Generating strings dynamically in Python

Question

I'm generating a URL (in string) that depends on 3 optional parameters, file, user and active.

From a base url: /hey I want to generate the endpoint, this means:

If file is specificied, my desired output would is: /hey?file=example
If file and user is specified, my desired output is: /hey?file=example&user=boo
If user and active are specified, my desired output is: /hey?user=boo&active=1
If no optional parameters are specified, my desired output is: /hey
and so on with all the combinations...

My code, which is working correctly, is as follows (change the None's at the top if you want to test it):

file = None
user = None
active = 1

ep = "/hey"
isFirst = True

if file:
  if isFirst:
    ep+= "?file=" + file;
    isFirst = False;
  else: ep += "&file=" + file;

if user:
  if isFirst:
    ep+= "?user=" + user;
    isFirst = False;
  else: ep += "&user=" + user;

if active:
  if isFirst:
    ep+= "?active=" + str(active);
    isFirst = False;
  else: ep += "&active=" + str(active);

print ep

Can someone give me a more python implementation for this? I can't use modules as requests.

Thanks in advance.

\$\begingroup\$ Lose the ;. makes python look ugly \$\endgroup\$

hjpotter92
– hjpotter92

2018-12-17 14:00:28 +00:00
Commented Dec 17, 2018 at 14:00 — hjpotter92
– hjpotter92, Commented Dec 17, 2018 at 14:00

Graipher · Accepted Answer · 2018-12-18 08:52:10Z

12

The most Pythonic version of this depends a bit on what you do with that URL afterwards. If you are using the requests module (which you probably should), this is already built-in by specifying the params keyword:

import requests

URL = "https://example.com/hey"

r1 = requests.get(URL, params={"file": "example"})
print(r1.url)
# https://example.com/hey?file=example

r2 = requests.get(URL, params={"file": "example", "user": "boo"})
print(r2.url)
# https://example.com/hey?file=example&user=boo

r3 = requests.get(URL, params={"user": "boo", "active": 1})
print(r3.url)
# https://example.com/hey?user=boo&active=1

r4 = requests.get(URL, params={})
print(r4.url)
# https://example.com/hey

If you do need a pure Python solution without any imports, this is what I would do:

def get_url(base_url, **kwargs):
    if not kwargs:
        return base_url
    params = "&".join(f"{key}={value}" for key, value in kwargs.items())
    return base_url + "?" + params

Of course this does not urlencode the keys and values and may therefore be a security risk or fail unexpectedly, but neither does your code.

Example usage:

print(get_url("/hey", file="example"))
# /hey?file=example

print(get_url("/hey", file="example", user="boo"))
# /hey?file=example&user=boo

print(get_url("/hey", user="boo", active=1))
# /hey?user=boo&active=1

print(get_url("/hey"))
# /hey

edited Dec 18, 2018 at 8:52

answered Dec 17, 2018 at 10:07

Graipher

41.7k7 gold badges70 silver badges134 bronze badges

\$\begingroup\$ Due to the implementation of the rest of the code, I need to do it everything without any requests module, just improving the code I posted using strings. \$\endgroup\$

Avión
– Avión

2018-12-17 10:09:18 +00:00
Commented Dec 17, 2018 at 10:09
2

\$\begingroup\$ @Avión: Just did. It captures all keyword arguments you pass to the function into one dictionary. \$\endgroup\$

Graipher
– Graipher

2018-12-17 10:13:13 +00:00
Commented Dec 17, 2018 at 10:13
4

\$\begingroup\$ Your code is good for illustrative purposes but it fails to URLencode the parameters and is therefore a potential security risk. \$\endgroup\$

Konrad Rudolph
– Konrad Rudolph

2018-12-17 14:35:32 +00:00
Commented Dec 17, 2018 at 14:35
1

\$\begingroup\$ @KonradRudolph Added a short disclaimer regarding that. \$\endgroup\$

Graipher
– Graipher

2018-12-17 14:38:32 +00:00
Commented Dec 17, 2018 at 14:38
1

\$\begingroup\$ It's not just that it's a security risk, it's also that if you have & in one of the values, it will fail to send the correct value (and most likely fail in general, unless there's also another = in the values). \$\endgroup\$

ChatterOne
– ChatterOne

2018-12-18 08:33:18 +00:00
Commented Dec 18, 2018 at 8:33

| Show 4 more comments

301_Moved_Permanently · Accepted Answer · 2018-12-17 10:12:13Z

You're pretty much reinventing urllib.parse.urlencode:

from urllib.parse import urlencode


def prepare_query_string(**kwargs):
    return urlencode([(key, value) for key, value in kwargs.items() if value is not None])

Usage being:

>>> prepare_query_string(active=1)
'active=1'
>>> prepare_query_string(active=1, user=None)
'active=1'
>>> prepare_query_string(active=1, user='bob')
'active=1&user=bob'
>>> prepare_query_string(file='foo.tar.gz', user='bob')
'file=foo.tar.gz&user=bob'
>>> prepare_query_string(file='foo.tar.gz', user='bob', active=None)
'file=foo.tar.gz&user=bob'
>>> prepare_query_string(file='foo.tar.gz', user='bob', active=1)
'file=foo.tar.gz&user=bob&active=1'

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