Let Bash trap DEBUG see pipe as one command

Question

I am using

trap 'echo -e "\e[92m$ $BASH_COMMAND\e[0m"' DEBUG

in a Bash script to print every command that is executed.

This works fine, except that it prints two commands when I pipe them together:

bzip2 -dc < dump.sql.bz2 | mysql test

Prints:

$ bzip2 -dc < dump.sql.bz2
$ mysql test

Can I make any change to the trap or to the line with the pipe so that the line will be printed as one?

Answer 1

Change your command to:

mysql test < <(bzip2 -dc < dump.sql.bz2)

Using a process substitution instead of a pipeline makes the command a simple command from Bash's perspective, which is the level of granularity at which the DEBUG trap and the built-in $BASH_COMMAND variable operate.

Background:

The DEBUG (pseudo) signal by design operates on the level of simple commands, as does $BASH_COMMAND.

bzip2 -dc < dump.sql.bz2 | mysql test is a pipeline composed of multiple simple commands.

Therefore, your trap statement cannot do what you want with a pipeline.

The closest thing to getting what you want with a compound command (e.g., a while loop or a command group ({ ...; ...; }) or a command list (simple commands or pipelines joined with operators ; && || &) is to use set -v, which echoes the entire command to stderr before it is executed, but I wouldn't know how to control the formatting of this output.

Answer 2

A crude workaround would be using a function, like this:

run () { eval $1; }
trap 'echo -e "\e[92m$ $BASH_COMMAND\e[0m"' DEBUG
run "bzip2 -dc < dump.sql.bz2 | mysql test"

Output:

$ run "bzip2 -dc < dump.sql.bz2 | mysql test"