CODEWITHSUNDEEP
pythonregexurlpython-2.7
Sangamesh Hs
Sangamesh Hs

Reputation: 1447

Extract a part of URL - python

I have an URL for example:

http://name.abc.wxyz:1234/Assts/asset.epx?id=F3F94D94-7232-4FA2-98EF-07sdfssfdsa3B5

From this Url I want to extract only 'asset.epx?id=F3F94D94-7232-4FA2-98EF-07sdfssfdsa3B5' how could i do that?

I am still learning regular expressions and I am not able to solve the above. Any suggestions would be appreciated.

Upvotes: 4

Views: 20601

Answers (3)

Blender
Blender

Reputation: 298532

In this specific example splitting the string is enough:

url.split('/')[-1]

If you have a more complex URL I would recommend the yarl library for parsing it:

>>> import yarl  # pip install yarl
>>> url = yarl.URL('http://name.abc.wxyz:1234/Assts/asset.epx?id=F3F94D94-7232-4FA2-98EF-07sdfssfdsa3B5')
>>> url.path_qs
'/Assts/asset.epx?id=F3F94D94-7232-4FA2-98EF-07sdfssfdsa3B5'

You could also use the builtin urllib.parse library but I find that it gets in the way once you start doing complex things like:

>>> url.update_query(asd='foo').with_fragment('asd/foo/bar')
URL('http://name.abc.wxyz:1234/Assts/asset.epx?id=F3F94D94-7232-4FA2-98EF-07sdfssfdsa3B5&asd=foo#asd/foo/bar')

Upvotes: 10

TerryA
TerryA

Reputation: 60024

You can use urlparse assuming asset.epx is the same:

>>> import urlparse
>>> url = 'http://name.abc.wxyz:1234/Assts/asset.epx?id=F3F94D94-7232-4FA2-98EF-07sdfssfdsa3B5'
>>> res = urlparse.urlparse(url)
>>> print 'asset.epx?'+res.query
asset.epx?id=F3F94D94-7232-4FA2-98EF-07sdfssfdsa3B5

This is useful if you ever need other information from the url (You can print res to check out the other info you can get ;))

If you're using Python 3 though, you'll have to do from urllib.parse import urlparse.

Upvotes: 10

Brett Lempereur
Brett Lempereur

Reputation: 815

Depending on the version of Python, you want either urlparse in Python 2.x (http://docs.python.org/2/library/urlparse.html) or urllib.parse in Python 3.x (http://docs.python.org/2/library/urlparse.html). In Python 3 (all I have available), the following snippet achieves what you need without resorting to regular expressions:

import urllib.parse

address = "http://name.abc.wxyz:1234/Assts/asset.epx?id=F3F94D94-7232-4FA2-98EF-07sdfssfdsa3B5"
parsed = urllib.parse.urlsplit(address)
print("{}?{}".format(parsed.path.split("/")[-1], parsed.query)

The output is "asset.epx?id=F3F94D94-7232-4FA2-98EF-07sdfssfdsa3B5" here.

Upvotes: 3

Related Questions