CODEWITHSUNDEEP
pythonregexparsing
TonyM
TonyM

Reputation: 728

Parsing hostname and port from string or url

I can be given a string in any of these formats:

I would like to extract the host and if present a port. If the port value is not present I would like it to default to 80.

I have tried urlparse, which works fine for the url, but not for the other format. When I use urlparse on hostname:port for example, it puts the hostname in the scheme rather than netloc.

I would be happy with a solution that uses urlparse and a regex, or a single regex that could handle both formats.

Upvotes: 32

Views: 52480

Answers (5)

Ishaan
Ishaan

Reputation: 1340

Method using urllib - 

    from urllib.parse import urlparse
    url = 'https://stackoverflow.com/questions'
    print(urlparse(url))

Output -

ParseResult(scheme='https', netloc='stackoverflow.com', path='/questions', params='', query='', fragment='')

Reference - https://www.tutorialspoint.com/urllib-parse-parse-urls-into-components-in-python

Upvotes: 5

dfostic
dfostic

Reputation: 1736

>>> from urlparse import urlparse   
>>> aaa = urlparse('http://www.acme.com:456')

>>> aaa.hostname  
'www.acme.com'

>>> aaa.port   
456
>>>

Upvotes: 20

Maksym Kozlenko
Maksym Kozlenko

Reputation: 10363

You can use urlparse to get hostname from URL string:

from urlparse import urlparse
print urlparse("http://www.website.com/abc/xyz.html").hostname # prints www.website.com

Upvotes: 57

claesv
claesv

Reputation: 2113

I'm not that familiar with urlparse, but using regex you'd do something like:

p = '(?:http.*://)?(?P<host>[^:/ ]+).?(?P<port>[0-9]*).*'

m = re.search(p,'http://www.abc.com:123/test')
m.group('host') # 'www.abc.com'
m.group('port') # '123'

Or, without port:

m = re.search(p,'http://www.abc.com/test')
m.group('host') # 'www.abc.com'
m.group('port') # '' i.e. you'll have to treat this as '80'

EDIT: fixed regex to also match 'www.abc.com 123'

Upvotes: 8

ntziolis
ntziolis

Reputation: 10221

The reason it fails for:

www.acme.com 456

is because it is not a valid URI. Why don't you just:

  1. Replace the space with a :
  2. Parse the resulting string by using the standard urlparse method

Try and make use of default functionality as much as possible, especially when it comes to things like parsing well know formats like URI's.

Upvotes: 6

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