How to get the complete calling command of a BASH script from inside the script (not just the arguments)

Question

I have a BASH script that has a long set of arguments and two ways of calling it:

my_script --option1 value --option2 value  ... etc

or

my_script val1 val2 val3 .....  valn 

This script in turn compiles and runs a large FORTRAN code suite that eventually produces a netcdf file as output. I already have all the metadata in the netcdf output global attributes, but it would be really nice to also include the full run command one used to create that experiment. Thus another user who receives the netcdf file could simply reenter the run command to rerun the experiment, without having to piece together all the options.

So that is a long way of saying, in my BASH script, how do I get the last command entered from the parent shell and put it in a variable? i.e. the script is asking "how was I called?"

I could try to piece it together from the option list, but the very long option list and two interface methods would make this long and arduous, and I am sure there is a simple way.

I found this helpful page:

BASH: echoing the last command run

but this only seems to work to get the last command executed within the script itself. The asker also refers to use of history, but the answers seem to imply that the history will only contain the command after the programme has completed.

Many thanks if any of you have any idea.

Answer 1

You can try the following:

myInvocation="$(printf %q "$BASH_SOURCE")$((($#)) && printf ' %q' "$@")"

$BASH_SOURCE refers to the running script (as invoked), and $@ is the array of arguments; (($#)) && ensures that the following printf command is only executed if at least 1 argument was passed; printf %q is explained below.

• While this won't always be a verbatim copy of your command line, it'll be equivalent - the string you get is reusable as a shell command.

• chepner points out in a comment that this approach will only capture what the original arguments were ultimately expanded to:

  • For instance, if the original command was my_script $USER "$(date +%s)", $myInvocation will not reflect these arguments as-is, but will rather contain what the shell expanded them to; e.g., my_script jdoe 1460644812

  • chepner also points that out that getting the actual raw command line as received by the parent process will be (next to) impossible. ^{Do tell me if you know of a way.}

    • However, if you're prepared to ask users to do extra work when invoking your script or you can get them to invoke your script through an alias you define - which is obviously tricky - there is a solution; see bottom.

Note that use of printf %q is crucial to preserving the boundaries between arguments - if your original arguments had embedded spaces, something like $0 $* would result in a different command.
printf %q also protects against other shell metacharacters (e.g., |) embedded in arguments.

printf %q quotes the given argument for reuse as a single argument in a shell command, applying the necessary quoting; e.g.:

 $ printf %q 'a |b'
 a\ \|b

a\ \|b is equivalent to single-quoted string 'a |b' from the shell's perspective, but this example shows how the resulting representation is not necessarily the same as the input representation.

^{Incidentally, ksh and zsh also support printf %q, and ksh actually outputs 'a |b' in this case.}

---

If you're prepared to modify how your script is invoked, you can pass $BASH_COMMANDas an extra argument: $BASH_COMMAND contains the raw^[1] command line of the currently executing command.
For simplicity of processing inside the script, pass it as the first argument (note that the double quotes are required to preserve the value as a single argument):

my_script "$BASH_COMMAND" --option1 value --option2

Inside your script:

# The *first* argument is what "$BASH_COMMAND" expanded to,
# i.e., the entire (alias-expanded) command line.
myInvocation=$1    # Save the command line in a variable...
shift              # ... and remove it from "$@".

# Now process "$@", as you normally would.

Unfortunately, there are only two options when it comes to ensuring that your script is invoked this way, and they're both suboptimal:

• The end user has to invoke the script this way - which is obviously tricky and fragile (you could however, check in your script whether the first argument contains the script name and error out, if not).

• Alternatively, provide an alias that wraps the passing of $BASH_COMMAND as follows:

  • alias my_script='/path/to/my_script "$BASH_COMMAND"'
  • The tricky part is that this alias must be defined in all end users' shell initialization files to ensure that it's available.
  • Also, inside your script, you'd have to do extra work to re-transform the alias-expanded version of the command line into its aliased form:

# The *first* argument is what "$BASH_COMMAND" expanded to,
# i.e., the entire (alias-expanded) command line.
# Here we also re-transform the alias-expanded command line to
# its original aliased form, by replacing everything up to and including
# "$BASH_COMMMAND" with the alias name.
myInvocation=$(sed 's/^.* "\$BASH_COMMAND"/my_script/' <<<"$1")
shift              # Remove the first argument from "$@".

# Now process "$@", as you normally would.

Sadly, wrapping the invocation via a script or function is not an option, because the $BASH_COMMAND truly only ever reports the current command's command line, which in the case of a script or function wrapper would be the line inside that wrapper.

---

^{[1] The only thing that gets expanded are aliases, so if you invoked your script via an alias, you'll still see the underlying script in $BASH_COMMAND, but that's generally desirable, given that aliases are user-specific.
All other arguments and even input/output redirections, including process substitutiions <(...) are reflected as-is.}

Answer 2

"$0" contains the script's name, "$@" contains the parameters.

Answer 3

Do you mean something like echo $0 $*?