Aaron Griffin wrote:
On Tue, Oct 20, 2009 at 10:17 PM, Dan McGee <dpmcgee@gmail.com> wrote:
On Tue, Oct 20, 2009 at 9:45 PM, Allan McRae <allan@archlinux.org> wrote:
Hi all,
This problem has been doing my head in... First a minimal example that reflects how makepkg does things:
--one.sh-- #!/bin/bash
echo "pass 1:" for arg in "$@"; do echo $arg done echo
ARGLIST="$@"
./two.sh $ARGLIST --end one.sh--
--two.sh-- #!/bin/bash
echo "pass 2:" for arg in "$@"; do echo $arg done --end two.sh--
then run: ./one.sh -f -h "foo bar" pass 1: -f -h foo bar
pass 2: -f -h foo bar
Note how in pass two, foo and bar are no longer in the one line. Of course, passing ./two.sh "$@" works, but the argument parsing in makepkg clears that, hence the need to save it to ARGLIST.
Any ideas?
Of course! I think I got it.
dmcgee@kilkenny /tmp $ ./one.sh -f -h "foo bar" pass 1: -f -h foo bar
pass 2: -f -h foo bar
dmcgee@kilkenny /tmp $ ./one-new.sh -f -h "foo bar" pass 1: -f -h foo bar
pass 2: -f -h foo bar
$ cat one-new.sh #!/bin/bash echo "pass 1:" for arg in "$@"; do echo $arg done echo
ARGLIST=("$@")
./two.sh "${ARGLIST[@]}"
Do I win a prize or anything? :P
It's worth noting that the following works two:
./two.sh "$@"
The reason being that "$@" is special in bash. It actually expands to the command line quoted as you passed it, with "foo bar" being one argument. The issue is with the assignment to a single bash variable
The reason that can not be used in makepkg is the option parsing uses "shift" and thus clears the value of $@ as it goes. Allan