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Converting %25C3%25BC in string to Umlaut python [duplicate]

In Python 2.7, given a URL like example.com?title=%D0%BF%D1%80%D0%B0%D0%B2%D0%BE%D0%B2%D0%B0%D1%8F+%D0%B7%D0%B0%D1%89%D0%B8%D1%82%D0%B0, how can I decode it to the expected result, example.com?title==правовая+защита?
I tried url=urllib.unquote(url.encode("utf8")), but it seems to give a wrong result.

The data is UTF-8 encoded bytes escaped with URL quoting, so you want to decode, with urllib.parse.unquote(), which handles decoding from percent-encoded data to UTF-8 bytes and then to text, transparently:
from urllib.parse import unquote
url = unquote(url)
Demo:
>>> from urllib.parse import unquote
>>> url = 'example.com?title=%D0%BF%D1%80%D0%B0%D0%B2%D0%BE%D0%B2%D0%B0%D1%8F+%D0%B7%D0%B0%D1%89%D0%B8%D1%82%D0%B0'
>>> unquote(url)
'example.com?title=правовая+защита'
The Python 2 equivalent is urllib.unquote(), but this returns a bytestring, so you'd have to decode manually:
from urllib import unquote
url = unquote(url).decode('utf8')

If you are using Python 3, you can use urllib.parse.unquote:
url = """example.com?title=%D0%BF%D1%80%D0%B0%D0%B2%D0%BE%D0%B2%D0%B0%D1%8F+%D0%B7%D0%B0%D1%89%D0%B8%D1%82%D0%B0"""
import urllib.parse
urllib.parse.unquote(url)
gives:
'example.com?title=правовая+защита'

You can achieve an expected result with requests library as well:
import requests
url = "http://www.mywebsite.org/Data%20Set.zip"
print(f"Before: {url}")
print(f"After: {requests.utils.unquote(url)}")
Output:
$ python3 test_url_unquote.py
Before: http://www.mywebsite.org/Data%20Set.zip
After: http://www.mywebsite.org/Data Set.zip
Might be handy if you are already using requests, without using another library for this job.

In HTML the URLs can contain html entities.
This replaces them, too.
#from urllib import unquote #earlier python version
from urllib.request import unquote
from html import unescape
unescape(unquote('https://v.w.xy/p1/p22?userId=xxxxxxxx-xxxx-xxxx-xxxx-xxxxxxxxxxxx&confirmationToken=7uAf%2fxJoxRTFAZdxslCn2uwVR9vV7cYrlHs%2fl9sU%2frix9f9CnVx8uUT%2bu8y1%2fWCs99INKDnfA2ayhGP1ZD0z%2bodXjK9xL5I4gjKR2xp7p8Sckvb04mddf%2fiG75QYiRevgqdMnvd9N5VZp2ksBc83lDg7%2fgxqIwktteSI9RA3Ux9VIiNxx%2fZLe9dZSHxRq9AA'))

I know this is an old question, but I stumbled upon this via Google search and found that no one has proposed a solution with only built-in features.
So I quickly wrote my own.
Basically a url string can only contain these characters: A-Z, a-z, 0-9, -, ., _, ~, :, /, ?, #, [, ], #, !, $, &, ', (, ), *, +, ,, ;, %, and =, everything else are url encoded.
URL encoding is pretty straight forward, just a percent sign followed by the hexadecimal digits of the byte values corresponding to the codepoints of illegal characters.
So basically using a simple while loop to iterate the characters, add any character's byte as is if it is not a percent sign, increment index by one, else add the byte following the percent sign and increment index by three, accumulate the bytes and decoding them should work perfectly.
Here is the code:
def url_parse(url):
l = len(url)
data = bytearray()
i = 0
while i < l:
if url[i] != '%':
d = ord(url[i])
i += 1
else:
d = int(url[i+1:i+3], 16)
i += 3
data.append(d)
return data.decode('utf8')
I have tested it and it works perfectly.

double url encoding for ascii charachters

What is the used library from python, or how in general the double URL encoding of characters can be done,
example :
character 'a' with URL encoding > %61
character 'a' with double URL encoding > %2561
how I can get the %2561 from 'a'?

Depends on your usecase. If you don't need a robust solution you can just use urllib
import urllib.parse as parser
encoded_string = "%2561"
decoded_string = parser.unquote(encoded_string) # this would be '%61'
double_decoded_string = parser.unquote(decoded_string) # this would be 'a'
You can also remove the decoded_string variable and instead call it like
double_encoded_string = parser.unquote(parser.unquote(encoded_string))
From there you could start with looping the decode while it contains encoded_chars etc.

How to read text file into pandas and convert utf8 chars to string / latin? [duplicate]

I have a list containing URLs with escaped characters in them. Those characters have been set by urllib2.urlopen when it recovers the html page:
http://www.sample1webpage.com/index.php?title=%E9%A6%96%E9%A1%B5&action=edit
http://www.sample1webpage.com/index.php?title=%E9%A6%96%E9%A1%B5&action=history
http://www.sample1webpage.com/index.php?title=%E9%A6%96%E9%A1%B5&variant=zh
Is there a way to transform them back to their unescaped form in python?
P.S.: The URLs are encoded in utf-8

Using urllib package (import urllib) :
Python 2.7
From official documentation :
urllib.unquote(string)
Replace %xx escapes by their single-character equivalent.
Example: unquote('/%7Econnolly/') yields '/~connolly/'.
Python 3
From official documentation :
urllib.parse.unquote(string, encoding='utf-8', errors='replace')
[…]
Example: unquote('/El%20Ni%C3%B1o/') yields '/El Niño/'.

