问题
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=$(rsync -r -t --out-format='%n%L' --delete -s /path/source2/ /path/target2 | tee /dev/stderr)
In the example above, I am calling rsync a few times and I want to see the file names as they are processed, but in the end I still want the output in a variable because I will be parsing it later.
Is there a 'cleaner' way of accomplishing this?
If it makes a difference, I am using Ubuntu 12.04, bash 4.2.24.
回答1:
Duplicate &1 in your shell (in my examle to 5) and use &5 in the subshell (so that you will write to stdout (&1) of the parent shell):
exec 5>&1
FF=$(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 $i; done | tee >(cat - >&5))
# use the value of VAR1
回答2:
Op De Cirkel's answer has the right idea. It can be simplified even more (avoiding use of cat):
exec 5>&1
FF=$(echo aaa|tee /dev/fd/5)
echo $FF
回答3:
Here's an example capturing both stderr and the command's exit code. This is building on the answer by Russell Davis.
exec 5>&1
FF=$(ls /taco/ 2>&1 |tee /dev/fd/5; exit ${PIPESTATUS[0]})
exit_code=$?
echo "$FF"
echo "Exit Code: $exit_code"
If the folder /taco/ exists, this will capture its contents. If the folder doesn't exist, it will capture an error message and the exit code will be 2.
If you omit 2>&1then only stdout will be captured.
回答4:
You can use more than three file descriptors. Try here:
http://tldp.org/LDP/abs/html/io-redirection.html
"Each open file gets assigned a file descriptor. [2] The file descriptors for stdin, stdout, and stderr are 0, 1, and 2, respectively. For opening additional files, there remain descriptors 3 to 9. It is sometimes useful to assign one of these additional file descriptors to stdin, stdout, or stderr as a temporary duplicate link."
The point is whether it's worth to make script more complicated just to achieve this result. Actually it's not really wrong, the way you do it.
回答5:
If by "the console" you mean your current TTY, try
variable=$(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).
回答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
来源：https://stackoverflow.com/questions/12451278/capture-stdout-to-a-variable-but-still-display-it-in-the-console