如何处理 urllib2.urlopen 中的 ®？

Question

我收到了一个网址：https://www.packtpub.com/virtualization-and-cloud/citrix-xenapp®-75-desktop-virtualization-solutions；它来自 BeautifulSoup。

url=u'https://www.packtpub.com/virtualization-and-cloud/citrix-xenapp\xae-75-desktop-virtualization-solutions'

我想再次反馈到 urllib2.urlopen。

import urllib2
source = urllib2.urlopen(url).read()

我得到的错误：

UnicodeEncodeError: 'gbk' codec can't encode character u'\xae' in position 43: illegal multibyte sequence

因此，我尝试了：

source = urllib2.urlopen(url.encode("utf-8")).read()

它有页面源，但它与原始网址不同。

originalUrl = 'https://www.packtpub.com/virtualization-and-cloud/citrix-xenapp®-75-desktop-virtualization-solutions'
originalSource = urllib2.urlopen(originalUrl).read()
originalSource == source

结果为假。有什么想法可以修复此网址吗？如何将 u'\xae' 转换成原来的®？

Answer 1

URL 必须是有效的字节串，并且非 ASCII 码点编码正确。您需要编码为 UTF-8，然后 url 引用您的 URL 路径：

import urllib
import urllib2
import urlparse

originalUrl = u'https://www.packtpub.com/virtualization-and-cloud/citrix-xenapp\xae-75-desktop-virtualization-solutions'
parsed_link = urlparse.urlsplit(originalUrl.encode('utf8'))
parsed_link = parsed_link._replace(path=urllib.quote(parsed_link.path))
encoded_link = parsed_link.geturl()
source = urllib2.urlopen(encoded_link).read()

演示：

>>> import urllib
>>> import urllib2 
>>> import urlparse
>>> originalUrl = u'https://www.packtpub.com/virtualization-and-cloud/citrix-xenapp\xae-75-desktop-virtualization-solutions'
>>> parsed_link = urlparse.urlsplit(originalUrl.encode('utf8'))
>>> parsed_link = parsed_link._replace(path=urllib.quote(parsed_link.path))
>>> encoded_link = parsed_link.geturl()
>>> encoded_link
'https://www.packtpub.com/virtualization-and-cloud/citrix-xenapp%C2%AE-75-desktop-virtualization-solutions'
>>> source = urllib2.urlopen(encoded_link).read()
>>> len(source)
68758