Extract version number of string

user_3142695 0
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I'm trying to get the version number out of a string which looks like this:

[7:38:06 PM] [semantic-release] › ℹ  Running semantic-release version 17.1.2
[7:38:09 PM] [semantic-release] › ✔  Published release 1.0.7 on default channel

As I need to do this in BusyBox, I can't use grep with the -P option:

grep -oP 'Published release \K.*? '

So I tried to use sed, but my attempt doesn't return anything:

sed -n 's/.*Published release (.*) .*/\1/p'

What is wrong with this command or is there a 'native' solution to get 1.0.7?

busybox shell regex
3 Answers

You could use Busybox awk:

$ busybox awk '
match($0,/Published release [0-9]+(\.[0-9]+)+/) {
    print substr($0,RSTART+18,RLENGTH-18)
}' file

Output:

1.0.7
user_41623560

You can use sed like this:

sed -n 's/.*Published release \([^ ]*\).*/\1/p' file

See an online demo:

s='[7:38:06 PM] [semantic-release] › ℹ  Running semantic-release version 17.1.2
[7:38:09 PM] [semantic-release] › ✔  Published release 1.0.7 on default channel'
sed -n 's/.*Published release \([^ ]*\).*/\1/p' <<< "$s"
# => 1.0.7

Details

  • -n - suppresses the default line output
  • .*Published release \([^ ]*\).* - matches
    • .* - any 0 or more chars
    • Published release - a literal text
    • \([^ ]*\) - Group 1: any zero or more chars other than a space
    • .* - any 0 or more chars
  • \1 - replaces with Group 1 value
  • p - prints the result of the substitution
user_38329700

If using sed, you must escape brackets:

sed -n 's/.*Published release \(\[.0-9\]+\).*/\1/'
user_2561960
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