CODEWITHSUNDEEP
bashcurlescapingquotes
Ondra Žižka
Ondra Žižka

Reputation: 46914

Bash - handling of (single) quotes in variables (for a Curl command)

I am writing a script which uses CURL.

There are some HTTP headers common to all requests. 

curl -X GET "..." -H  'accept: */*' -H "api-key: $API_KEY"

So I would like to put those to a variable:

HEADERS="-H  'accept: */*' -H \"api-key: $API_KEY\""

curl -X GET "..." $HEADERS      ### <-- This is what I want to achieve.

As you can see, it needs a mix of

I have tried quite a few combinations of escaping, mixing ' and ", building the string gradually, but always I end up with Bash adding unwanted escapes or leaving those which I need.

How can I achieve to have both -H params (i.e. all 4 arguments) in a variable which I can then use in a command?

There are a few similar questions but they are specific for other cases.

Upvotes: 0

Views: 901

Answers (1)

Alex O
Alex O

Reputation: 8174

What you want to do is to store a list of words (like, -H or accept: */*) where each word may contain whitespace characters. As bash delimits words in a string by whitespace (by default), it's difficult to use a string variable for a list of words that possibly contain whitespace characters.

A simple solution is to use a list variable. Each list member is a word (and it can thus contain whitepace as needed). When expanding that list, ensure proper quoting. In your case:

declare -a HEADERS
HEADERS=(
  '-H' 'accept: */*'
  '-H' 'api-key: $API_KEY'
)
curl -X GET "..." "${HEADERS[@]}"

Mind the "..." characters when expanding the array. They will ensure that each list item is expanded as a single, dedicated word.

Upvotes: 2

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