I want to get the information of a torrent file in a desired format through python

Question

I am writing code to parse tracker information in torrent file using python.

import bencoder
import sys

target = './'+sys.argv[1]

with open(target, 'rb') as torrent_file:
    torrent = bencoder.decode(torrent_file.read())

i=0
while True:
    try:
        print(torrent[b'announce-list'][i])
        i+=1
    except:
        break
    

The output is as follows.

[b'udp://tracker.openbittorrent.com:80/announce']

[b'udp://tracker.opentrackr.org:1337/announce']

I want to parse the value in the form below.

["tracker.openbittorrent.com", 80]

["tracker.opentrackr.org", 1337]

How should I parse it?

Answer 1

You might use urllib.parse.urlparse for this as follows

from urllib.parse import urlparse
url1 = b'udp://tracker.openbittorrent.com:80/announce'
url2 = b'udp://tracker.opentrackr.org:1337/announce'
c1 = urlparse(url1)
c2 = urlparse(url2)
hostport1 = c1.netloc.rsplit(b':',1)
hostport2 = c2.netloc.rsplit(b':',2)
hostport1[0] = hostport1[0].decode()
hostport1[1] = int(hostport1[1])
hostport2[0] = hostport2[0].decode()
hostport2[1] = int(hostport2[1])
print(hostport1)
print(hostport2)

output

['tracker.openbittorrent.com', 80]
['tracker.opentrackr.org', 1337]

Explanation: I extract netloc, then split at most once at first from right b':', then apply .decode to host port to convert bytes into str and int to convert bytes into int.

EDIT: After more careful reading, I noticed that you might access .hostname and .port which allow much more concise code to do that task, that is

from urllib.parse import urlparse
url1 = b'udp://tracker.openbittorrent.com:80/announce'
url2 = b'udp://tracker.opentrackr.org:1337/announce'
c1 = urlparse(url1)
c2 = urlparse(url2)
hostport1 = [c1.hostname.decode(), c1.port]
hostport2 = [c2.hostname.decode(), c2.port]
print(hostport1)
print(hostport2)

gives same output as code above.