Retrieving part of URL in python [duplicate] - python

This question already has answers here:
Python: Get URL path sections
(7 answers)
Closed 5 years ago.
I'm on Python 3.6.
How I can extract part of URL so I can use it as variable.
For example, the URL is http://example.com/comp/project.sec and I would like to get the project part without .sec as variable.
How can I achieve that?


url = "http://example.com/comp/project.sec"
project = url.split("/")[-1].split(".")[0]


Here's an alternative, which removes split requirement.
from os.path import splitext, basename
splitext(basename('http://example.com/comp/project.sec'))[0]


import urllib
urllib.parse.urlparse("http://example.com/comp/project.sec").path.split("/")[-1].split(".")[-1]

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How to get an absolute file path in Python
(11 answers)
Closed 2 years ago.
Let's say I have this path
D:\something\something1\from_here_I_now\stuff\stuff2.
So, I know that \from_here_I_now\stuff\stuff2 is a permanent path, but the beginning is different, like I know that D:\something\something1\ may be different for someone else. How can I find the D:\something\something1\ knowing only \from_here_I_now\stuff\stuff2?

Try something like this:
import os
filestr = '\from_here_I_now\stuff\stuff2'
fullstr = os.path.abspath(filestr)
print(fullstr)
>>> 'D:\something\something1\from_here_I_now\stuff\stuff2'
print(fullstr[:len(filestr)])
>>> 'D:\something\something1'

Hex to plain ASCII? Python 2.7.13 [duplicate]

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Decode escaped characters in URL
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Closed 2 years ago.
Im trying to turn this:
%73%6c%61%70%72%69%73%65%40%6c%69%65%6e%6d%75%6c%74%69%6d%65%64%69%61%2e%63%6f%6d
into this:
slaprise#lienmultimedia.com
and my brain is exploding..
Any help would be appreciated.
Thank you

Python 2.7.17 (should work for Python 2.7.13)
import urllib2
url = urllib2.unquote("%73%6c%61%70%72%69%73%65%40%6c%69%65%6e%6d%75%6c%74%69%6d%65%64%69%61%2e%63%6f%6d")
print(url)
# slaprise#lienmultimedia.com

You are trying to URL-decode that string, use urllib:
from urllib import unquote
url = unquote("%73%6c%61%70%72%69%73%65%40%6c%69%65%6e%6d%75%6c%74%69%6d%65%64%69%61%2e%63%6f%6d")
# url = slaprise#lienmultimedia.com

regex getting url link [duplicate]

This question already has answers here:
Why can't Python parse this JSON data? [closed]
(3 answers)
Closed 5 years ago.
I am trying to extract the link I am getting from a curl command. Curl command throws back of type string.
{"success":true,"key":"Syv77d","link":"https://file.io/Syv77d","expiry":"14 days"}
In my below code this gets https://file.io/Syv77d","expiry":"14 days"}
link = re.search('https://.*$',fileIO)
What I wanted was just https://file.io/Syv77d
The link would vary so i would need the url without the double-qoutes. I think I am missing something in my regex.

Convert the string object to a JSON object.
Ex:
import json
jData = json.loads('{"success":true,"key":"Syv77d","link":"https://file.io/Syv77d","expiry":"14 days"}')
jData["link"]

Python: search for html tag starting with characters [duplicate]

This question already has an answer here:
Python BeautifulSoup: wildcard attribute/id search
(1 answer)
Closed 5 years ago.
I am using the following code to find a tag with class "product-category".
soup0.findAll("a",{"class":"product-category"})
How do i write if I need to find a tag whose class starts with product-category.
I tried using
soup0.findAll("a",{"class":%"product-category"}) ##This doesnt work.
Is there any way?
Thanks,

From the documentation, you can do something like so:
import re
tagsStartingWithB = soup.findAll(re.compile('^b'))
[tag.name for tag in tagsStartingWithB]
# [u'body', u'b', u'b']
See more here.

Not receiving the parameters containing '?' properly. The part after '?' is getting trimmed. How do I avoid it? [duplicate]

This question already has answers here:
How to get URL of current page, including parameters, in a template?
(7 answers)
Closed 7 years ago.
I am writing a Django application. In my 'urls.py' I have written a URL pattern like this:
url(r'^rest/post/(.*)/$', rest_post),
Now when I am passing some URL like:
http://www.google.com/a?b
In my rest_post view I am getting only: http://www.google.com/a
I want to get the full URL. How do I do it?

simply
request.get_full_path()
in your views.
see https://docs.djangoproject.com/en/dev/ref/request-response/#django.http.HttpRequest.get_full_path

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