How to Continuously Iterate Through a Linked List in Python Using urllib

[Arman777]

import urllib.request

import urllib.request
request_url = urllib.request.urlopen(
    'http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=12345')
print(request_url.read())

I have a code somthing like this which prints

"and the next nothing is 44827"

Now is there a way to get that number and put it into the url ? becasue in that case I ll obtain another number and I need to also put that in that url ?

 

[Greg Bernhardt]

I would use the Requests library instead. urllib is a little outdated.

There are many ways to handle variables in strings (look into f strings), but in a pinch you can always use concatenation.

idnum = 12345

request_url = urllib.request.urlopen(
    'http://www.pythonchallenge.com/pc/def/linkedlist.php?nothing=' + idnum)

 

[pbuk]

What a strange website: it purports to teach Python but it seems that any language that can make requests can be used, and the site itself is written in PHP!

Given that solving the puzzles requires making repeated requests I strongly echo @Greg Bernhardt's recommendation of Requests - it makes things like passing query parameters much easier.

import requests
url = 'http://www.pythonchallenge.com/pc/def/linkedlist.php'
vars = { 'idnum': 12345 }
req = requests.get(url, params = vars)

# http://www.pythonchallenge.com/pc/def/linkedlist.php?idnum=12345.
print(req.url)
# The text body of the response.
print(req.text)

 

[pbuk]

What a strange website: it purports to teach Python but it seems that any language that can make requests can be used, and the site itself is written in PHP!

... and it's 15 years old which explains why the model answers use urllib and no doubt other things that are now deprecated or broken. I'd stay away from the model answers and the wiki.