"Force psuedo-tty allocation" doesn't translate to pssh - ssh

Supposedly, pssh's -x option passes along extra SSH command-line arguments. SSH's "-t" option should have taken care of the "Pseudo-terminal" error. Is there another pssh/ssh option that should be used?
# pssh -i -H ec2-user#xxx.xxx.xxx.xx1 -H ec2-user#xxx.xxx.xxx.xx2 -x "-t -i /tmp/key.pem" 'sudo hostname'
[1] 13:46:54 [FAILURE] ec2-user#xxx.xxx.xxx.xx1 Exited with error code 1
Stderr: Pseudo-terminal will not be allocated because stdin is not a terminal.
sudo: sorry, you must have a tty to run sudo
[2] 13:46:54 [FAILURE] ec2-user#xxx.xxx.xxx.xx1 Exited with error code 1
Stderr: Pseudo-terminal will not be allocated because stdin is not a terminal.
sudo: sorry, you must have a tty to run sudo
Without the "-t" and the "sudo", the command works fine, but I need to run some commands as sudo.
# pssh -i -H ec2-user#xxx.xxx.xxx.xx1 -H ec2-user#xxx.xxx.xxx.xx2 -x "-i /tmp/key.pem" 'hostname'
[1] 14:08:35 [SUCCESS] ec2-user#xxx.xxx.xxx.xx1
ip-10-0-0-140
[2] 14:08:35 [SUCCESS] ec2-user#xxx.xxx.xxx.xx2
ip-10-0-0-139

Try running pssh like this, so that the "-t" option is specified twice:
pssh -i -H ec2-user#... -x "-t -t -i /tmp/key.pem" 'sudo hostname'
^^^^^
The ssh man page says this about "-t" (emphasis added):
-t
Force pseudo-terminal allocation. This can be used to execute arbitrary
screen-based programs on a remote machine, which can be very
useful, e.g. when implementing menu services. Multiple -t options
force tty allocation, even if ssh has no local tty.
You are apparently running pssh in such a way that ssh has no local tty. So you have to arrange for ssh to be run with "-t" specified twice. This forces ssh to request a remote tty, despite not having a local tty.

Related

Escaping karate.fork Commands

I am trying to run the following command in karate using karate.fork
ssh -o ProxyCommand="ssh -W %h:%p -i ~/.ssh/id_rsa root#myjumphost" -i ~/.ssh/id_rsa -o StrictHostKeyChecking=no -o PasswordAuthentication=no root#finaldest echo test
I have broken this up into an array to pass to karate.fork like so:
[
ssh,
-o,
ProxyCommand="ssh -W %h:%p -i ~/.ssh/id_rsa root#myjumphost",
-i,
~/.ssh/id_rsa,
-o,
StrictHostKeyChecking=no,
-o,
PasswordAuthentication=no,
root#finaldest,
echo test
]
Then run the command like so:
* karate.fork(args) where args is the array mentioned above
The command works when I paste it into the terminal and run it manually, however when run with karate.fork I get
zsh:1: no such file or directory: ssh -W finaldest:22 -I ~/.ssh/id_rsa root#myjumphost
kex_exchange_identification: Connection closed by remote host
I have tried adding a few backslashes before the " in the ProxyCommand but no amount of back slashes fixes this issue. I think I am misunderstanding what karate.fork is doing to run the command, is there some internal parsing or manipulating of the given input? I was able to get this command to work when I used useShell: true however this option breaks other tests for me so I would really like to avoid it.
I had to remove the double quotes, seems like they didn't play well with karate.fork and the command still runs without them
[
ssh,
-o,
ProxyCommand=ssh -W %h:%p -i ~/.ssh/id_rsa root#myjumphost,
-i,
~/.ssh/id_rsa,
-o,
StrictHostKeyChecking=no,
-o,
PasswordAuthentication=no,
root#finaldest,
echo test
]

How to deal with "Pseudo-terminal will not be allocated because stdin is not a terminal."

ssh -t remotehost vim /tmp/x.txt
I know that I can run a command like the above.
But I would like to be able to run any local bash code in a remote machine. For this reason, I'd like to call the remote 'bash -s' so that can process any local bash code.
ssh -t remotehost 'bash -s' <<< vim /tmp/x.txt
However, the above example shows "Pseudo-terminal will not be allocated because stdin is not a terminal." Is there any way to let ssh take local bash code via stdin and run it via the remote 'bash -s'? Thanks.
ssh -t remotehost 'bash -s' <<< vim /tmp/x.txt
You're getting the "Pseudo-terminal will not be allocated..." message because you're running ssh with a single -t option, when the standard input to the ssh process isn't a TTY. ssh prints that message specifically in this case. The documentation for -t says:
-t
Force pseudo-terminal allocation. This can be used to execute arbitrary screen-based programs on a remote machine, which can be very useful, e.g. when implementing menu services. Multiple -t options force tty allocation, even if ssh has no local tty.
The -t command-line option is related to the ssh configuration option RequestTTY:
RequestTTY
Specifies whether to request a pseudo-tty for the session. The argument may be one of: no (never request a TTY), yes (always request a TTY when standard input is a TTY), force (always request a TTY) or auto (request a TTY when opening a login session). This option mirrors the -t and -T flags for ssh(1).
A single -t is equivalent to "RequestTTY yes", while two of them is equivalent to "RequestTTY force".
If you want your remote command(s) to run with a TTY, then specify -t twice:
ssh -tt remotehost 'bash -s' <<< vim /tmp/x.txt
or
ssh -t -t remotehost 'bash -s' <<< vim /tmp/x.txt
ssh will allocate a TTY for the remote system and it won't print that message.
If the command(s) being run on the remote system don't require a TTY, you can leave the -t option out:
ssh remotehost 'bash -s' <<< vim /tmp/x.txt
I believe the following might suit your purposes:
vim /tmp/x.txt ; ssh remotehost 'bash -s' < /tmp/x.txt
The first expression (vim ...) allows you to specify the commands you want to execute remotely as a local file called /tmp/x.txt; the second expression (ssh ...) calls the remote bash interpreter, and sends the contents of your local file to it for execution. Note that you do not need the -t option for ssh in this case (which gave rise to the pseudo-terminal warning), and that you do not need to use a here string (<<<) but can use the normal file input operator (<).
This solution seems to work for, e.g., the following file contents:
echo These commands are being executed on $HOSTNAME
echo This is a second command

