Retrieve version number from file follow string

Question

I want to be able to retrieve a version number (n.n.n) from a file (yaml) that follows a string.

sdk: ">=0.2.4 <3.1.1"
version: ^2.3.1
sdk: flutter
version: 1.0.0+1

sed -n '/version:/p sample.yaml

which returns the line for the string provided "version:", I then piped this into another command grep to pattern match the line for the version number:

sed -n '/version:/p sample.yaml | grep -oP '(\[0-9\]+).(\[0-9\]+).(\[0-9\]+)'

I would like to see if there is a more efficient way to do this, and if string = 'sdk' from the example the returned value returns both when I would like to return a single value.

Regarding I would like to return a single value - which value and based on what criteria? — Ed Morton
– Ed Morton, Commented Apr 1, 2022 at 2:30
@HaiVu expected output: "I want to be able to retrieve a version number (n.n.n) from a file (yaml)" — confused-nerd
– confused-nerd, Commented Apr 2, 2022 at 15:14

Barmar · Accepted Answer · 2022-03-31 20:23:43Z

1

Use awk

awk '/version:/ { 
    pos=match($2, /[0-9]+\.[0-9]+\.[0-9]+/);
    print substr($2, pos, RLENGTH); }' sample.yaml

match() returns the position in the field where the regexp is found, and sets RLENGTH to its length.

edited Mar 31, 2022 at 20:23

answered Mar 31, 2022 at 20:06

Barmar

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1 Comment

confused-nerd Over a year ago

thank you, this didn't work for version: 1.0.0+1 as it returned 1.0.01. Apologies I didn't realise i had missed the : in the last version in the sample section.

Wiktor Stribiżew · Accepted Answer · 2022-03-31 20:11:25Z

You can use

sed -nE 's/.*version:[^0-9]*([0-9][0-9.]*).*/\1/p' file
grep -oP 'version:\D*\K\d[\d.]*' file

See an online demo:

#!/bin/bash
s='sdk: ">=0.2.4 <3.1.1"
version: ^2.3.1
sdk: flutter
version 1.0.0+1'
sed -nE 's/.*version:[^0-9]*([0-9][0-9.]*).*/\1/p' <<< "$s"
# => 2.3.1
grep -oP 'version:\D*\K\d[\d.]*' <<< "$s"
# => 2.3.1

The version:\D*\K\d[\d.]* grep regex matches version: substring, then any zero or more non-digits (with \D*), then omits the matched text (with \K) and then matches and returns a digit and then any zero or more digits or dots (with \d[\d.]*).

The sed POSIX ERE (see -E option enabling this regex flavor) pattern matches

.* - any text
version: - a fixed substring
[^0-9]* - zero or more non-digits
([0-9][0-9.]*) - Group 1 (\1): a digit and then zero or more digits or dots
.* - any text.

The whole match is replaced with Group 1 value, the p flag prints the result of the successful substitution (-n suppresses default line output).

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