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

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?

CodePudding user response：

is that what you wanted?

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

CodePudding user response：

Use

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

EXPLANATION

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