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regexexpr

How to use `expr` when there's an optional substring in the regex?

I want to find the go[^ ]+ inside these two strings using expr. The output should be 1.17.6 and 1.18-becaeea119.

go version go1.17.6 linux/amd64
go version devel go1.18-becaeea119 Tue Dec 14 17:43:51 2021 +0000 linux/amd64

However, the devel part is optional and I can't figure out a way to properly ignore it with expr.

expr "$(go version)" : ".*go version go\([^ ]*\) .*"
expr "$(go version)" : ".*go version devel go\([^ ]*\) .*"

Using normal regexes, I would just (?: devel)? it, but expr doesn't support ? for some reason.

Is there any way to achieve this using expr in one command?

Solution

  • Use

    .*go version.* go\([^[:space:]]*\) .*

    EXPLANATION

    --------------------------------------------------------------------------------
  .*                       any character (0 or more times)
--------------------------------------------------------------------------------
  go version               'go version'
--------------------------------------------------------------------------------
  .*                       any character (0 or more times)
--------------------------------------------------------------------------------
   go                      ' go'
--------------------------------------------------------------------------------
  \(                       group and capture to \1:
--------------------------------------------------------------------------------
    [^[:space:]]*            any character except: whitespace
                             characters (0 or more times)
--------------------------------------------------------------------------------
  \)                       end of \1
--------------------------------------------------------------------------------
                           ' '
--------------------------------------------------------------------------------
  .*                       any character (0 or more times)