removing command line parameters to shell a script

[Srini Jay]

Hi All,
I have two shell scripts, say 'a' and 'b'.
Contents are as given below:

# file 'a'
echo "in a: $*"
. b
----------------------
# file 'b'
echo "in b: $*"
----------------------


When I invoke 'a' with some command line parameters, those are seen by 'b' also (sourcing effect); but I do not want 'b' to be aware of the command line parameters to 'a'.

Since 'b' sets environment variables for 'a', I have to source it as given above.

Also note that I cannot edit script 'b'.

Any ideas how to unset those command line parameters?

[James R. Ferguson · Accepted solution]

Hi:

Use 'shift' to chop off the number of arguments before sourcing the second script:

# file 'a'
echo "in a: $*"
shift $#
. b

Regards!

...JRF...

[Srini Jay]

Thanks James!
That works!!!

[A. Clay Stephenson]

Note that because "b" is invoked via "." (dot), that all of this is actually one process "a". There is no child process "b".

what you can do is:
set ""
. b

The set "" will clear all positional variables.

If you still need access to these variables in "a" after "b" has been invoked then you can save thenm in a temporary variable (which will be visible to "b" since it's the same process but presumably "b" will have no knowledge of your data hiding).

XX=${@}
set "" # positional variables are cleared
. b
set ${XX} # positional variables are now restored

[Rodney Hills]

When you call b do-

. b ""

This will set the command parameters to empty.

Or you could use "shift $#" in a.

HTH

-- Rod Hills