How to pass a parameter in the string in python using format-CodePudding

There as a url link which i have in settings.py file :

KAVENEGAR_URL = "https://api.kavenegar.com/v1/{key}/verify/lookup.json?receptor={phone}&token={otp}".format(key,phone,otp)

I want to use it in service.py file inside send_otp_sms method I dont know how to pass key, phone, otp variables in it

from django.conf import settings

def send_otp_sms(key, phone, otp):
    kavenegar_url = settings.KAVENEGAR_URL.format(key, phone, otp)
    response = requests.post(kavenegar_url)
    if response.status_code != 200:
        raise APICallError
    return response

I used .format as you see in the codes but it pass error in settings.py which Unresolved reference 'key' Unresolved reference 'phone' Unresolved reference 'otp'

CodePudding user response：

The problem is this that when i use parameters in the string like this :

KAVENEGAR_URL = "https://api.kavenegar.com/v1/{key}/verify/lookup.json?receptor={phone}&token={otp}"

I have to use .format like this :

.format(key=key, phone=phone, otp=otp)

and if i use empty braces like this :

KAVENEGAR_URL = "https://api.kavenegar.com/v1/{}/verify/lookup.json?receptor={}&token={}"

I have to use .format like this :

.format(key, phone, otp)

CodePudding user response：

Tente Isso:

from django.conf import settings

def send_otp_sms(key, phone, otp):
  kavenegar_url = f"https://api.kavenegar.com/v1/{key}/verify/lookup.json?receptor={phone}&token={otp}"
  response = requests.post(kavenegar_url)
  if response.status_code != 200:
    raise APICallError
  return response