Escape charactors getting removed in echo statement

Question

Original string getting mutated while printing with echo statement.

#!/bin/bash

response='{\\\"test\\\":\\\"data\\\"}'
echo $response;

Actual Output - {\\"test\\":\\"data\\"}

Expected output - {\\\"test\\\":\\\"data\\\"}

How are you running the script? Your actual output is what I would expect from a POSIX-compliant implementation of `echo`, which `bash` does not (by default) provide. — chepner, Jul 04 '19 at 18:33
For anyone saying it "works for me": different versions of `echo` have different behavior. Some interpret escapes, some don't, some only interpret them if given the `-e` option, and some will print the "-e" as part of their output. It's all hopelessly inconsistent. According to the [POSIX spec](https://pubs.opengroup.org/onlinepubs/9699919799/utilities/echo.html), "It is not possible to use `echo` portably across all POSIX systems unless both `-n` (as the first argument) and escape sequences are omitted." So use `printf` instead. — Gordon Davisson, Jul 04 '19 at 19:46

score 2 · Accepted Answer · answered Jul 04 '19 at 18:32

2

e.g.:

$ response='{\\\"test\\\":\\\"data\\\"}'
$ printf '%s\n' "$response"
{\\\"test\\\":\\\"data\\\"}

answered Jul 04 '19 at 18:32

Ed Morton

score 0 · Answer 2 · answered Jul 04 '19 at 18:32

This works as expected in bash, but you are instead running it with sh. See: Why does my bash code fail when I run it with sh?

However, when you want to print a string exactly as is, use printf:

response='{\\\"test\\\":\\\"data\\\"}'
printf '%s\n' "$response"

This works correctly for all values in all shells, including response='*' reponse='-n' and response='foo bar'

2 Answers2