linux - unix: how to tell if a string matches a regex -

trying out fish shell, i'm translating bash functions. problem in 1 case, i'm using bash regexes check if string matches regex. can't figure out how translate fish.

here example.

if [[ "$arg" =~ ^[0-9]+$ ]] ... 

• i looked sed, don't see way set exit status based on whether regex matches.
• i looked delegating ruby, again, getting exit status set based on match requires making ugly (see below).
• i looked delegating bash, despite trying maybe 3 or 4 ways, never got match.

so, there way in *nix check if string matches regex, can drop conditional?

---

here have works, unhappy with:

# kill jobs job number, or range of job numbers # example: k 1 2 5 # example: k 1..5 # example: k 1..5 7 10..15 # example: k 1-5 7 10-15 function k   arg in $argv     if ruby -e "exit ('$arg' =~ /^[0-9]+\$/ ? 0 : 1)"       kill -9 %$arg     else       set _start (echo "$arg" | sed 's/[^0-9].*$//')       set _end   (echo "$arg" | sed 's/^[0-9]*[^0-9]*//')        n in (seq $_start $_end)         kill -9 %"$n"       end     end   end end 

the standard way use grep:

if echo "$arg" | grep -q -e '^[0-9]+$'   kill -9 %$arg