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

Answer 2

Op De Cirkel's answer has the right idea. It can be simplified even more (avoiding use of cat):

exec 5>&1
FF=$(echo aaa|tee /dev/fd/5)
echo $FF

Answer 3

Here's an example capturing both stderr and the command's exit code. This is building on the answer by Russell Davis.

exec 5>&1
FF=$(ls /taco/ 2>&1 |tee /dev/fd/5; exit ${PIPESTATUS[0]})
exit_code=$?
echo "$FF"
echo "Exit Code: $exit_code"

If the folder /taco/ exists, this will capture its contents. If the folder doesn't exist, it will capture an error message and the exit code will be 2.

If you omit 2>&1then only stdout will be captured.

Answer 4

If by "the console" you mean your current TTY, try

variable=$(command with options | tee /dev/tty)

This is slightly dubious practice because people who try to use this sometimes are surprised when the output lands somewhere unexpected when they don't have a TTY (cron jobs etc).

Answer 5

You can use more than three file descriptors. Try here:

http://tldp.org/LDP/abs/html/io-redirection.html

"Each open file gets assigned a file descriptor. [2] The file descriptors for stdin, stdout, and stderr are 0, 1, and 2, respectively. For opening additional files, there remain descriptors 3 to 9. It is sometimes useful to assign one of these additional file descriptors to stdin, stdout, or stderr as a temporary duplicate link."

The point is whether it's worth to make script more complicated just to achieve this result. Actually it's not really wrong, the way you do it.

Answer 6

Alternative to using /dev/tty, or an additional file descriptor as suggested by the other answers, you can also flip it and simply use a temp file. This is arguably easier to read and more portable in certain situations.

tmpFile=$(mktemp)  # mak-a de temp
rsync /a /b | tee $tmpFile # sync my b*tch up
if grep "U F'd up" $tmpFile; then
  rm -rf / #Seppuku
fi