问题
I am trying to figure out a sane way to do a NOT clause in a case. The reason I am doing this is for transcoding when a case is met, aka if I hit an avi, there's no reason to turn it into an avi again, I can just move it out of the way (which is what the range at the base of my case should do). Anyway, I have some proto code that I wrote out that kind of gives the gist of what I am trying to do.
#!/bin/bash
for i in $(seq 1 3); do
echo "trying: $i"
case $i in
! 1) echo "1" ;; # echo 1 if we aren't 1
! 2) echo "2" ;; # echo 2 if we aren't 2
! 3) echo "3" ;; # echo 3 if we aren't 3
[1-3]*) echo "! $i" ;; # echo 1-3 if we are 1-3
esac
echo -e "\n"
done
expected results would be something like this
2 3 ! 1
1 3 ! 2
1 2 ! 3
Help is appreciated, thanks.
回答1:
This is contrary to the design of case, which executes only the first match. If you want to execute on multiple matches (and in your design, something which is 3 would want to execute on both 1 and 2), then case is the wrong construct. Use multiple if blocks.
[[ $i = 1 ]] || echo "1"
[[ $i = 2 ]] || echo "2"
[[ $i = 3 ]] || echo "3"
[[ $i = [1-3]* ]] && echo "! $i"
Because case only executes the first match, it only makes sense to have a single "did-not-match" handler; this is what the *) fallthrough is for.
回答2:
You can do this with the extglob extension.
$ shopt -s extglob
$ case foo in !(bar)) echo hi;; esac
hi
$ case foo in !(foo)) echo hi;; esac
$
来源：https://stackoverflow.com/questions/21919182/bash-not-in-case