pipe stdout and stderr to two different processes in shell script?

Question

Use another file descriptor

{ command1 2>&3 | command2; } 3>&1 1>&2 | command3

You can use up to 7 other file descriptors: from 3 to 9.
If you want more explanation, please ask, I can explain ;-)

Test

{ { echo a; echo >&2 b; } 2>&3 | sed >&2 's/$/1/'; } 3>&1 1>&2 | sed 's/$/2/'

output:

b2
a1

Example

Produce two log files:
1. stderr only
2. stderr and stdout

{ { { command 2>&1 1>&3; } | tee err-only.log; } 3>&1; } > err-and-stdout.log

If command is echo "stdout"; echo "stderr" >&2 then we can test it like that:

$ { { { echo out>&3;echo err>&1;}| tee err-only.log;} 3>&1;} > err-and-stdout.log
$ head err-only.log err-and-stdout.log
==> err-only.log <==
err

==> err-and-stdout.log <==
out
err

The accepted answer results in the reversing of stdout and stderr. Here's a method that preserves them (since Googling on that purpose brings up this post):

{ command 2>&1 1>&3 3>&- | stderr_command; } 3>&1 1>&2 | stdout_command

Notice:

• 3>&- is required to prevent fd 3 from being inherited by command. (As this can lead to unexpected results depending on what command does inside.)

Parts explained:

1. Outer part first:

   1. 3>&1 -- fd 3 for { ... } is set to what fd 1 was (i.e. stdout)
   2. 1>&2 -- fd 1 for { ... } is set to what fd 2 was (i.e. stderr)
   3. | stdout_command -- fd 1 (was stdout) is piped through stdout_command

2. Inner part inherits file descriptors from the outer part:

   1. 2>&1 -- fd 2 for command is set to what fd 1 was (i.e. stderr as per outer part)
   2. 1>&3 -- fd 1 for command is set to what fd 3 was (i.e. stdout as per outer part)
   3. 3>&- -- fd 3 for command is set to nothing (i.e. closed)
   4. | stderr_command -- fd 1 (was stderr) is piped through stderr_command

Example:

foo() {
    echo a
    echo b >&2
    echo c
    echo d >&2
}

{ foo 2>&1 1>&3 3>&- | sed -u 's/^/err: /'; } 3>&1 1>&2 | sed -u 's/^/out: /'

Output:

out: a
err: b
err: d
out: c

(Order of a -> c and b -> d will always be indeterminate because there's no form of synchronization between stderr_command and stdout_command.)

Using process substitution:

command1 > >(command2) 2> >(command3)

See http://tldp.org/LDP/abs/html/process-sub.html for more info.

Simply redirect stderr to stdout

{ command1 | command2; } 2>&1 | command3

Caution: commnd3 will also read command2 stdout (if any).
To avoid that, you can discard commnd2 stdout:

{ command1 | command2 >/dev/null; } 2>&1 | command3

However, to keep command2 stdout (e.g. in the terminal),
then please refer to my other answer more complex.

Test

{ { echo -e "a\nb\nc" >&2; echo "----"; } | sed 's/$/1/'; } 2>&1 | sed 's/$/2/'

output:

a2
b2
c2
----12

Pipe stdout as usual, but use Bash process substitution for the stderr redirection:

some_command 2> >(command of stderr) | command of stdout

Header: #!/bin/bash

Zsh Version

I like the answer posted by @antak, but it doesn't work correctly in zsh due to multios. Here is a small tweak to use it in zsh:

{ unsetopt multios; command 2>&1 1>&3 3>&- | stderr_command; } 3>&1 1>&2 | stdout_command

To use, replace command with the command you want to run, and replace stderr_command and stdout_command with your desired pipelines. For example, the command ls / /foo will produce both stdout output and stderr output, so we can use it as a test case. To save the stdout to a file called stdout and the stderr to a file called stderr, you can do this:

{ unsetopt multios; ls / /foo 2>&1 1>&3 3>&- | cat >stderr; } 3>&1 1>&2 | cat >stdout

See @antak's original answer for full explanation.