curl with variables - Bash

Question

I'm trying to do a post using curl in the following loop:

for name in "${first_name[@]}"; do
        password=$(</dev/urandom tr -dc '12345!@#$%qwertQWERTasdfgASDFGzxcvbZXCVB' | head -c16; echo "")
        curl -X POST -H "Content-Type: application/json" -H "Authorization: Token ${token}" -d '{"username": "${username[$name]}", "password": "${password}", "email": "${email[$name]}",  "first_name": "${name}", "last_name": "${last_name[$name]}", "is_staff": "${staff[$name]}", "is_active": "${active[$name]}", "groups": ["${group_id}"]' https://app.com/api/users/users/
done

With this syntax I am getting the following error:

{"detail":"JSON parse error - Expecting ',' delimiter: line 1 column 242 (char 241)"}

Each of the variables have a value as I was printing them inside the loop with an echo.

It looks like a syntax error. Any ideas what it is wrong?

Thank you.

You single-quoted the argument to `-d`, so none of your parameters get expanded. — chepner, Feb 22 '22 at 16:13
Use `jq` to generate the payload, as it takes care of safely escaping any parameter values to generate valid JSON for you. — chepner, Feb 22 '22 at 16:13
there is also a missing `}` at the end of your "json" object. — MarcoLucidi, Feb 22 '22 at 16:14
you may want to give [`jo`](https://github.com/jpmens/jo) a try. — MarcoLucidi, Feb 22 '22 at 16:16
The highly-ranked answers on the linked duplicate are going to give you a more robust solution than the answer that's currently directly added here; I strongly recommend linking through and reading them. — Charles Duffy, Feb 22 '22 at 20:48

score 0 · Answer 1 · answered Feb 22 '22 at 20:43

Try this :

for name in "${first_name[@]}"; do
    password=$(</dev/urandom tr -dc '12345!@#$%qwertQWERTasdfgASDFGzxcvbZXCVB' |
               head -c16; echo "")
    curl -X POST \
        -H "Content-Type: application/json" \
        -H "Authorization: Token ${token}" \
        -d "$(cat << EOF
{"username": "${username[$name]}",
 "password": "${password}",
 "email": "${email[$name]}",
 "first_name": "${name}",
 "last_name": "${last_name[$name]}",
 "is_staff": "${staff[$name]}",
 "is_active": "${active[$name]}",
 "groups": ["${group_id}"]
}
EOF
        )" https://app.com/api/users/users/
done

This doesn't fix any data that isn't valid inside a JSON string without escaping. Consider a user who goes by `Joe ("Johnny") Hill` -- the `"`s need backslashes added. A tool like `jq` will do that automatically. — Charles Duffy, Feb 22 '22 at 20:46

curl with variables - Bash

1 Answers1