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.

Solution 1

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

Solution 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.

Comments

[Miloš Đakonović]

Yes, working, thanks. Will accept answer in a minute

[Hauke Laging]

@user1934428 That's not how I understand "TAROUTPUT (with both stdout&stderr)"