bash - Capture stdout to a variable but still display it in the console

Question

I have a bash script which calls several long-running processes. I want to capture the output of those calls into variables for processing reasons. However, because these are long running processes, I would like the output of the rsync calls to be displayed in the console in real-time and not after the fact.

To this end, I have found a way of doing it but it relies on outputting the text to /dev/stderr. I feel that outputting to /dev/stderr is not a good way of doing things.

VAR1=$(for i in {1..5}; do sleep 1; echo $i; done | tee /dev/stderr)

VAR2=$(rsync -r -t --out-format='%n%L' --delete -s /path/source1/ /path/target1 | tee /dev/stderr)

VAR3=$(rsync -r -t --out-format='%n%L' --delete -s /path/source2/ /path/target2 | tee /dev/stderr)

In the example above, I am calling rsync a few times and I want to see the file names as they are processed, but in the end I still want the output in a variable because I will be parsing it later.

Is there a 'cleaner' way of accomplishing this?

If it makes a difference, I am using Ubuntu 12.04, bash 4.2.24.

Answer 1

Duplicate &1 in your shell (in my examle to 5) and use &5 in the subshell (so that you will write to stdout (&1) of the parent shell):

exec 5>&1
FF=$(echo aaa|tee >(cat - >&5))
echo $FF

Will print aaa two times, ones because of the echo in the subshell, and second time print the value of the variable.

In your code:

exec 5>&1
VAR1=$(for i in {1..5}; do sleep 1; echo $i; done | tee >(cat - >&5))
# use the value of VAR1