Bash script: Using “script” command from a bash script for logging a session

Your shell script did not terminate. It is still running. You are getting a prompt because script is spawning a new shell. the prompt you see is the prompt from the spawned shell.

The normal use case for script is something like this:

1. start script. this spawns a new shell.
2. do commands in the new shell.
3. exit the shell and drop to previous shell
4. examine the logfile created by script

So basically script is working as expected. You will have to find another way to achieve what you want.

You can log the execution of your script like this:

#! /bin/bash
exec > logfile 2>&1
set -x
FOO=BAR
echo $FOO

Explanation:

• exec > logfile 2>&1 redirects stdout and stderr to logfile
• set -x makes bash print every command before executing it

Example:

$ ./foo.sh
  # (no output here because everything goes to logfile)
$ cat logfile 
+ FOO=BAR
+ echo BAR
BAR

Disadvantage of this method is that the script prints no output for humans to see. Everything goes to the logfile.

Alternatively you can do it like this:

#! /bin/bash
# nothing special here
FOO=BAR
echo $FOO

Then execute like this:

$ script -c "bash -x foo.sh"
Script started, file is typescript
+ FOO=BAR
+ echo BAR
BAR
Script done, file is typescript

now output is directly visible and also saved to logfile (default name of logfile is typescript)

$ cat typescript 
Script started on Mi 18 Mai 2011 01:05:29 CEST
+ FOO=BAR
+ echo BAR
BAR

Script done on Mi 18 Mai 2011 01:05:29 CEST