Bash: reference the first part of '||' or pipe command

Question

Is a command somehow saved/possible to be referenced to in a pipe? For example:

cd /tmp/non_existing_dir || echo "could not execute $PREVIOUS_PIPE_CMD

without pre-saving the original command. $PREVIOUS_PIPE_CMD is an example variable.

Possible duplicate of: http://stackoverflow.com/questions/1221833/bash-pipe-output-and-capture-exit-status — user2182349, Oct 13 '15 at 12:29
But, It is not pipe. It is `||` control operator. Is it typo? — sat, Oct 13 '15 at 12:30
@sat thanks, I don't know why I said that this is a pipe. Will edit. — Zloj, Oct 13 '15 at 12:45

score 1 · Accepted Answer · answered Oct 13 '15 at 12:37

No. Piping works by black-boxing and relaying IO streams and statuses. There is no way for the command that has been piped into to know the downstream history without relaying it/saving it and passing it around.

Also: your example is not using piping, it is using ||, which is OR, not a pipe.

cd /tmp/non_existing_dir || echo "could not execute $PREVIOUS_PIPE_CMD"

is the same as

if [[ $(cd foo) ]]; then
  :
else
  echo "could not execute $PREVIOUS_PIPE_CMD
fi

Given your example you would want:

tmpdir="/tmp/non_existing_dir"
cd ${tmpdir} 2>/dev/null || echo "could not cd into '${tmpdir}'"

Bash: reference the first part of '||' or pipe command

1 Answers1