当正则表达式中有可选的 substring 时，如何使用“expr”？

Question

I want to find the go[^ ]+ inside these two strings using expr . The output should be go1.17.6 and go1.18-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?

Answer 1

is that what you wanted?

.*go version [a-w ]*go\([^ ]*\) .*

Answer 2

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)