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I have a user-provided bash script that creates a directory in some way and outputs its path. It can work in many ways, for example, it can clone a git repository, do some setting up, and then output the path. Something like this:

git clone ...repo --quiet && echo "...path"

I run it using command substitution, wrapping with $(...) and using the resulting path. Something like this:

path=$(...)

The command itself is user-provided, so it can be anything, I can't anticipate it will always be git clone.

It works well, but if the commands before the final echo outputs anything, it will mess up the result. I could redirect stdout to stderr for all preceding commands and only use stdout at the last step, but I feel it's a hack.

Is there any way to leave stdout and stderr to the caller's stdout and stderr, but have a separate stream for the result?

Kamil Maciorowski
  • 81,893
Tamás Sallai
  • 128

1 Answers1

3

This will save the last line of output:

path=$(whatever | tail -n 1)

If you need all the output to be utilized in some way:

  • use tee to print the output to the tty

    path=$(whatever | tee /dev/tty | tail -n 1)

  • or to save it

    path=$(whatever | tee ./file | tail -n 1)

  • or to process it (and print to the tty or a file, not to stdout which feeds the variable)

    path=$(whatever | tee >(grep foo | wc -c >/dev/tty) | tail -n 1)

  • You can even send (fork) the output to many destinations:

    path=$(whatever | tee /dev/tty ./file /some/named/fifo >(tool1) | tail -n 1)

Obviously a newline within the echoed path will break the solution.

If you can modify the final echo command in the user-provided script then please see this answer of mine for an approach that uses temporary files.

Kamil Maciorowski
  • 81,893