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

score 11 · Answer 1 · answered May 14 '12 at 08:42

11

You can use grep:

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

answered May 14 '12 at 08:42

codaddict

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score 5 · Answer 2 · answered May 14 '12 at 13:27

5

In pure Bash:

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

answered May 14 '12 at 13:27

Dennis Williamson

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score 1 · Answer 3 · edited May 23 '17 at 10:31

1

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.

edited May 23 '17 at 10:31

Community

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answered May 14 '12 at 08:42

keyser

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`echo` does not and `find` is not applicable to the question. – Dennis Williamson May 14 '12 at 13:24
@DennisWilliamson I never said they were. And you can use patterns with echo. Example: echo *ocu?ent? would return "Documents" if you're in a regular home folder. I realize you meant regex though, I'll edit that :) – keyser May 14 '12 at 14:09
That's not a regex, that's globbing. – Dennis Williamson May 14 '12 at 14:15
@DennisWilliamson which is pattern matching. – keyser May 14 '12 at 14:16

score 1 · Answer 4 · answered May 14 '12 at 09:13

1

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

answered May 14 '12 at 09:13

tripleee

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That would be `$@`. There's a difference. – Dennis Williamson May 14 '12 at 13:24
When you access the command line, there certainly is a difference between `$*` and `$@`; but I specifically chose to call the variable which contains the `argv` array by its original Bourne name because that's the name many beginner-level expositions use; and in this context, when you want the values split into whitespace-separated tokens, that's what a script would use. I agree that if you have to refer to the positional parameters directly in a script, `"$@"` is almost always what you want. – tripleee May 14 '12 at 13:32
With, for example, `set -- 'abc def' ghi`, `for arg in "$@"; do echo "$arg"; done` does the right thing. These rarely do the right thing: `$*` or `"$*"` (rarely = only when you want to flatten the arguments). There's no point in continuing to repeat beginner-level expositions which are wrong. – Dennis Williamson May 14 '12 at 13:40
`for token; do` does the right thing here, regardless of how you got the things in there; what's to argue about? – tripleee May 14 '12 at 13:58

score 0 · Answer 5 · edited May 14 '12 at 23:01

0

The following example using your input data string should solve the problem using sed.

$ echo abcderwer 123123 10,200 asdfasdf iopjjop | sed -ne 's/^.*\([0-9,]\{6\}\).*$/\1/p'
10,200

edited May 14 '12 at 23:01

octopusgrabbus

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answered May 14 '12 at 09:17

bos

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score 0 · Answer 6 · answered May 14 '12 at 14:22

A slightly non-typical solution:

< input tr -cd [0-9,\ ] | tr \  '\012' | grep '^..,...$'

(The first tr removes everything except commas, spaces, and digits. The second tr replaces spaces with newlines, putting each "number" on a separate line, and the grep discards everything except those that match your criterion.)

Bash - Extract numbers from String

6 Answers6