Use grep to extract some text from file based on regex

Question

I have this file:

header:
  title: hello
  version: 1.2.3

I want to extract the version number.

My original attempt was

grep ^\s+version:\s+(\d\.\d\.\d) file.txt

but that produced empty output. After suggestions in the comments, I tried

grep -P '^\s+version:\s+(\d\.\d\.\d)' file.txt

but I get " version: 1.2.3" instead of "1.2.3".

What am I doing wrong?

Answer 1

grep uses Posix Basic Regex (BRE) by default which does not support your notation.

Use grep -E to use Posix Extended Regex (ERE) and grep -P to use Perl Compatible Regex (PCRE) if available.

Your notation works with grep -P:

grep -P '^\s+version:\s+(\d\.\d\.\d)' file.txt

This works with BRE:

grep '^ \+version: \+\([0-9]\.[0-9]\.[0-9]\)' file.txt

Output:

  version: 1.2.3

Note, that the capture group is not necessary here, as grep doesn't do anything with it.

---

If you want the version nr only., use \K and -o option:

grep -Po '^\s+version:\s+\K\d\.\d\.\d' file.txt

Output:

1.2.3

With BRE, this is not possible, you will need to chain two grep commands:

grep 'version: ' file.txt | grep -o '[0-9]\.[0-9]\.[0-9]'

or use sed (credits @Kusalananda):

sed -n 's/.*version: //p' file.txt

Answer 2

According the comment of pLumo and How to use grep to get anything just after name a working command could be

 grep -oP "(?<=version: )(\d\.\d\.\d)" file.txt