Bash variable in 2 quotes

There are several ways to do it ex. given

$ set -- 'foo bar'

(to assign value foo bar to the shell's first positional parameter, as if in a script invoked like myscript "foo bar") then for example

$ echo '{"asd:":"'"$1"'"}'
{"asd:":"foo bar"}

$ echo {\"asd:\":\""$1"\"}
{"asd:":"foo bar"}

However you may find it cleaner to use the shell's printf builtin to create a formatted string that you can assign to a new variable:

$ printf -v data '{"asd": "%s"}' "$1"
$ echo "$data"
{"asd": "foo bar"}

which you can then use as

curl some parameters and headers --data-binary "$data"

Alternatively, since you appear to be trying to pass a JSON object to the curl command, you could consider using jq in place of printf:

$ jq -nc --arg x "$1" '{asd: $x}'
{"asd":"foo bar"}

or similarly using the built-in $ARGS array

$ jq -nc --arg asd "$1" '$ARGS.named'
{"asd":"foo bar"}

if you want to pass both the name and value to the constructor.

I tend to use a heredoc and -d@- (read input data from stdin) for composing JSON request bodies:

cat <<EOF |
{
    "secret": "$SECRET"
}
EOF
curl ... \
    -X POST \
    -H "Content-Type: application/json" \
    "${BASE_URL%%/}/the/path" -d@-

Yes, the syntax is a little weird.

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