Duplicate command output, print un-post-treated copy to stdout and store post-treated copy into a file

Question

I'm trying to duplicate a command's output so I apply a post-treatment to one stream, then store it to a file while printing the original one on stdout.

Closest I could come with is:

command | tee /dev/stdout | sed 's/foo/bar/g' > out.txt

which does not work as the > captures both outputs.

I'd like to avoid using a temporary file if possible. Any clue?

You could store the output in a bash variable with var=$(yor command) — terence hill, Nov 19 '21 at 16:03
Does this answer your question? [How can I send the stdout of one process to multiple processes using (preferably unnamed) pipes in Unix (or Windows)?](https://stackoverflow.com/questions/60942/how-can-i-send-the-stdout-of-one-process-to-multiple-processes-using-preferably) — pjh, Nov 19 '21 at 18:00

score 3 · Accepted Answer · answered Nov 19 '21 at 17:47

3

Try

command | tee >(sed 's/foo/bar/g' > out.txt)

See ProcessSubstitution - Greg's Wiki for an explanation of >(...). In short, it causes tee to write to a path that pipes the written data as input to the sed 's/foo/bar/g' > out.txt command.

answered Nov 19 '21 at 17:47

pjh

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Brilliant! 10 years in programming and never used process substitution... – wecx Nov 22 '21 at 17:42

Duplicate command output, print un-post-treated copy to stdout and store post-treated copy into a file

1 Answers1