在python中更改网址

Question

how can I change the activeOffset in this url? I am using Python and a while loop

https://www.dieversicherer.de/versicherer/auto---reise/typklassenabfrage#activeOffset=10&orderBy=kh&orderDirection=ASC

It first should be 10, then 20, then 30 ...

I tried urlparse but I don't understand how to just increase the number

Thanks!

Answer 1

If this is a fixed URL, you can write activeOffset={} in the URL then use format to replace {} with specific numbers:

url = "https://www.dieversicherer.de/versicherer/auto---reise/typklassenabfrage#activeOffset={}&orderBy=kh&orderDirection=ASC"

for offset in range(10,100,10):
  print(url.format(offset))

If you cannot modify the URL (because you get it as an input from some other part of your program), you can use regular expressions to replace occurrences of activeOffset=... with the required number ( reference ):

import re

url = "https://www.dieversicherer.de/versicherer/auto---reise/typklassenabfrage#activeOffset=10&orderBy=kh&orderDirection=ASC"

query = "activeOffset="
pattern = re.compile(query + "\\d+") # \\d+ means any sequence of digits

for offset in range(10,100,10):
  # Replace occurrences of pattern with the modified query
  print(pattern.sub(query + str(offset), url))

If you want to use urlparse , you can apply the previous approach to the fragment part returned by urlparse :

import re

from urllib.parse import urlparse, urlunparse

url = "https://www.dieversicherer.de/versicherer/auto---reise/typklassenabfrage#activeOffset=10&orderBy=kh&orderDirection=ASC"

query = "activeOffset="
pattern = re.compile(query + "\\d+") # \\d+ means any sequence of digits

parts = urlparse(url)

for offset in range(10,100,10):
  fragment_modified = pattern.sub(query + str(offset), parts.fragment)
  parts_modified = parts._replace(fragment = fragment_modified)
  url_modified = urlunparse(parts_modified)
  print(url_modified)