bash Cookbook

Getting Input from Another Machine

Problem

Your script needs to get input from another machine, perhaps to check if a file exists or a process is running.

Solution

Use SSH with public keys and command substitution. To do this, set up SSH so that you do not need a password, as described in Using SSH Without a Password. Next, tailor the command that SSH runs to output exactly what your script needs as input. Then simply use command substitution.

#!/usr/bin/env bash
# cookbook filename: command_substitution

REMOTE_HOST='host.example.com'  # Required
REMOTE_FILE='/etc/passwd'       # Required
SSH_USER='user@'                # Optional, set to '' to not use
#SSH_ID='-i ~/.ssh/foo.id'       # Optional, set to '' to not use
SSH_ID=''

result=$(
    ssh $SSH_ID $SSH_USER$REMOTE_HOST \
      "[ -r $REMOTE_FILE ] && echo 1 || echo 0"
) || { echo "SSH command failed!" >&2; exit 1; }

if [ $result = 1 ]; then
    echo "$REMOTE_FILE present on $REMOTE_HOST"
else
    echo "$REMOTE_FILE not present on $REMOTE_HOST"
fi

Discussion

We do a few interesting things here. First, notice how both $SSH_USER and $SSH_ID work. They have an effect when they have a value, but when they are empty they interpolate to the empty set and are ignored. This allows us to abstract the values in the code, which lends itself to putting those values in a configuration file, putting the code into a function, or both.

# Interpolated line of the variables have values:
ssh -i ~/.ssh/foo.id [e-mail] [...]

# No values:
ssh host.example.com [...]

Next, we set up the command ...