Aaron Griffin wrote:
On Tue, Oct 20, 2009 at 10:17 PM, Dan McGee <[email protected]> wrote:
On Tue, Oct 20, 2009 at 9:45 PM, Allan McRae <[email protected]> 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.
dmc...@kilkenny /tmp
$ ./one.sh -f -h "foo bar"
pass 1:
-f
-h
foo bar
pass 2:
-f
-h
foo
bar
dmc...@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 "${argli...@]}"
Do I win a prize or anything? :P
It's worth noting that the following works two:
./two.sh "$@"
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
