[PYTHON] Add query to url parsed url and return to original url

Overview

Assuming that there is a URL like this, https://api.twitter.com/1.1/statuses/user_timeline.json&screen_name=drwtsn64 After ʻurllib.parse.urlparse () How to addcount = 200` to the query part and return to the original URL.

verification code

import urllib.parse

url = 'https://api.twitter.com/1.1/statuses/user_timeline.json?screen_name=drwtsn64'
parsed = urllib.parse.urlparse(url)

print(parsed)

query = urllib.parse.parse_qs(parsed.query, True)

print(query)

query['count'] = 200

print(query)

encoded = urllib.parse.urlencode(query, True)

print(encoded)

reversed = urllib.parse.ParseResult(parsed.scheme, parsed.netloc, parsed.path, parsed.params, encoded, parsed.fragment).geturl()

print(reversed)

output

ParseResult(scheme='https', netloc='api.twitter.com', path='/1.1/statuses/user_timeline.json', params='', query='screen_name=drwtsn64', fragment='')
{'screen_name': ['drwtsn64']}
{'screen_name': ['drwtsn64'], 'count': 200}
screen_name=drwtsn64&count=200
https://api.twitter.com/1.1/statuses/user_timeline.json?screen_name=drwtsn64&count=200

important point

Since the query passed to ʻurlencode ()` is a dictionary, the order of the query characters is not guaranteed.