + carry;
if (temp > 10)
{
testArr[k] = temp % 10;
carry = temp / 10;
}
else
{
testArr[k] = temp;
carry = 0;
}
}
if (carry > 10)
{
testArr[k] = carry % 10;
k++;
testArr[k] = carry / 10;
digits = k + 1;
carry = 0;
}
else if (carry > 0)
{
testArr[k] = carry;
digits = k + 1;
carry = 0;
}
}
}
for (k = (digits - 1); k >= 0; k--)
{
printf("%d", testArr[k]);
}
printf("\n");
}
return 0;
}