ash: Remove last param

Question

I'm in a busybox environment which only has sh and ash available.

Now I'm doing a script in which I need to pass all but the last param to ln.

So far it looks like this:

#!/bin/ash
TARGET="/some/path/"

for last; do true; done

ln $@ $TARGET$last

Obviously now I pass the last param twice, first unmodified then modified with $TARGET in front of it.

How can I get rid of the last param in $@?

Try modifying the solution in this answer here https://stackoverflow.com/questions/9057387/process-all-arguments-except-the-first-one-in-a-bash-script. — namokarm, Jun 09 '18 at 10:12
@namokarm echo "${@:2}" just cuts off the first characters of the first param, how would I modify it to remove the last param? — MetalSnake, Jun 09 '18 at 10:24
I can't try it atm, but could you try this echo "${@:1:$#-1}" — namokarm, Jun 09 '18 at 10:28
Did you try running this echo "${@:1:$#-1}" or as a part of your script? — namokarm, Jun 09 '18 at 10:34

score 1 · Answer 1 · answered Jun 09 '18 at 21:12

1

You can try this way

last_arg () {
shift $(($#-1))
echo "$@"
}
last=$(last_arg "$@")
echo "all but last = ${@%$last}"

answered Jun 09 '18 at 21:12

ctac_

this had weird effects, with some param lists it would cut of the last 2 plus the last character of the param before that, and with others it did nothing. But it made me think of a working solution. – MetalSnake Jun 10 '18 at 08:44
@Diskutant give sample of param lists where it cut the last char and another where it do nothing. – ctac_ Jun 10 '18 at 09:14

score 0 · Answer 2 · answered Jun 10 '18 at 08:46

0

Got a working solution now, it's not that nice, and it shifts the params, but until a better solution comes up this will do it.

for last; do true; done
while [[ "$1" != "$last" ]] ; do
    args="$args $1"
    shift
done

echo $args

answered Jun 10 '18 at 08:46

MetalSnake

2 Answers2