Bash - Extract numbers from String

Question

I got a string which looks like this:

"abcderwer 123123 10,200 asdfasdf iopjjop"

Now I want to extract numbers, following the scheme xx,xxx where x is a number between 0-9. E.g. 10,200. Has to be five digit, and has to contain ",".

How can I do that?

Thank you

Answer 1

You can use grep:

$ echo "abcderwer 123123 10,200 asdfasdf iopjjop" | egrep -o '[0-9]{2},[0-9]{3}'
10,200

Answer 2

In pure Bash:

pattern='([[:digit:]]{2},[[:digit:]]{3})'
[[ $string =~ $pattern ]]
echo "${BASH_REMATCH[1]}"

Answer 3

Simple pattern matching (glob patterns) is built into the shell. Assuming you have the strings in $* (that is, they are command-line arguments to your script, or you have used set on a string you have obtained otherwise), try this:

for token; do
  case $token in
    [0-9][0-9],[0-9][0-9][0-9] ) echo "$token" ;;
  esac
done

Answer 4

Check out pattern matching and regular expressions.

Links:

Bash regular expressions

Patterns and pattern matching

SO question

and as mentioned above, one way to utilize pattern matching is with grep. Other uses: echo supports patterns (globbing) and find supports regular expressions.