How to run SSH commands in a function and store output in a variable?

Question

I am trying to find (and store to a variable) the path of a file that is stored on another server by running the following command in one of my functions (included in .bashrc):

FILE_PATH=$(ssh -T user@host 'find <directory> -name *<filename>*')

However this is not returning any output. I have checked that there should be a file in the location that is grabbed, and I have also been able to replicate the command and store output to variable on the command line, but it does not work when running the function.

Does anyone know what's going on and why the command isn't working?

EDIT:

Here is a representation of what I am seeing:

function get_path {
    FILE=$1
    FILE_PATH=$(ssh -T user@hostB 'find /home/daverbuj -name *${FILE}*')
    echo "Here is the file: ${FILE_PATH}"
}

[daverbuj@hostA]$ FILE_A=$(ssh -T user@hostB 'find /home/daverbuj -name foo.bar')
[daverbuj@hostA]$ echo $FILE_A
/home/daverbuj/foo.bar

[daverbuj@hostA]$ get_path foo.bar
Here is the file:

I am seeing what I expect when I run from the command line, but not when I run the function.

Answer 1

You can reroute stderr to stdout using

OUT=$(ssh -tt -vv [email protected] "remote command" 2>&1 )

you can edit something like this in your function and it should work!

Answer 2

get_path() {
  local name=$1
  readarray -td '' reply < <(
    LC_FILE_NAME=$name ssh -o SendEnv=LC_FILE_NAME -n -T user@hostB '
      LC_ALL=C find /home/daverbuj -name "*$LC_FILE_NAME*" -print0')
  (( ${#reply[@]} )) &&
    echo "File path${reply[2]+s} with that name:" &&
    printf ' - %s\n' "${reply[@]}"
}

• no need to use the ksh style of function definition (function f {...}), bash supports the standard f() compound-command as well.
• there could be more than one matching file. To separate them out reliably, you need to use NUL as delimiter (-print0). bash variables can't hold NULs, but since version 4.4, its readarray builtin can split a NUL delimited input stream into elements of an array with -d ''.
• * is a glob operator, so it needs to be quoted for both the local shell and remote shell so it be passed literally to find for it's -name name pattern matching operator.
• You don't want to have the local shell expand a variable into the code passed to the remote shell as that would constitute some form of arbitrary command injection vulnerability, instead here, we've passing the $name value to the remote shell via a LC_FILE_NAME environment variable (assumes sshd on the remote host is configured with AcceptEnv LC_* which is common). In your code, you had the remote shell expand a $FILE variable (though had forgotten to quote it as well as the *), but you hadn't passed that variable from the local shell to the remote shell.
• $REPLY for scalar and $reply for array is a common convention for a variable name where we return a result to the caller.
• You can't use echo for arbitrary data (for data that you can't guarantee won't contain backslashes or start with -).
• -name '*foo*' could fail to match all files with foo in their name in locales using multi-byte charsets such as UTF-8, as failure to decoding the file names into that charset would trip the fnmatch() used by find's -name. Hence the LC_ALL=C to fix to a single-byte locale.
• here, we're returning a failure exit status unless at least one file was found and it was printed properly.
• We're using the -n option of ssh to avoid it consuming input from stdin. You may also want to add a -o BatchMode=yes for more tuning relevant to processing meant to be unattended.

More reading on those here:

• How to execute an arbitrary simple command over ssh without knowing the login shell of the remote user?
• Why is looping over find's output bad practice?
• When is double-quoting necessary?
• Why is printf better than echo?