Suppose I'm using a shell like bash or zsh, and also suppose that I have a command which writes to stdout. I want to capture the output of the command into a shell variable, and also to capture the command's return code into another shell variable.
I know I can do something like this ...
command >tempfile
rc=$?
output=$(cat tempfile)
Then, I have the return code in the 'rc' shell variable and the command's output in the 'output' shell variable.
However, I want to do this without using any temporary files.
Also, I could do this to get the command output ...
output=$(command)
... but then, I don't know how to get the command's return code into any shell variable.
Can anyone suggest a way to get both the return code and the command's output into two shell variables without using any files?
Thank you very much.
Just capture $? as you did before. Using the executables true and false, we can demonstrate that command substitution does set a proper return code:
$ output=$(true); rc=$?; echo $rc
0
$ output=$(false); rc=$?; echo $rc
1
If more than one command substitution appears in an assignment, the return code of the last command substitution determines the return code of the assignment:
$ output="$(true) $(false)"; rc=$?; echo $rc
1
$ output="$(false) $(true)"; rc=$?; echo $rc
0
From the section of man bash describing variable assignment:
If one of the expansions contained a command substitution, the exit status of the command is the exit status of the last command substitution performed. If there were no command substitutions, the command exits with a status of zero.