How to pass dynamic parameters in curl request - shell

Question

I am passing dynamic value to testing method and executing the curl request. There is an issue with $PARAMETERS.
When I execute the following method, I get error as below
Error:-
curl: option -F: requires parameter
curl: try 'curl --help' or 'curl --manual' for more information
Function:-
testing() {
local URL=$1
local HTTP_TYPE=$2
local PARAMETERS=$3
# store the whole response with the status at the and
HTTP_RESPONSE=$(curl -w "HTTPSTATUS:%{http_code}" -X $HTTP_TYPE $URL $PARAMETERS)
echo $HTTP_RESPONSE
}
URL='https://google.com'
HTTP_TYPE='POST'
PARAMETERS="-F 'name=test' -F 'query=testing'"
result=$(testing $URL $HTTP_TYPE $PARAMETERS)
FYI, above is a sample method and am using shell script. Kindly tell me how to solve this?

Answer 1

Since you are passing multiple command arguments in your 3rd argument, shell will do splitting as you are using unquoted variable inside your function.
You should better use shell array to store PARAMETERS argument.
Have your function as:
testing() {
local url="$1"
local httpType="$2"
shift 2 # shift 2 times to discard first 2 arguments
local params="$#" # store rest of them in a var
response=$(curl -s -w "HTTPSTATUS:%{http_code}" -X "$httpType" "$url" "$params")
echo "$response"
}
and call it as:
url='https://google.com'
httpType='POST'
params=(-F 'name=test' -F 'query=testing')
result=$(testing "$url" "$httpType" "${params[#]}")
Additionally you should avoid using all-caps variable names in your shell script to avoid clashes with env variables.
In case you're not using bash, you can also use:
params="-F 'name=test' -F 'query=testing'"
result=$(testing "$url" "$httpType" "$params")
Code Demo