[SOLVED] Remove sequence of a specific character from the end of a string in Bash

Issue

Input:

i="Item1;Item2;Item3;;;;;;;;"

Desired output:

i="Item1;Item2;Item3"

How do I get rid of the last few semicolons?

I know this one way to do it using 'sed':

sed 's/;$//'

However, it only removes the last semicolon. Running it repeatedly doesn't seem practical.

Solution

You can use

sed 's/;*$//'

The main point here is to add the * quantifier (that means zero or more) after ; to make the regex engine match zero or more semi-colons.

Synonymic sed commands can look like

sed 's/;;*$//'    # POSIX BRE "one ; and then zero or more ;s at the end of string"
sed 's/;\+$//'    # GNU sed POSIX BRE "one or more semi-colons at the end of string"
sed -E 's/;+$//'  # POSIX ERE "one or more semi-colons at the end of string"

Answered By - Wiktor Stribiżew