How to pass bash variable to JSON - json

Question

I'm trying to write a sample script where I'm generating names like 'student-101...student-160'. I need to post JSON data and when I do, I get a JSON parse error.
Here's my script:
name="student-10"
for i in {1..1}
do
r_name=$name$i
echo $r_name
curl -i -H 'Authorization: token <token>' -d '{"name": $r_name, "private": true}' "<URL>" >> create_repos_1.txt
echo created $r_name
done
I always get a "Problems parsing JSON" error. I've tried various combination of quotes, etc but nothing seems to work!
What am I doing wrong?

Answer 1

First, your name property is a string, so you need to add double quotes to it in your json.
Second, using single quotes, bash won't do variable expansion: it won't replace $r_name with the variable content (see Expansion of variable inside single quotes in a command in bash shell script for more information).
In summary, use:
-d '{"name": "'"$r_name"'", "private": true}'

Answer 2

Another option is to use printf to create the data string:
printf -v data '{"name": "%s", "private": true}' "$r_name"
curl -i -H 'Authorization: token <token>' -d "$data" "$url" >> create_repos_1.txt

Answer 3

Don't; use jq (or something similar) to build correctly quoted JSON using variable inputs.
name="student-10"
for i in {1..1}
do
r_name=$name$i
jq -n --arg r_name "$r_name" '{name: $r_name, private: true}' |
curl -i -H 'Authorization: token <token>' -d #- "<URL>" >> create_repos_1.txt
echo created $r_name
done
The #- argument tells curl to read data from standard input (via the pipe from jq) to use for -d.
Something like "{\"name\": \"$r_name\", \"private\": true}" may work, but it is ugly and will also fail if r_name contains any character which needs to be quoted in the resulting JSON, such as double quotes or ASCII control characters.