86

我在 Fedora 中有两个用户：

  1. 瓦尼
  2. 根（很明显！）

我的用户 Wani 的 .bashrc 的内容是：

# .bashrc
echo "Hello"
# Source global definitions
if [ -f /etc/bashrc ]; then
    . /etc/bashrc
fi

# User specific aliases and functions

现在登录到 root 后，我​​输入以下命令：

[root@Dell Wani]# touch try.txt
[root@Dell Wani]# service sshd start
[root@Dell Wani]# scp try.txt Wani@localhost:~/
Wani@localhost's password: 
Hello
[root@Dell Wani]#

现在我登录到 Wani，然后输入：

[Wani@Dell ~]$ cat try.txt
cat: try.txt: No such file or directory
[Wani@Dell ~]$

现在我再次登录到 root 并输入相同的命令-v：

[root@Dell Wani]# scp -v morph.log Wani@localhost:
Executing: program /usr/bin/ssh host localhost, user Wani, command scp -v -t -- .
OpenSSH_5.6p1, OpenSSL 1.0.0j-fips 10 May 2012
debug1: Reading configuration data /etc/ssh/ssh_config
debug1: Applying options for *
debug1: Connecting to localhost [127.0.0.1] port 22.
debug1: Connection established.
debug1: permanently_set_uid: 0/0
debug1: identity file /root/.ssh/id_rsa type -1
debug1: identity file /root/.ssh/id_rsa-cert type -1
debug1: identity file /root/.ssh/id_dsa type -1
debug1: identity file /root/.ssh/id_dsa-cert type -1
debug1: Remote protocol version 2.0, remote software version OpenSSH_5.6
debug1: match: OpenSSH_5.6 pat OpenSSH*
debug1: Enabling compatibility mode for protocol 2.0
debug1: Local version string SSH-2.0-OpenSSH_5.6
debug1: SSH2_MSG_KEXINIT sent
debug1: SSH2_MSG_KEXINIT received
debug1: kex: server->client aes128-ctr hmac-md5 none
debug1: kex: client->server aes128-ctr hmac-md5 none
debug1: SSH2_MSG_KEX_DH_GEX_REQUEST(1024<1024<8192) sent
debug1: expecting SSH2_MSG_KEX_DH_GEX_GROUP
debug1: SSH2_MSG_KEX_DH_GEX_INIT sent
debug1: expecting SSH2_MSG_KEX_DH_GEX_REPLY
debug1: Host 'localhost' is known and matches the RSA host key.
debug1: Found key in /root/.ssh/known_hosts:2
debug1: ssh_rsa_verify: signature correct
debug1: SSH2_MSG_NEWKEYS sent
debug1: expecting SSH2_MSG_NEWKEYS
debug1: SSH2_MSG_NEWKEYS received
debug1: Roaming not allowed by server
debug1: SSH2_MSG_SERVICE_REQUEST sent
debug1: SSH2_MSG_SERVICE_ACCEPT received
debug1: Authentications that can continue: publickey,gssapi-keyex,gssapi-     with-mic,password
debug1: Next authentication method: gssapi-keyex
debug1: No valid Key exchange context
debug1: Next authentication method: gssapi-with-mic
debug1: Unspecified GSS failure.  Minor code may provide more information
Credentials cache file '/tmp/krb5cc_0' not found

debug1: Unspecified GSS failure.  Minor code may provide more information
Credentials cache file '/tmp/krb5cc_0' not found

