How to extract a value from a string using regex and a shell?

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I am in shell and I have this string: 12 BBQ ,45 rofl, 89 lol

Using the regexp: \d+ (?=rofl), I want 45 as a result.

Is it correct to use regex to extract data from a string? The best I have done is to highlight the value in some of the online regex editor. Most of the time it remove the value from my string.

I am investigating expr, but all I get is syntax errors.

How can I manage to extract 45 in a shell script?

shell regex
7 Answers

Using ripgrep's replace option, it is possible to change the output to a capture group:

rg --only-matching --replace '$1' '(\d+) rofl'
  • --only-matching or -o outputs only the part that matches instead of the whole line.
  • --replace '$1' or -r replaces the output by the first capture group.
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