how does fabric execute commands?

i am wondering how does fabric execute commands.
Let's say I give him env.user=User, env.host=HOST. Then i ask him to sudo('ls')
Is that equivalent to me typing in a shell : ssh User#host 'sudo(/bin/ls)'
or it's more : ssh User#host in a first time, then sudo ls commande in a seconde time ?
I'm asking that because sometimes using a shell, if the TTY has a bad configuration (I am a bit blurry on this), ssh User#Host 'sudo /bin/ls'
return : sudo: no tty present and no askpass program specified
but you can first log in with ssh User#Host then sudo ls and it works.
I don't know how to replicate the no tty error, but I know it can occurs. Would this block the sudo commande from Fabric?
Basically how it works is:
First a connection is established (equivalent as doing ssh User#host)
Over this connection a command is executed as follows:
sudo -S -p 'sudo password:' /bin/bash -l -c "your_command"
You can also allow Fabric not to request a pty with either pty=False argument, env.always_use_pty=False or --no-pty commandline option.

Sudo over SSH mixes up password tty and stdin

Setup:
Local *nix machine with a SQL script script.sql (Postgres).
Remote machine remote (Debian 7) with Postgres.
I can SSH in as some_user, who is a sudoer.
Anything with Postgres needs to be done as postgres user.
The server only listens on localhost:5432.
How do I execute script.sql on remote without copying it there first?
This works well:
ssh -t some_user#remote 'sudo -u postgres psql -c "COMMANDS FOO BAR"'
The -t flag means that sudo will ask for some_user's password correctly on the local terminal.
One thing remains, to be able to pipe script.sql to psql. This does not work:
ssh -t some_user#remote 'sudo -u postgres psql' < script.sql
It fails with the message:
Pseudo-terminal will not be allocated because stdin is not a terminal.
sudo: no tty present and no askpass program specified
Edit: simplified example
Postgres and psql don't seem to figure much in the problem. The following code has the same issues:
ssh some_user#remote xargs sudo ls < input_file
The problem seems to be: we need to send 2 inputs to sudo, both the password using a tty, and the stdin to pass to ls.
Edit: even simpler
ssh localhost xargs sudo ls < input_file
sudo: no tty present and no askpass program specified
Adding -t does not work:
$ ssh -t localhost xargs sudo ls < input_file
Pseudo-terminal will not be allocated because stdin is not a terminal.
sudo: no tty present and no askpass program specified
Adding another -t does not work either:
$ ssh -t -t localhost xargs sudo ls < input_file
<content of input_file>
<waiting on a prompt>
ssh -T some_user#remote "sudo -u postgres psql -f-" < script.sql
"-f-" will read the script from STDIN. Just redirect the file in there, and there you go.
Don't bother with -t option to ssh, you don't need a full terminal for this.
ssh -T ${user}#${ip} sudo DEBIAN_FRONTEND=noninteractive postgres psql -f- < test.sql
Use DEBIAN_FRONTEND=noninteractive for resolve no tty present or equivalent of your distribution.

Running ssh command and keeping connection

Is there a way to execute a command before accessing a remote terminal
When I enter this command:
bash
$> ssh user#server.com 'ls'
The ls command is executed on the remote computer but ssh quits and I cannot continue in my remote session.
Is there a way of keeping the connection? The reason that I am asking this is that I want to create a setup for ssh session without having to modify the remote .bashrc file.
I would force the allocation of a pseudo tty and then run bash after the ls command:
syzdek#host1$ ssh -t host2.example.com 'ls -l /dev/null; bash'
-rwxrwxrwx 1 root other 27 Apr 1 2005 /dev/null
bash-4.1$
You can try using process subsitution on the init file of bash. In the example below, I define a function myfunc:
myfunc () {
echo "Running myfunc"
}
which I transform to a properly-escaped one-liner echoed in the <(...) construct for process subsitution for the --init-file argument of bash:
$ ssh -t localhost 'bash --init-file <( echo "myfunc() { echo \"Running myfunc\" ; }" ) '
Password:
bash-3.2$ myfunc
Running myfunc
bash-3.2$ exit
Note that once connected, my .bashrc is not sourced but myfunc is defined and properly usable in an interactive session.
It might prove a little difficult for more complex bash functions, but it works.