Streaming command substitution output to terminal w/o squelching color

Question

Is there a way to show the output of a command substitution, in color, while it is executing, while still storing it to a variable? This would mainly be for testing/debugging purposes, and the output would not be shown normally.

If I do this:

output=$(some_command)
if [ "$debug" ]; then
  echo "$output"
fi

...then this is close, but not what I want in a few ways:

• Even if the command would have otherwise emitted output to the terminal in color, it's printed in black-and-white.
• The output is only shown after the command is finished, rather than streaming as it runs.

How can I conditionally stream output to a terminal without squelching color?

Answer 1

To store a command's results to a variable as well as streaming its output to the console:

var=$(some_command | tee /dev/stderr)

If you want to force your command to think it's outputting directly to a TTY, and thus to enable color output if it would do so when not in a pipeline, use the tool unbuffer, shipped with expect:

var=$(unbuffer some_command | tee /dev/stderr)

---

All that said: If you only want to show debugging conditionally for a long script, it makes sense to put that conditional up front at the top of your script rather than scattering it around everywhere. For instance:

# put this once at the top of your script
set -o pipefail
if [[ $debug ]]; then
  exec 3>/dev/stderr
else
  exec 3>/dev/null
fi

# define a function that prepends unbuffer only if debugging is enabled
maybe_unbuffer() {
  if [[ $debug ]]; then
    unbuffer "$@"
  else
    "$@"
  fi
}

# if debugging is enabled, pipe through tee; otherwise, use cat
capture() {
  if [[ $debug ]]; then
    tee >(cat >&3)
  else
    cat
  fi
}

# ...and thereafter, when you want to hide a command's output unless debug is enabled:
some_command >&3

# ...or, to capture its output while still logging to stderr without squelching color...
var=$(maybe_unbuffer some_command | capture)