Grep using a regular expression and capturing using groups

Question

I have multiple text files containing lines as below: """Version : 3.4.0.0 xxx xxx xxx""" I'm trying to capture 3.4.0.0 from each text file and display on the terminal. the command i've written is this :

grep -m1 'Version[\s\S]*?(\d[\s\S]+?)\n' *.txt

grep -m1 'Version[\s\S]*?(\d[\s\S]+?)\n' *.txt

I'm unable to get any results using the above code. Can someone assist me on this?

Answer 1

Your pattern is a PCRE compliant regex, [\s\S] matches any char in a PCRE pattern, but not with POSIX BRE regex flavor that you are using with your grep since you did not use-P option.

If you want to do it with GNU grep use

grep  -oPm 1 'Version\s*:\s*\K\d+(?:\.\d+)+' *.txt

See this online demo

Details

• Version - Version string
• \s*:\s* - a colon enclosed with 0+ whitespaces
• \K - match reset operator
• \d+ - 1+ digits and then
• (?:\.\d+)+ - 1 or more repetitions of . and 1+ digits.

You may do that with awk:

awk '/^Version : [0-9]+/{print $3; exit}' *.txt

See the online awk demo:

s="Text
Version : 3.4.0.0 xxx xxx xxx
More text
Version : 5.6.0.0 xxx xxx xxx"
awk '/^Version : [0-9]+/{print $3; exit}' <<< "$s"
# => 3.4.0.0

Details

• ^Version : [0-9]+ finds a line that starts with Version : <1 or more digits>
• {print $3; exit} outputs the value of Field 3 and stops processing getting you just the first match.