bash encapsulate command options in variables - bash

Question

I have a bunch of these to test my RESTful API
$CURL \
-v \
-H 'Content-Type: application/json' \
-H 'Accept: application/json' \
-X POST \
-d '{"user":{"email":"user#example.com","password":"secret"}}' \
$URL/$PATH/sessions
I kinda want to shorten it to something like
CURLOPTS="-v -H 'Content-Type: application/json' -H 'Accept: application/json'"
$CURL \
$CURLOPTS \
-X POST \
-d '{"user":{"email":"user#example.com","password":"secret"}}' \
$URL/$PATH/sessions
but the options don't seem to be passed in. Any clues?

Answer 1

Short answer: see BashFAQ #50:'m trying to put a command in a variable, but the complex cases always fail!.
Long answer: Putting commands (or parts of commands) into variables and then getting them back out intact is complicated. The reason your script doesn't work is because of the order in which the shell parses the command line: it parses (and removes) quotes and escapes, then replaces variable values. By the time $CURLOPTS gets replaced, it's too late for the quotes to have their intended effect; instead, they're passed to curl as part of the arguments, which confuses curl greatly.
The solution: store the options in an array rather than a plain string:
CURLOPTS=(-v -H 'Content-Type: application/json' -H 'Accept: application/json')
$CURL \
"${CURLOPTS[#]}" \
-X POST \
-d '{"user":{"email":"user#example.com","password":"secret"}}' \
"$URL/$PATH/sessions"

Answer 2

You can use an array and trigger word splitting
$ set -x
$ CURLOPTS=(-v -H 'Content-Type: application/json' -H 'Accept: application/json')
$ : curl "${CURLOPTS[#]}"
+ : curl -v -H 'Content-Type: application/json' -H 'Accept: application/json'