Get part of string that matches regex

Question

I'm writing a bash script and trying to parse a file to get a software version. The output is :

Version 13.0.R7.0 - Built on Tue Feb 09 18:47:29 EST 2016

I want to be able to pull out 13.0 (I will later cut off the .0) with the regex ^1[0-9]\.0, and pull out the R7 with R[0-9]{1}.

I know how to match them, I just don't know how to put the value that matches the regex into a variable, is such a thing possible?

Does that mean that you could use sed or something similar? — Michael Vehrs
– Michael Vehrs, Commented Jul 27, 2016 at 7:41

Cyrus · Accepted Answer · 2016-07-26 19:32:32Z

2

With GNU bash:

echo 'Version 13.0.R7.0 - Built on Tue Feb 09 18:47:29 EST 2016' | while IFS=" ." read -r a b c d rest; do echo $b $c $d; done

or

while IFS=" ." read -r a b c d rest; do echo $b $c $d; done < filename

Output:

 13 0 R7

answered Jul 26, 2016 at 19:32

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