Create a curl url in bash - bash

Question

I need to have curl url formatted like this in order for it to be successful.
$instanceUrl/sobjects/$sObject/describe -H 'Authorization: Bearer $apiToken'
This is my script so far.
#!/bin/bash
CURL='/usr/bin/curl'
CURLARGS='-H'
echo "welcome to sf-db-mock"
echo "please enter the instance url of the Salesforce org database you'd like to mock, followed by [ENTER]:"
# get rest endpoint
read instanceUrl
echo "please enter your api token, followed by [ENTER]:"
# get api token from users current session
read apiToken
# get the objects the users wants described
echo "please enter each sObject you'd like to mock by entering its API Name, using a space after each sObject, followed by [ENTER]:"
read -a sObjects
for sObject in ${sObjects[#]}
do
describe="$('sobjects'\/$sObject\/'describe '$CURLARGS ' Authorization: Bearer '$apiToken)"
echo $instanceUrl$describe
done
The error I get while running it is
./mock.txt: line 25: sobjects/survey/describe -H: No such file or directory'
My bash script is executable.

Answer 1

The notation $(a b c) means run command a with arguments b and c.
What you want is to remove the parenthesis:
describe="sobjects/$sObject/describe $CURLARGS 'Authorization: Bearer '$apiToken'"