I am not sure about the necessity for the 'dir' argument in Recipes.
Right now, I am trying to update the Foomatic recipe for 4.0, and it
has a mess in it about figuring out what the $dir should be.
It would seem reasonable to examine the archive file to find out where
the files will be unpacked, and just set $dir to that; and if there's
more than one top-level name, create a dir and unpack inside it. Are
there any problems with this approach?
Here is some bash code which works just fine for me. It can be
considered pseudo-code, though, since I don't know the proper names
for "$archiveDir", "$ProgramName", or so forth (but I trust their
meaning is self-evident.) Also, I assume there's some methodology in
place for dealing with different types of archives, and you'd want to
plug into that instead of using this case statement.
function ziplist() {
unzip -l $1 | sed -re '0,/( +-+){4}/d; /( +-+){2}/,$d; s/ *[0-9]*
*[0-9-]* *[0-9:]* *//'
}
unset multifile ddd dir list_cmd extract_cmd
case $archive_file in
*.tar|*.tar.*|*.t[gbxl7]z|*.tbz2)
list_cmd="tar tf" # Did you realize this just works for
gzipped, bzipped, xzipped, lzopped, etc files?
extract_cmd="tar xf" ;; # I didn't! But it does!
*.zip)
list_cmd=ziplist
extract_cmd=unzip ;;
*)
if which atool > /dev/null
then list_cmd="atool -l"
extract_cmd="atool -x"
else echo "Agh"
fi ;;
esac #is ridiculous
for ddd in `$list_cmd $archiveDir/$archive_file`
do ddd="${ddd#./}"
ddd="${ddd%%/*}"
if [[ -n "$dir" && "$ddd" != "$dir" ]]
then multifile=1
elif [[ -z "$dir" ]]
then dir="$ddd"
fi
done
if [[ "$multifile" == 1 ]]
then dir="$ProgramName-$ProgramVersion"
mkdir $dir
cd $dir
$extract_cmd $archiveDir/$archive_file
else
$extract_cmd $archiveDir/$archive_file
cd $dir
fi
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