如何处理urllib2.urlopen的in？

Question

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

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

我想再回到urllib2. urllib2.urlopen。

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

我所犯的错误：

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

因此，我尝试：

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

它有页面源，但是它不同于原始的url。

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

结果是假的。有什么办法修复这个网址吗？如何将u'\xae‘转换为原始®

Answer 1

URL必须是有效的字节串，非ASCII码点必须正确编码.您需要将代码编码到UTF-8，然后url引用URL的路径：

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