curl command can not be formed dynamically due to single quote

March 16, 2022

I want to have a curl command like below

curl --location --request POST 'https://abcd.com/api/v4/projects/<projectId>/triggers' \
--header 'PRIVATE-TOKEN: <your_access_token>' \
--form 'description="my description"'

Now I wrote a shell script function to generate it dynamically

it need projectId, token, and description as a pramter

callApi(){
while IFS="," read -r -a users; do
for u in "${users[@]}"
do
url="'https://abcd.com/api/v4/projects/$1/triggers'"
echo $url

header="'PRIVATE-TOKEN: $2'"
echo $header

desc="'description=$u token'"
echo $desc

tk=$(curl --location --request POST $url \
--header $header \
--form $desc)

echo $tk

done
done <<< $(cat $3)
}


callApi "<projectId>" "<token>" ./users.csv

It echo perfectly
But
It thorws error

>Solution :

Don’t use both double and single quotes like that. You are adding literal single quotes to the url (and other variables) which, as you have discovered, breaks.

Use double quotes if you need to allow command or parameter substitution, single quotes otherwise. Double quote your variables everywhere you dereference them.

Use indentation for readability.

Useless use of cat.

callApi() {
  while IFS="," read -r -a users; do
    for u in "${users[@]}"; do
      url='https://abcd.com/api/v4/projects/$1/triggers'
      echo "$url"

      header="PRIVATE-TOKEN: $2"
      echo "$header"

      desc="description=$u token"
      echo "$desc"

      tk=$(
        curl --location \
             --request POST \
             --header "$header" \
             --form "$desc" \
             "$url"
      )
      echo "$tk"
    done
  done < "$3"
}