unix: how to tell if a string matches a regex

Question

Trying out fish shell, so I'm translating my bash functions. The problem is that in one case, I'm using bash regexes to check if a string matches a regex. I can't figure out how to translate this into fish.

Here is my example.

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

• I looked into sed, but I don't see a way to get it to set its exit status based on whether the regex matches.
• I looked into delegating to Ruby, but again, getting the exit status set based on the match requires making this really ugly (see below).
• I looked into delegating back to bash, but despite trying maybe three or four ways, never got that to match.

So, is there a way in *nix to check if a string matches a regex, so I can drop it into a conditional?

---

Here is what I have that currently works, but which I am unhappy with:

# kill jobs by 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
  for 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]*//')

      for n in (seq $_start $_end)
        kill -9 %"$n"
      end
    end
  end
end

Answer 1

I would suggest to use the built-in string match subcommand

if string match -r -q  '^[0-9]+$' $arg
  echo "number!"
else
  echo "not a number"
end

Answer 2

The standard way is to use grep:

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