bash - How to use STDOUT inside /etc/ssh/sshrc without breaking SCP

Question

I want to call a program when any SSH user logs in that prints a welcome message. I did this by editing the /etc/ssh/sshrc file:

#!/bin/bash

ip=`echo $SSH_CONNECTION | cut -d " " -f 1`
echo $USER logged in from $ip

For simplicity, I replaced the program call with a simple echo command in the example

The problem is, I learned SCP is sensitive to any script that prints to stdout in .bashrc or, apparently, sshrc. My SCP commands failed silently. This was confirmed here: https://stackoverflow.com/a/12442753/2887850

Lots of solutions offered quick ways to check if the user is in an interactive terminal:

• if [[ $- != *i* ]]; then return; fi link
  • Fails becase [ is not linked
• case $- in *i* link
  • Fails because in is not recognized?
• Use tty program (same as above)
  • tty gave me a bizarre error code when executed from sshrc

While all of those solutions could work in a normal BASH environment, none of them work in the sshrc file. I believe that is because PATH (and I suspect a few other things) aren't actually available when executing from sshrc, despite specifying BASH with a shebang. I'm not really sure why this is the case, but this link is what tipped me off to the fact that sshrc is running in a limited environment.

So the question becomes: is there a way to detect interactive terminal in the limited environment that sshrc executes in?

Answer 1

Use test to check $SSH_TTY (final solution in this link):

test -z $SSH_TTY || echo $USER logged in from $ip