And if you are using Python3 you could use:
import urllib.parse
urllib.parse.unquote(url)

or urllib.unquote_plus
>>> import urllib
>>> urllib.unquote('erythrocyte+membrane+protein+1%2C+PfEMP1+%28VAR%29')
'erythrocyte+membrane+protein+1,+PfEMP1+(VAR)'
>>> urllib.unquote_plus('erythrocyte+membrane+protein+1%2C+PfEMP1+%28VAR%29')
'erythrocyte membrane protein 1, PfEMP1 (VAR)'

You can use urllib.unquote

import re
def unquote(url):
return re.compile('%([0-9a-fA-F]{2})',re.M).sub(lambda m: chr(int(m.group(1),16)), url)

Unicode strings to byte strings without the addition of backslashes

I'm learning python by doing the python challenge using python3.3 and I'm on question eight. There's a comment in the markup providing you with two bz2-compressed unicode strings outputting byte strings, one for username and one for password. There's also a link where you need the decompressed credentials to enter.
One way to easily solve this is just to manually copy the strings and assign it to two variables as byte strings and then just use the bz2 library to decompress it:
>>>un=b'BZh91AY&SYA\xaf\x82\r\x00\x00\x01\x01\x80\x02\xc0\x02\x00 \x00!\x9ah3M\x07<]\xc9\x14\xe1BA\x06\xbe\x084'
>>>print(bz2.decompress(un).decode('utf-8'))
huge
But that's not for me since I want the answer by just running my python file.
My code like this:
>>>import bz2, re, requests
>>>url = requests.get('http://www.pythonchallenge.com/pc/def/integrity.html')
>>>un = re.findall(r'un: \'(.*)\'',url.text)[0]
>>>correct=b'BZh91AY&SYA\xaf\x82\r\x00\x00\x01\x01\x80\x02\xc0\x02\x00 \x00!\x9ah3M\x07<]\xc9\x14\xe1BA\x06\xbe\x084'
>>>print(un,un is correct,sep='\n')
b'BZh91AY&SYA\\xaf\\x82\\r\\x00\\x00\\x01\\x01\\x80\\x02\\xc0\\x02\\x00 \\x00!\\x9ah3M\\x07<]\\xc9\\x14\\xe1BA\\x06\\xbe\\x084'
False
The problem is that when it converts from unicode string to byte string the escaping backslash gets added so that it cannot be read by bz2 module. I have tried everything I know and what got up when I searched.
How do I get it from unicode to byte so that it doesn't get changed?

Here it is a solution:
import urllib
import bz2
import re
def decode(line):
out = re.search(r"\'(.*?)\'",''.join(line)).group()
out = eval("b%s" % out)
return bz2.decompress(out)
#read lines that contain the encoded message
page = urllib.urlopen('http://www.pythonchallenge.com/pc/def/integrity.html').readlines()[20:22]
print "Click on the bee and insert: "
User_Name = decode(page[0])
print "User Name is: " + User_Name
Password = decode(page[1])
print "Password is: " + Password

The backslashes are present in the HTML source, so it's not surprising that the requests module preserves them. I don't have requests installed on my Python 3 environment, so I haven't been able to exactly replicate your situation, but it looks to me like if you start capturing the surrounding ' characters, you can use ast.literal_eval to parse the character sequence into a bytes array:
>>> test
"'BZh91AY&SYA\\xaf\\x82\\r\\x00\\x00\\x01\\x01\\x80\\x02\\xc0\\x02\\x00 \\x00!\\x9ah3M\\x07<]\\xc9\\x14\\xe1BA\\x06\\xbe\\x084'"
>>> import ast
>>> res = ast.literal_eval("b%s" % test)
>>> import bz2
>>> len(bz2.decompress(res))
4
There are probably other ways, but why not use Python's built in knowledge that the byte sequence b'\\xaf' can be parsed into a bytes array?

Unescape Python Strings From HTTP

I've got a string from an HTTP header, but it's been escaped.. what function can I use to unescape it?
myemail%40gmail.com -> myemail#gmail.com
Would urllib.unquote() be the way to go?

I am pretty sure that urllib's unquote is the common way of doing this.
>>> import urllib
>>> urllib.unquote("myemail%40gmail.com")
'myemail#gmail.com'
There's also unquote_plus:
Like unquote(), but also replaces plus signs by spaces, as required for unquoting HTML form values.

In Python 3, these functions are urllib.parse.unquote and urllib.parse.unquote_plus.
The latter is used for example for query strings in the HTTP URLs, where the space characters () are traditionally encoded as plus character (+), and the + is percent-encoded to %2B.
In addition to these there is the unquote_to_bytes that converts the given encoded string to bytes, which can be used when the encoding is not known or the encoded data is binary data. However there is no unquote_plus_to_bytes, if you need it, you can do:
def unquote_plus_to_bytes(s):
if isinstance(s, bytes):
s = s.replace(b'+', b' ')
else:
s = s.replace('+', ' ')
return unquote_to_bytes(s)
More information on whether to use unquote or unquote_plus is available at URL encoding the space character: + or %20.

Yes, it appears that urllib.unquote() accomplishes that task. (I tested it against your example on codepad.)