[SOLVED] How can I use a variable in a curl call within a Bash script?

How can I use a variable in a curl call within a Bash script?

I have this simple task, and I've spent a few hours already trying to figure out how can I use a variable inside a curl call within my Bash script:

message="Hello there"
curl -X POST -H 'Content-type: application/json' --data '{"text": "${message}"}'

This is outputting ${message}, literally because it's inside a single quote. If I change the quotes and put a double quote outside and a single quote inside, it says:

command not found: Hello

and then

command not found: there.

How can I make this work?

Solution

Variables are not expanded within single quotes. Rewrite using double quotes:

curl -X POST -H 'Content-type: application/json' --data "{\"text\": \"${message}\"}"

Just remember that double quotes within double quotes have to be escaped.

Another variation could be:

curl -X POST -H 'Content-type: application/json' --data '{"text": "'"${message}"'"}'

This one breaks out of the single quotes, encloses ${message} within double quotes to prevent word splitting, and then finishes with another single-quoted string. That is:

... '{"text": "'"${message}"'"}'
    ^^^^^^^^^^^^
    single-quoted string


... '{"text": "'"${message}"'"}'
                ^^^^^^^^^^^^
                double-quoted string


... '{"text": "'"${message}"'"}'
                            ^^^^
                            single-quoted string

However, as the other answer and @Charles Duffy pointed out in a comment, this is not a robust solution, because literal " and other characters in $message may break the JSON content.

Use the other solution, which passes the content of $message to jq in a safe way, and jq takes care of correct escaping.