Piping both stdout and stderr in bash?
Bash has a shorthand for 2>&1 |
, namely |&
, which pipes both stdout and stderr (see the manual):
cmd-doesnt-respect-difference-between-stdout-and-stderr |& grep -i SomeError
This was introduced in Bash 4.0, see the release notes.
(Note that &>>file
appends to a file while &>
would redirect and overwrite a previously existing file.)
To combine stdout
and stderr
you would redirect the latter to the former using 1>&2
. This redirects stdout (file descriptor 1) to stderr (file descriptor 2), e.g.:
$ { echo "stdout"; echo "stderr" 1>&2; } | grep -v std
stderr
$
stdout
goes to stdout, stderr
goes to stderr. grep
only sees stdout
, hence stderr
prints to the terminal.
On the other hand:
$ { echo "stdout"; echo "stderr" 1>&2; } 2>&1 | grep -v std
$
After writing to both stdout and stderr, 2>&1
redirects stderr back to stdout and grep
sees both strings on stdin, thus filters out both.
You can read more about redirection here.
Regarding your example (POSIX):
cmd-doesnt-respect-difference-between-stdout-and-stderr 2>&1 | grep -i SomeError
or, using >=bash-4
:
cmd-doesnt-respect-difference-between-stdout-and-stderr |& grep -i SomeError