debug1: Unspecified GSS failure.  Minor code may provide more information


debug1: Unspecified GSS failure.  Minor code may provide more information


debug1: Next authentication method: publickey
debug1: Trying private key: /root/.ssh/id_rsa
debug1: Trying private key: /root/.ssh/id_dsa
debug1: Next authentication method: password
Wani@localhost's password: 
debug1: Authentication succeeded (password).
Authenticated to localhost ([127.0.0.1]:22).
debug1: channel 0: new [client-session]
debug1: Requesting no-more-sessions@openssh.com
debug1: Entering interactive session.
debug1: Sending environment.
debug1: Sending env XMODIFIERS = @im=none
debug1: Sending env LANG = en_US.UTF-8
debug1: Sending command: scp -v -t -- .
Hello
[root@Dell Wani]# debug1: client_input_channel_req: channel 0 rtype exit-status      reply      0
debug1: channel 0: free: client-session, nchannels 1
debug1: fd 0 clearing O_NONBLOCK
debug1: fd 1 clearing O_NONBLOCK
Transferred: sent 1664, received 1976 bytes, in 0.1 seconds
Bytes per second: sent 22961.5, received 27266.8
debug1: Exit status 0

（在我按 Enter 后）

[root@Dell Wani]#

谁能解释一下这里到底发生了什么？为什么文件没有从 root 复制到 Wani？

4

10 回答 10

81

使用echoin a .bashrcwill break scp，正如scp预期通过标准输入/标准输出通道看到它的协议数据。有关此问题的更多讨论，请参阅https://bugzilla.redhat.com/show_bug.cgi?id=20527。

有一些解决方法可用：

  • 'interactive' 标志的条件（例如case $- in *i*，由 Tripleee 建议）
  • 使用该tty实用程序检测交互式 shell（例如if tty > /dev/null或if [ -t 0 ]）
  • 检查值$SSH_TTY

我想你应该使用适合你的任何一个。不幸的是，我不知道最好的（最便携/最可靠的）选项是什么。

于 2012-09-15T21:39:58.683 回答
20

要添加到 nneonneo 的选项，您还可以使用交互式标志设置条件

if [[ $- =~ "i" ]]

我认为这可能是 bash 中最清晰的方法。

于 2013-06-12T16:51:01.327 回答
17

这对我有用，
将.bashrc第一行添加为：

if [ -z "$PS1" ]; then
    return
fi

https://superuser.com/questions/690735/can-i-tell-if-im-in-an-scp-session-in-my-bashrc

于 2014-08-05T07:05:49.720 回答
12

默认的Ubuntu.bashrc包含以下代码片段，它已经解决了这个问题：

# If not running interactively, don't do anything
case $- in
    *i*) ;;
    *) return;;
esac
于 2017-06-24T19:02:40.237 回答
4

在.bashrc中，改为使用 STDERR 作为输出：

echo "# Important Notice" >&2

更新：不要使用它！echo我们最近遇到了一个问题，一个（封闭源代码）工具由于.bashrc. 该工具（使用rcp）完全不期望任何输出，无论是在 STDOUT 还是 STDERR 上。当它得到回声时它卡住了。经验教训：为人类和机器（脚本）分别创建帐户，或者停止通过.bashrc.

于 2015-04-13T15:01:54.533 回答
1

测试交互式 shell 的最便携的方法似乎是：

test -t 0
if [ $? -eq 0 ]
then
    # interactive
    ;
else
    # non-interactive
    ;
fi
于 2018-06-01T18:58:20.597 回答
0

nneonneo 的解决方案也对我有用。但由于我的默认 shell 是 TCSH，我不得不稍微编辑修复如下（在 .tcshrc 中）：

if ( $?SSH_TTY ) then
    exec /bin/bash
endif

只是想我会为了大家的利益分享。

于 2017-01-12T05:42:01.770 回答
0

如果您使用的是 Red Hat Enterprise Linux (RHEL) 或变体，请将执行echo. 或任何您想要的脚本放入 /etc/profile.d/

于 2018-10-16T21:14:36.740 回答
0

如果您还想要 echo 语句，来自@Blauhirn 的回答，您可以继续将您的 echo 语句放在 case 条件之后。

case $- in
    *i*) ;;
    *) return;;
esac

echo "Your Greeting/Warning Message/s here!"
于 2019-10-05T12:56:45.770 回答
-1 
if [ 0 -eq $(shopt -q login_shell; echo $?) ]; then
  echo "do something?"
fi

来源

于 2018-03-23T18:42:05.570 回答