[SOLVED] How to replace a specific character in bash

Issue

I want to replace '_v' with a whitespace and the last dot . into a dash "-". I tried using

sed 's/_v/ /' and tr '_v' ' '

Original Text

src-env-package_v1.0.1.18

output

src-en -package 1.0.1.18

Expected Output

src-env-package 1.0.1-18

Solution

With your shown samples please try following sed code. Using sed's capability to store matched regex values into temp buffer(called capturing groups) here. Also using -E option here to enable ERE(extended regular expressions) for handling regex in better way.

Here is the Online demo for used regex.

sed -E 's/^(src-env-package)_v([0-9]+\..*)\.([0-9]+)$/\1 \2-\3/' Input_file

OR if its a variable value on which you want to run sed command then use following:

var="src-env-package_v1.0.1.18"
sed -E 's/^(src-env-package)_v([0-9]+\..*)\.([0-9]+)$/\1 \2-\3/' <<<"$var"

src-env-package 1.0.1-18

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Bonus solution: Adding a perl one-liner solution here, using capturing groups concept(as explained above) in perl and getting the values as per requirement.

perl -pe 's/^(src-env-package)_v((?:[0-9]+\.){1,}[0-9]+)\.([0-9]+)$/\1 \2-\3/'  Input_file

Answered By - RavinderSingh13

Answer Checked By - Candace Johnson (WPSolving Volunteer)