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bashgitshell
johnbumble
johnbumble

Reputation: 700

Remove Double Quotes from a String in bash

I am trying to fetch a single file from Git. When I run my git command with the hardcoded value it works. However, when I run it with a variable it breaks. I suspect it is due to double quotes being placed around my variable somehow. I have tried many commands to remove the double quotes from the string and none of them seem to work.

I am basically trying to use the filePath variable in my git command with no double quotes around it.

#!/usr/bin/env bash
 #reads the JSON value of the file-path key. 
 #file path key is "home/docs" (quotes included)
filePath=$(grep -o '"file-path": *"[^"]*"' ../package.json | grep -o '"[^"]*"$')
git archive [email protected]_URL.com:help/docs.git HEAD $filePath | tar -x

Upvotes: 1

Views: 5545

Answers (3)

stark
stark

Reputation: 13187

I ran into the problem recently that bash only removes quotes that did not result from variable expansion. So to get bash to remove double quotes that are in a string use eval to process the string after the variable expansion:

foo=\"abc\"\ \"def\"
echo $foo
"abc" "def"
eval echo $foo
abc def

Of course, don't use eval on unsanitised strings from users, as it is susceptible to ';', '&&', etc. to execute arbitrary bash.

Upvotes: 0

l0b0
l0b0

Reputation: 58978

This works:

$ printf '{"file-path": "/some/path"}' | jq --raw-output '."file-path"'
/some/path

So in your case:

filePath=$(jq --raw-output '."file-path"' ../package.json)

You have to quote the key because it contains a hyphen.

Upvotes: 5

Jack
Jack

Reputation: 6198

The following will work:

filePath=$(grep -o '"file-path": *"[^"]*"' ./package.json | cut -d: -f2 | tr -d ' "')

Upvotes: 0

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