Is there a way to ignore an optional query field in an API url?

Question

I'll try to clarify what I mean.

Let's say I have this url:

https://test-api.com/users?age=17&sex=M

There's 2 fields: age and sex. The age field is required but the sex field is optional.

Let's say I want to make a bunch of tests and I use this code:

import requests

def testUserQuery(user_age, user_sex):
    url = f'https://test-api.com/users?age={user_age}&sex={user_sex}'
    response = requests.get(url)

test_query = testUserQuery(17)

Now, assuming that I can't go into the actual code of the API itself and change how empty fields are interpreted...

How can I make this test and leave the user_sex field blank?

In other words, is there a special universal symbol (like "&" which means "and" for every URL in the world) that I can put for user_sex that'll force the API to ignore the field and not cause errors?

Otherwise, I would have to do this:

import requests

def testUserQuery(user_age, user_sex=None):
    if user_sex is None:
       url = f'https://test-api.com/users?age={user_age}'
    elif user_sex is not None:
       url = f'https://test-api.com/users?age={user_age}&sex={user_sex}'
    response = requests.get(url)

test_query = testUserQuery(17)

Imagine if I'm dealing with 10 optional fields. I don't think it would be very efficient to make multiple elif statements to change the URL for every single possible case where an optional field is empty.

I hope I'm being clear, sorry if this sounds confusing.

score 1 · Answer 1 · answered Jul 24 '22 at 06:16

One way to dynamically achieve this is by changing testUserQuery to accept its arguments as **kwargs then using urllib.parse.urlencode to dynamically build the query string.

from urllib.parse import urlencode


def testUserQuery(base_url='https://test-api.com/users', **kwargs):
    params = urlencode({k: v for k, v in kwargs.items() if v is not None})
    url = f"{base_url}{'?' + params if params else ''}"
    print(url)


testUserQuery()
testUserQuery(a=1)
testUserQuery(a=1, b=2)

This outputs

https://test-api.com/users
https://test-api.com/users?a=1
https://test-api.com/users?a=1&b=2

score 1 · Accepted Answer · answered Jul 24 '22 at 06:32

1

Here's a simple way to do this by utilising the params parameter:

import requests

URL = 'https://test-api.com/users'

def testUserQuery(**params):
    return requests.get(URL, params=params)

testUserQuery(age=21, sex='male')
testUserQuery(age=21)

In other words, all you have to do is match the parameter names with those that are understood by the API. No need to manipulate the URL

answered Jul 24 '22 at 06:32

DarkKnight

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This works, thanks. However, how can I assign default values for arguments? Imagine that if no age is entered, I want age=19. – Yen Jul 24 '22 at 16:03
NVM. I figured it out. If 'param name' not in params --> params['params name'] = default value. – Yen Jul 24 '22 at 16:09
@Yen Take a look at [setdefault](https://docs.python.org/3/library/stdtypes.html#dict.setdefault) – DarkKnight Jul 24 '22 at 18:56

Is there a way to ignore an optional query field in an API url?

2 Answers2