Regex to extract version from string

问题

I have a string which has below value

String1=winos-app-1.2.0-4.20120308.InLinuxOS.x86_64

I need to extract only version from this string which is "1.2.0-4"

I have tried regular expression as mentioned below with sed

sed -ne 's/[^0-9]*\(\([0-9]\.\)\{0,1\}[0-9][^.]\).*/\1/p'

but I am only getting result "1.2.0-", missing number after "-" which is 4. I tried correcting it but it is returning null.

Kindly, advise

回答1:

how about grep:

grep -Po "(?<=-)[\d.-]*(?=.\d{8})"

回答2:

This should work, and should also account for any of the version numbers being more than one digit long.

String1=winos-app-1.2.0-4.20120308.InLinuxOS.x86_64
sed -n 's/[^0-9]*\(\([0-9]\+\.\)\{0,2\}[0-9]\+-[0-9]\+\).*/\1/p' <<< "$String1"
1.2.0-4

BTW, this will also match version strings like (which from your question is not clear if you want this behavior or not):

1-1
1.2-1

If you want to enforce w.x.y-z, you could use this:

sed -n 's/[^0-9]*\(\([0-9]\+\.\)\{2\}[0-9]\+-[0-9]\+\).*/\1/p' <<< "$String1"

回答3:

sed -n 's/^.*-\([0-9.]*-[0-9]*\)\..*$/\1/p'

回答4:

Can you try this?

[^0-9]*\(\([0-9]\.\)\{0,2\}[0-9][^.]\).*

回答5:

You could also try parameter expansion:

string1=winos-app-1.2.0-4.20120308.InLinuxOS.x86_64
string2=${string1%".${string1#*-*-*-*.}"}
version=${string2#*-*-}
printf "%s\n" "$version"

来源：https://stackoverflow.com/questions/15305065/regex-to-extract-version-from-string