How to replace a specific character in bash

Question

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

This is not "a specific character", and thus tr is not a suitable tool for this. It seems you are seeking to replace a specific sequence of characters. — tripleee
– tripleee, Commented Aug 15, 2022 at 7:37
_v is a string, not a character. _ is a character, as is v. — Ed Morton
– Ed Morton, Commented Aug 15, 2022 at 19:11

potong · Accepted Answer · 2022-08-15 07:43:23Z

2

This might work for you (GNU sed):

sed -E 's/(.*)_v(.*)\./\1 \2-/' file

Use the greed of the .* regexp to find the last occurrence of _v and likewise . and substitute a space for the former and a - for the latter.

If one of the conditions may occur but not necessarily both, use:

sed -E 's/(.*)_v/\1 /;s/(.*)\./\1-/' file

edited Aug 15, 2022 at 7:43

answered Aug 15, 2022 at 7:27

potong

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RavinderSingh13 · Accepted Answer · 2022-08-15 00:42:06Z

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

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

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