how to remove trailing characters with bash regex-CodePudding

In bash I have the following code:

a='and/or fox-----'
a=${a//[\/\_ ]/-}
a=${a//- $/ }
echo $a

With the 3rd line I want to replace the trailing '-' with an empty string, and I know in some contexts the $ means "one or more at the end of the string". However, the result I'm getting is and-or-fox-----. Does anyone know how to get a to be and-or-fox?

CodePudding user response：

I would remove the hyphens first:

a='and/or fox-----'
a=${a//-/}
echo $a
a=${a//[\/\_ ]/-}
echo $a

This yields and-or-fox.

CodePudding user response：

You can enable bash extended globbing (if not already enabled) and use it with a shell parameter expansion

#!/bin/bash
shopt -s extglob

a='and-or-fox-----'
echo "${a%%*(-)}"

output:

and-or-fox

CodePudding user response：

Replace with Bash's replace variable expansion and extract with Bash Regex all within same statement:

#!/usr/bin/env bash

a='and/or fox-----'

# Extracting with bash regex
[[ ${a//[\/\_]/-} =~ (.*[^-]) ]] || :
a=${BASH_REMATCH[1]}

printf %s\\n "$a"