Copy stdout and stderr to a log file and leave them on the console within the script itself

Question

Using bash, how do I copy stderr and stdout to a log file and also leave them displayed on the console?

I would like to do this within the script itself using an exec.

I tried with

exec &>> log.out

echo "This is stdout"
echo "This is stderr" >&2

But the above prints nothing on the console. How can I achieve this in bash?

Answer 1

You are looking for tee.

See man tee for details.

To combine it with exec, you have to use process substitution. (See man bash for details.)

exec &> >(tee  log.out)
echo "This is stdout"
echo "This is stderr" >&2

Answer 2

You can do:

: > log # empty log file if necessary
{ { {

  ...the script

} 3>&- | tee -a log >&3 3>&-
exit "${PIPESTATUS[0]}"
} 2>&1 | tee -a log 3>&-
} 3>&1
exit "${PIPESTATUS[@]}"

You could also write it as:

: > log # empty log file if necessary
exec 2> >(tee -a log >&2) > >(tee -a log)

...the script

But because bash is not waiting for those processes started with >(...), that has the nasty effect of sometimes outputting something to the terminal after the command has returned which can have even nastier effects (like silently discarding that output) if the terminal "tostop" attribute is on.

In any case, by making stdout a pipe in both solutions, and because two commands independently output the output and error messages, this will affect output buffering and the order the output and error messages are displayed.