Use a posix equivalent solution instead of the pattern substitution bashism.

Signed-off-by: Cristian Ionescu-Idbohrn <cristian.ionescu-idbohrn@axis.com>
Signed-off-by: Dirk Hohndel <dirk@hohndel.org>
This commit is contained in:
Cristian Ionescu-Idbohrn 2013-02-24 10:48:04 +01:00 committed by Dirk Hohndel
parent 77a6b18a71
commit 9ceb65a3ce

View file

@ -32,7 +32,9 @@ case $os in
darwin|win) darwin|win)
# just the dots in the version string - this way we can # just the dots in the version string - this way we can
# count them # count them
dots="${v0//[^.]}" IFS=.
set -- $v0 # split $v0 using $IFS separator
dots=$(($# - 1)) # use positional argument count
# split version string using a '-' separator # split version string using a '-' separator
IFS='-' IFS='-'
set -- $v0 set -- $v0
@ -40,7 +42,7 @@ case $os in
# do we need to add another digit? # do we need to add another digit?
# We know there are 1 or 2 dots in $v, so if it's just one # We know there are 1 or 2 dots in $v, so if it's just one
# or we are trying to get to 4, add one digit # or we are trying to get to 4, add one digit
if [ ${#dots} -eq 1 ] || [ $os = win ]; then if [ $dots -eq 1 ] || [ $os = win ]; then
if [ $# -gt 1 ]; then if [ $# -gt 1 ]; then
v=$v.$2 v=$v.$2
else else
@ -48,7 +50,7 @@ case $os in
fi fi
fi fi
# and if it was just one dot and we want 4, at another 0 # and if it was just one dot and we want 4, at another 0
if [ ${#dots} -eq 1 ] && [ $os = win ]; then if [ $dots -eq 1 ] && [ $os = win ]; then
v=$v.0 v=$v.0
fi fi
;; ;;