Execute command and store everything to variable in bash

Question

In bash script I'm developing I'm trying to execute command and capture in variable(s):

• stdout
• stderr
• status code

how to achieve that? The command is tar, if it is of any significance.

I tried the most standard approach:

TAROUTPUT=$(tar -cf arch.tar /path/to/dir) 

Based on some work I did (I haven't actually produced tar failure) I get only stdout from this, stderr is not stored to variable. The perfect solution has TAROUTPUT (with both stdout&stderr) and TARSTATUS variables.

Thanks in advance.

Answer 1

TAROUTPUT=$(tar -cf arch.tar /path/to/dir 2>&1) this_is_the_tar_exit_code=$? 

Answer 2

If you want to separate stdout from stderr:

craft@engine:~$ tar -cf arch.tar /path/to/dir 1>/tmp/tar_stdout 2>/tmp/tar_stderr; RETCODE=$( echo ${?} ); craft@engine:~$ stdout_var=$( cat /tmp/tar_stdout ) craft@engine:~$ stderr_var=$( cat /tmp/tar_stderr ) craft@engine:~$ echo -e "STDOUT : ${stdout_var}\nSTDERR : ${stderr_var}\nCommand Status: ${RETCODE}" 

• 1>/tmp/tar_stdout : save the stdout output to a temp file.
• 2>/tmp/tar_stderr : save stderr output to a file.
• Return code of the command (exit status) is saved into the ${?} variable.