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

Question

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

Answer 1

There's a simpler and more efficient way, using the native shell prefix/suffix removal feature:

temp="${opt%\"}"
temp="${temp#\"}"
echo "$temp"

${opt%\"} will remove the suffix " (escaped with a backslash to prevent shell interpretation).

${temp#\"} will remove the prefix " (escaped with a backslash to prevent shell interpretation).

Another advantage is that it will remove surrounding quotes only if there are surrounding quotes.

BTW, your solution always removes the first and last character, whatever they may be (of course, I'm sure you know your data, but it's always better to be sure of what you're removing).

Using sed:

echo "$opt" | sed -e 's/^"//' -e 's/"$//'

(Improved version, as indicated by jfgagne, getting rid of echo)

sed -e 's/^"//' -e 's/"$//' <<<"$opt"

So it replaces a leading " with nothing, and a trailing " with nothing too. In the same invocation (there isn't any need to pipe and start another sed. Using -e you can have multiple text processing).

Answer 2

Use tr to delete ":

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

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

Answer 3

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

Answer 4

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.

Answer 5

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 ;-))

Answer 6

You can do it with only one call to sed:

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

Answer 7

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.

Answer 8

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

Answer 9

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.

Answer 10

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'

Answer 11

If you come here for aws cli --query, try --output text.

Answer 12

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

Answer 13

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}"

Answer 14

There is another way to do it. Like:

echo ${opt:1:-1}

Answer 15

Linux=`cat /etc/os-release | grep "ID" | head -1 | awk -F= '{ print $2 }'`

echo $Linux
Output:
"amzn"

Simplest ways to remove double quotes from variables are

Linux=`echo "$Linux" | tr -d '"'` 

Linux=$(eval echo $Linux)

Linux=`echo ${Linux//\"/}`

Linux=`echo $Linux | xargs`

All provides the Output without double quotes:

echo $Linux

amzn

Answer 16

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