linux - SSH parameter help on Ubuntu Server

07
2014-07

inquisitor

I have a variable setup like this:

SSH_EXEC="ssh -X -o ControlPath=~/.ssh/master-$$ -o ControlPersist=60"

The parameter in question is the -X, because if I then call this command within my local script

$SSH_EXEC user@server "./server_script.sh $aFile"

which in a nutshell does the following:

if [ -e /path/"$1".name ] || [ -e /path/"$1" ]
then 
   do something
else
   error
   echo "/path/"$1".name"
   exit
fi

Everything works! But if I replace -X with -t, my server_script fails to the else on the test. I'd prefer not use x11, but I'm not sure what the difference is that is causing it to fail in one instance, and pass in the other.

EDIT

So I just did more troubleshooting and decided to echo the value of /path/$1.name on the server and it is garbled junk. If my $aFile name happened to be hello.name the result of the echo shows .nameello and also rids the beginning of the path when I use -t in place of -X.

What could be causing the corruption of my variables?

Answers

Hasturkun

The variable $aFile probably has a trailing Carriage Return character, causing the terminal to return to the start of the line when encountered.

As an example, the following:

echo "/path/"hello^M".name"

outputs:

/.namehello

This sort of thing might happen due to a shell script with DOS line endings, causing a line like

aFile = "hello"

to be interpreted as

aFile = "hello"^M

If that's the case, you should be able to convert your script to use Unix line endings using the dos2unix utility.

Related Answers

Tzury Bar Yochay

I would have say to use RSA key or try this one: http://bash.cyberciti.biz/security/expect-ssh-login-script/ bare in mind that you need to install expect tool.

But since you have say NO to RSA and to expect, the only way is to write your own script utilizing Twisted's conch (http://twistedmatrix.com/projects/conch/documentation/examples/) or any other platform available which offer SSH implementation such as ruby or Perl

tangens

You could try host-based authentication to log in without a password. In this case you don't need a private key for the user. But you still need a host key. Would this be a possible solution for you?

The man page of ssh sais:

Host-based authentication works as follows: If the machine the user logs in from is listed in /etc/hosts.equiv or /etc/ssh/shosts.equiv on the remote machine, and the user names are the same on both sides, or if the files ~/.rhosts or ~/.shosts exist in the user's home directory on the remote machine and contain a line containing the name of the client machine and the name of the user on that machine, the user is considered for login. Additionally, the server must be able to verify the client's host key (see the description of /etc/ssh/ssh_known_hosts and ~/.ssh/known_hosts, below) for login to be permitted. This authentication method closes security holes due to IP spoofing, DNS spoofing, and routing spoofing. [Note to the administrator: /etc/hosts.equiv, ~/.rhosts, and the rlogin/rsh protocol in general, are inherently insecure and should be disabled if security is desired.]

geek

If you are familiar with Python, you can try paramiko.

Ryan Thompson

In debian and derivatives, there's a package called sshpass (website here) that's designed to do exactly that.

nilbus

You can also use The dropbear SSH client, which allows specifying the password on the command line, like this: Dropbear for automatic login for shell scripts

Jimmy Hedman

Have a look at the ControlMaster option in ssh_config. This allows you to open a authenticated tunnel by hand that all following ssh connections uses.

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linux - SSH parameter help on Ubuntu Server