CODEWITHSUNDEEP
regexpython-3.x
Georges Kayembe
Georges Kayembe

Reputation: 121

Python Regex solution?

I have a string:

line = "https://dbwebb.se/kunskap/uml#sequence, ftp://bth.com:32/files/im.jpeg, file://localhost:8585/zipit, http://v2-dbwebb.se/do%hack"

I want to get a result like this : 

[('https', 'dbwebb.se', ''), ('ftp', 'bth.com', '32'), ('file', 'localhost', '8585'), ('http', 'v2-dbwebb.se', '')]

I tried like :

match = re.findall("([fh]t*ps?|file):[\\/]*(.*?)(:\d+|(?=[\\\/]))", line)

And than i got :

[["https", "dbwebb.se", ""], ["ftp", "bth.com", ":32"], ["file", "localhost", ":8585"], ["http", "v2-dbwebb.se", ""]]

There is one diffrence, you can se ":32" and ":8585". How can i do to get just "32" and "8585" and not the stupid ":" Thanx!

Upvotes: 0

Views: 76

Answers (3)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626802

I suggest

import re
line = line = "https://dbwebb.se/kunskap/uml#sequence, ftp://bth.com:32/files/im.jpeg, file://localhost:8585/zipit, http://v2-dbwebb.se/do%hack"
match = re.findall(r"([fh]t*ps?|file)://([^/]*?)(?::(\d+))?(?:/|$)", line)
print(match)

See the Python demo

The main point is (?::(\d+))?(?:/|$ part where : and 1+ digits part is optional ((?...)? matches 1 or 0 times) and (?:/|$) matches a / or end of string.

Details

  • ([fh]t*ps?|file) - Group 1 (the first item in the tuple): a literal
    • [fh]t*ps? - f or h, zero or more t, p and 1 or 0 ss
    • | - or
    • file - file substring
  • :// - a literal substring
  • ([^/]*?) - Group 2 (the second item in the tuple): any 0 or more chars other than /
  • (?::(\d+))? - an optional sequence of:
    • : - a colon
    • (\d+) - Group 2 (the third item in the tuple): one or more digits
  • (?:/|$) - a : or end of string.

Upvotes: 1

Jon Clements
Jon Clements

Reputation: 142146

Instead of a regex, why not split on the , and then use Python's urllib.parse.urlparse, eg:

from urllib.parse import urlparse

line = "https://dbwebb.se/kunskap/uml#sequence, ftp://bth.com:32/files/im.jpeg, file://localhost:8585/zipit, http://v2-dbwebb.se/do%hack"
output = [urlparse(url) for url in line.split(', ')]

Gives you:

[ParseResult(scheme='https', netloc='dbwebb.se', path='/kunskap/uml', params='', query='', fragment='sequence'),
 ParseResult(scheme='ftp', netloc='bth.com:32', path='/files/im.jpeg', params='', query='', fragment=''),
 ParseResult(scheme='file', netloc='localhost:8585', path='/zipit', params='', query='', fragment=''),
 ParseResult(scheme='http', netloc='v2-dbwebb.se', path='/do%hack', params='', query='', fragment='')]

Then filter out the elements you want:

wanted = [(url.scheme, url.hostname, url.port or '') for url in output]

Which gives you:

[('https', 'dbwebb.se', ''),
 ('ftp', 'bth.com', 32),
 ('file', 'localhost', 8585),
 ('http', 'v2-dbwebb.se', '')]

Upvotes: 1

Casimir et Hippolyte
Casimir et Hippolyte

Reputation: 89557

Regex isn't the good tool to parse urls, there's a dedicated library to do this complicated task urllib:

from urllib.parse import urlparse

line = "https://dbwebb.se/kunskap/uml#sequence, ftp://bth.com:32/files/im.jpeg, file://localhost:8585/zipit, http://v2-dbwebb.se/do%hack"

result = []
for i in line.split(', '):
    o = urlparse(i)
    result.append([o.scheme, o.hostname, o.port])

Upvotes: 1

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