I need to fetch data from a URL with non-ascii characters but urllib2.urlopen refuses to open the resource and raises:
UnicodeEncodeError: 'ascii' codec can't encode character u'\u0131' in position 26: ordinal not in range(128)
I know the URL is not standards compliant but I have no chance to change it.
What is the way to access a resource pointed by a URL containing non-ascii characters using Python?
edit: In other words, can / how urlopen open a URL like:
http://example.org/Ñöñ-ÅŞÇİİ/
Strictly speaking URIs can't contain non-ASCII characters; what you have there is an IRI.
To convert an IRI to a plain ASCII URI:
non-ASCII characters in the hostname part of the address have to be encoded using the Punycode-based IDNA algorithm;
non-ASCII characters in the path, and most of the other parts of the address have to be encoded using UTF-8 and %-encoding, as per Ignacio's answer.
So:
import re, urlparse
def urlEncodeNonAscii(b):
return re.sub('[\x80-\xFF]', lambda c: '%%%02x' % ord(c.group(0)), b)
def iriToUri(iri):
parts= urlparse.urlparse(iri)
return urlparse.urlunparse(
part.encode('idna') if parti==1 else urlEncodeNonAscii(part.encode('utf-8'))
for parti, part in enumerate(parts)
)
>>> iriToUri(u'http://www.a\u0131b.com/a\u0131b')
'http://www.xn--ab-hpa.com/a%c4%b1b'
(Technically this still isn't quite good enough in the general case because urlparse doesn't split away any user:pass@ prefix or :port suffix on the hostname. Only the hostname part should be IDNA encoded. It's easier to encode using normal urllib.quote and .encode('idna') at the time you're constructing a URL than to have to pull an IRI apart.)
import urllib.parse instead of urlparse, decode b in urlEncodeNonAscii: b.decode('utf-8') and leave the idna part out of the iriToUri: return urllib.parse.urlunparse([url_encode_non_ascii(part.encode('utf-8')) for part in parts])
In python3, use the urllib.parse.quote function on the non-ascii string:
>>> from urllib.request import urlopen
>>> from urllib.parse import quote
>>> chinese_wikipedia = 'http://zh.wikipedia.org/wiki/Wikipedia:' + quote('首页')
>>> urlopen(chinese_wikipedia)
Python 3 has libraries to handle this situation. Use
urllib.parse.urlsplit to split the URL into its components, and
urllib.parse.quote to properly quote/escape the unicode characters
and urllib.parse.urlunsplit to join it back together.
>>> import urllib.parse
>>> url = 'http://example.com/unicodè'
>>> url = urllib.parse.urlsplit(url)
>>> url = list(url)
>>> url[2] = urllib.parse.quote(url[2])
>>> url = urllib.parse.urlunsplit(url)
>>> print(url)
http://example.com/unicod%C3%A8
Based on @darkfeline answer:
from urllib.parse import urlsplit, urlunsplit, quote
def iri2uri(iri):
"""
Convert an IRI to a URI (Python 3).
"""
uri = ''
if isinstance(iri, str):
(scheme, netloc, path, query, fragment) = urlsplit(iri)
scheme = quote(scheme)
netloc = netloc.encode('idna').decode('utf-8')
path = quote(path)
query = quote(query)
fragment = quote(fragment)
uri = urlunsplit((scheme, netloc, path, query, fragment))
return uri
It is more complex than the accepted @bobince's answer suggests:
This is how all browsers work; it is specified in https://url.spec.whatwg.org/ - see this example. A Python implementation can be found in w3lib (this is the library Scrapy is using); see w3lib.url.safe_url_string:
from w3lib.url import safe_url_string
url = safe_url_string(u'http://example.org/Ñöñ-ÅŞÇİİ/', encoding="<page encoding>")
An easy way to check if a URL escaping implementation is incorrect/incomplete is to check if it provides 'page encoding' argument or not.
For those not depending strictly on urllib, one practical alternative is requests, which handles IRIs "out of the box".
For example, with http://bücher.ch:
>>> import requests
>>> r = requests.get(u'http://b\u00DCcher.ch')
>>> r.status_code
200
urllib.request.urlopen(url).read() The code after the change to requests: requests.get(url).contentEncode the unicode to UTF-8, then URL-encode.
unicode(url, 'utf-8') raises TypeError: decoding Unicode is not supported. also which function do you suggest for encoding url? urlencode for example is for building query string. but mine is only a path on the server.
url.encode('utf-8') (assuming url is a unicode object).Use iri2uri method of httplib2. It makes the same thing as by bobin (is he/she the author of that?)
urllib2.urlopen(httplib2.iri2uri("http://домены.рф"), timeout=15) returns urlopen error [Errno -2] Name or service not knownAnother option to convert an IRI to an ASCII URI is to use furl package:
gruns/furl: 🌐 URL parsing and manipulation made easy. - https://github.com/gruns/furl
Python's standard urllib and urlparse modules provide a number of URL related functions, but using these functions to perform common URL operations proves tedious. Furl makes parsing and manipulating URLs easy.
http://国立極地研究所.jp/english/ (Japanese National Institute of Polar Research website)
import furl
url = 'http://国立極地研究所.jp/english/'
furl.furl(url).tostr()
'http://xn--vcsoey76a2hh0vtuid5qa.jp/english/'
https://ja.wikipedia.org/wiki/日本語 ("Japanese" article in Wikipedia)
import furl
url = 'https://ja.wikipedia.org/wiki/日本語'
furl.furl(url).tostr()
'https://ja.wikipedia.org/wiki/%E6%97%A5%E6%9C%AC%E8%AA%9E'
I could not avoid from this strange characters, but at the end I come through it.
import urllib.request
import os
url = "http://www.fourtourismblog.it/le-nuove-tendenze-del-marketing-tenere-docchio/"
with urllib.request.urlopen(url) as file:
html = file.read()
with open("marketingturismo.html", "w", encoding='utf-8') as file:
file.write(str(html.decode('utf-8')))
os.system("marketingturismo.html")
