[string] Shell script - remove first and last quote (") from a variable

Below is the snippet of a shell script from a larger script. It removes the quotes from the string that is held by a variable. I am doing it using sed, but is it efficient? If not, then what is the efficient way?

#!/bin/sh

opt="\"html\\test\\\""
temp=`echo $opt | sed 's/.\(.*\)/\1/' | sed 's/\(.*\)./\1/'`
echo $temp

My version

strip_quotes() {
    while [[ $# -gt 0 ]]; do
        local value=${!1}
        local len=${#value}
        [[ ${value:0:1} == \" && ${value:$len-1:1} == \" ]] && declare -g $1="${value:1:$len-2}"
        shift
    done
}

The function accepts variable name(s) and strips quotes in place. It only strips a matching pair of leading and trailing quotes. It doesn't check if the trailing quote is escaped (preceded by \ which is not itself escaped).

In my experience, general-purpose string utility functions like this (I have a library of them) are most efficient when manipulating the strings directly, not using any pattern matching and especially not creating any sub-shells, or calling any external tools such as sed, awk or grep.

var1="\"test \\ \" end \""
var2=test
var3=\"test
var4=test\"
echo before:
for i in var{1,2,3,4}; do
    echo $i="${!i}"
done
strip_quotes var{1,2,3,4}
echo
echo after:
for i in var{1,2,3,4}; do
    echo $i="${!i}"
done

You can do it with only one call to sed:

$ echo "\"html\\test\\\"" | sed 's/^"\(.*\)"$/\1/'
html\test\

Use tr to delete ":

 echo "$opt" | tr -d '"'

Note: This removes all double quotes, not just leading and trailing.

In Bash, I would use the following one-liner:

[[ "${str}" == \"*\" || "${str}" == \'*\' ]] && str="${str:1:-1}"

This will remove surrounding quotes (both single and double) while keeping quoting characters inside the string intact. Also, it won't do anything if there's only a single leading quote or only a single trailing quote, which is usually what you want in my experience.

Wrapped in a function:

#!/usr/bin/env bash

# Strip surrounding quotes from string [$1: variable name]
function strip_quotes() {
    local -n var="$1"
    [[ "${var}" == \"*\" || "${var}" == \'*\' ]] && var="${var:1:-1}"
}

str="$*"
echo "Before: ${str}"
strip_quotes str
echo "After:  ${str}"

There is a straightforward way using xargs:

> echo '"quoted"' | xargs
quoted

xargs uses echo as the default command if no command is provided and strips quotes from the input. See e.g. here.

I know this is a very old question, but here is another sed variation, which may be useful to someone. Unlike some of the others, it only replaces double quotes at the start or end...

echo "$opt" | sed -r 's/^"|"$//g'

If you're using jq and trying to remove the quotes from the result, the other answers will work, but there's a better way. By using the -r option, you can output the result with no quotes.

$ echo '{"foo": "bar"}' | jq '.foo'
"bar"

$ echo '{"foo": "bar"}' | jq -r '.foo'
bar

There is another way to do it. Like:

echo ${opt:1:-1}

The shortest way around - try:

echo $opt | sed "s/\"//g"

It actually removes all "s (double quotes) from opt (are there really going to be any more double quotes other than in the beginning and the end though? So it's actually the same thing, and much more brief ;-))

If you try to remove quotes because the Makefile keeps them, try this:

$(subst $\",,$(YOUR_VARIABLE))

Based on another answer: https://stackoverflow.com/a/10430975/10452175

This is the most discrete way without using sed:

x='"fish"'
printf "   quotes: %s\nno quotes:  %s\n" "$x" "${x//\"/}"

Or

echo $x
echo ${x//\"/}

Output:

   quotes: "fish"
no quotes:  fish

I got this from a source.

Update

A simple and elegant answer from Stripping single and double quotes in a string using bash / standard Linux commands only:

BAR=$(eval echo $BAR) strips quotes from BAR.

=============================================================

Based on hueybois's answer, I came up with this function after much trial and error:

function stripStartAndEndQuotes {
    cmd="temp=\${$1%\\\"}"
    eval echo $cmd
    temp="${temp#\"}"
    eval echo "$1=$temp"
}

If you don't want anything printed out, you can pipe the evals to /dev/null 2>&1.

Usage:

$ BAR="FOO BAR"
$ echo BAR
"FOO BAR"
$ stripStartAndEndQuotes "BAR"
$ echo BAR
FOO BAR

This will remove all double quotes.

echo "${opt//\"}"

The easiest solution in Bash:

$ s='"abc"'
$ echo $s
"abc"
$ echo "${s:1:-1}"
abc

This is called substring expansion (see Gnu Bash Manual and search for ${parameter:offset:length}). In this example it takes the substring from s starting at position 1 and ending at the second last position. This is due to the fact that if length is a negative value it is interpreted as a backwards running offset from the end of parameter.