Different result from a bash function

Question

I am trying to code a script to log all commands executed on a server, currently I have 2 versions:

(simplified version of the scripts)

Version 1:

function fc
{
  fc -ln -0
}
trap fc DEBUG

Version 2:

function fc
{
 cmd=`fc -ln -0`
 echo $cmd
}
trap fc DEBUG

The script is inluded in /etc/bash.bashrc : source script.sh

The problem is that when I run this scripts I don't get the same result:

When I run the first verion I get the command that triggered the trap :

SERVER#
SERVER#
SERVER# test
SERVER# ls
ls

But when I run the second I get the last-1 command :

SERVER#
SERVER#
SERVER# test
SERVER# ls
test

Thanks in advance.

In the 2nd version, you're executing fc in a different process, so you won't have the same history. — glenn jackman
– glenn jackman, Commented Jan 6, 2015 at 16:43
@glennjackman Thanks, I have tried to run this directly in bash cmd=fc -ln -0; echo $cmd and cmd=fc -ln -0`, I had the same result. — ob_dev
– ob_dev, Commented Jan 7, 2015 at 14:53

anubhava · Accepted Answer · 2015-01-07 11:55:31Z

1

If you're using a function for trap DEBUG then you can make use of special BASH variable BASH_COMMAND.

dbg() { echo "$BASH_COMMAND" >> ~/cmd.log }

trap 'dbg' DEBUG

Make sure this trap command is very last line in your ~/.bashrc (or last line in ~/.bash_profile, if that exists).

answered Jan 6, 2015 at 21:24

anubhava

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4 Comments

is it going to impact general system performance if I put trap debug in /etc/bashrc?

Not really, I have even used trap debug and PROMPT_COMMAND together.

Ok. When I connect to the server I see some commands from others scripts (~/.bashrc, ..) it there a way to ignore those scripts. thanks

So do you just want to store each and every executed command in a file?