mjhjmtu

I have the following function:

awk 'gsub(""", "\""); print' ORS='\n' "$1" 

It takes a string such as:

query GetAnimal 
 getAnimal(id: "bear") 
 name
 

and return:

query GetAnimal n getAnimal(id: "bear") n namen n

It works! But when I run it as following:

variable="$(awk 'gsub(""", "\""); print' ORS='\n' "$1")"

It doesn't work. I think it's something to do with the quotation marks, but I've tried escaping them, and I can't figure out any working syntax.

Why is this not working?

While the `...` ancient form of command substitution does some extra backslash processing, the $(...) one doesn't (one of the reasons why it should be preferred over `...`.

As long as the output of cmd doesn't end in newline characters (as is the case in your example) and (except in zsh) doesn't contain NUL character and (in yash) doesn't contain byte sequences not forming valid characters

cmd

Should produce the same output¹ (standard output) as:

output=$(cmd); printf %s "$output"

$output will contain the exact output of cmd stripped of all trailing newline characters.

Here, most likely, I think you ran echo "$variable" to check the content of the variable and saw that \ were turned in and n into newline characters.

But that's not $(...)'s fault, it would be echo's. echo does expand the escape sequences in its arguments (though some implementations only do it when passed a non-standard -e option).

Generally, you can't use echo to output arbitrary data, use printf instead.

^{¹ Strictly speaking, when using command substitution, cmd's stdout become a pipe. cmd could decide to produce a different output when its stdout is not a terminal or is not seekable...}