How to use grep and awk together to get specific part of a url/string

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So I have a svn repo URL like this--

svn info
Path: .
URL: svn://xyz-repo/svn/ccsmp/branches/features/cre40_jenkins
Repository Root: svn://xyz-repo/svn/ccsmp
Repository UUID: 5de8f2f4-0d3b-0410-8f2c-d418a2640d16
Revision: 23430

I want to grab only this part--

branches/features/cre40_jenkins

So far I have tried something like this

def BRANCH_NAME = sh(script: " svn info | grep -Po 'Relative URL: \\^/\\K.*' ", returnStdout: true , trim: true)

But this didn't worked as I am using a very old Jenkins version with old plugins and svn.

Is there any other way to do this. Please help any suggestions would be highly appreciated.

What I have tried:

def BRANCH_NAME = sh(script: " svn info | grep -Po 'Relative URL: \\^/\\K.*' ", returnStdout: true , trim: true)

Posted 16-Feb-22 11:03am

Member 14030165

Updated 16-Feb-22 22:05pm

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Peter_in_2780 · Answer 1 · 2022-02-16T22:05:00

Solution 1

Run it through something like this:

Bash

svn info | awk 'BEGIN {FS="/"} /^URL:/ {print $6 "/" $7 "/" $8}'

[edit] fixed typo [/edit]

Posted 16-Feb-22 22:05pm

Peter_in_2780

Updated 17-Feb-22 9:58am

v2