我可以在Python 3上提供指向lxml.etree.parse的URL吗? [英] Can I supply a URL to lxml.etree.parse on Python 3?

问题描述

文档说我可以:

lxml可以从本地文件，HTTP URL或FTP URL进行解析.它也是 自动检测并读取gzip压缩的XML文件(.gz).

(来自解析器"下的 http://lxml.de/parsing.html )

但是快速实验似乎暗示了其他情况:

Python 3.4.1 (v3.4.1:c0e311e010fc, May 18 2014, 10:45:13) [MSC v.1600 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> from lxml import etree
>>> parser = etree.HTMLParser()
>>> from urllib.request import urlopen
>>> with urlopen('https://pypi.python.org/simple') as f:
...   tree = etree.parse(f, parser)
...
>>> tree2 = etree.parse('https://pypi.python.org/simple', parser)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "lxml.etree.pyx", line 3299, in lxml.etree.parse (src\lxml\lxml.etree.c:72655)
  File "parser.pxi", line 1791, in lxml.etree._parseDocument (src\lxml\lxml.etree.c:106263)
  File "parser.pxi", line 1817, in lxml.etree._parseDocumentFromURL (src\lxml\lxml.etree.c:106564)
  File "parser.pxi", line 1721, in lxml.etree._parseDocFromFile (src\lxml\lxml.etree.c:105561)
  File "parser.pxi", line 1122, in lxml.etree._BaseParser._parseDocFromFile (src\lxml\lxml.etree.c:100456)
  File "parser.pxi", line 580, in lxml.etree._ParserContext._handleParseResultDoc (src\lxml\lxml.etree.c:94543)
  File "parser.pxi", line 690, in lxml.etree._handleParseResult (src\lxml\lxml.etree.c:96003)
  File "parser.pxi", line 618, in lxml.etree._raiseParseError (src\lxml\lxml.etree.c:95015)
OSError: Error reading file 'https://pypi.python.org/simple': failed to load external entity "https://pypi.python.org/simple"
>>>

我可以使用urlopen方法，但是文档似乎暗示传递URL会更好.另外，如果文档不准确，尤其是当我开始需要做更复杂的事情时，我会担心依赖lxml.

从已知URL用lxml解析HTML的正确方法是什么?我应该在哪里查看该文件的记录?

更新:如果我使用http URL而不是https URL，则会出现相同的错误.

解决方案

问题是lxml不支持HTTPS网址，并且http://lxml.de/parsing.html under "Parsers")

but a quick experiment seems to imply otherwise:

Python 3.4.1 (v3.4.1:c0e311e010fc, May 18 2014, 10:45:13) [MSC v.1600 64 bit (AMD64)] on win32
Type "help", "copyright", "credits" or "license" for more information.
>>> from lxml import etree
>>> parser = etree.HTMLParser()
>>> from urllib.request import urlopen
>>> with urlopen('https://pypi.python.org/simple') as f:
...   tree = etree.parse(f, parser)
...
>>> tree2 = etree.parse('https://pypi.python.org/simple', parser)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "lxml.etree.pyx", line 3299, in lxml.etree.parse (src\lxml\lxml.etree.c:72655)
  File "parser.pxi", line 1791, in lxml.etree._parseDocument (src\lxml\lxml.etree.c:106263)
  File "parser.pxi", line 1817, in lxml.etree._parseDocumentFromURL (src\lxml\lxml.etree.c:106564)
  File "parser.pxi", line 1721, in lxml.etree._parseDocFromFile (src\lxml\lxml.etree.c:105561)
  File "parser.pxi", line 1122, in lxml.etree._BaseParser._parseDocFromFile (src\lxml\lxml.etree.c:100456)
  File "parser.pxi", line 580, in lxml.etree._ParserContext._handleParseResultDoc (src\lxml\lxml.etree.c:94543)
  File "parser.pxi", line 690, in lxml.etree._handleParseResult (src\lxml\lxml.etree.c:96003)
  File "parser.pxi", line 618, in lxml.etree._raiseParseError (src\lxml\lxml.etree.c:95015)
OSError: Error reading file 'https://pypi.python.org/simple': failed to load external entity "https://pypi.python.org/simple"
>>>

I can use the urlopen method, but the documentation seems to imply that passing a URL is somehow better. Also, I'm a bit concerned about relying on lxml if the documentation is inaccurate, particularly if I start needing to do anything more complex.

What is the correct way to parse HTML with lxml, from a known URL? And where should I be looking to see that documented?

Update: I get the same error if I use a http URL rather than a https one.

解决方案

The issue is that lxml does not support HTTPS urls, and http://pypi.python.org/simple redirects to a HTTPS version.

So for any secure website, you need to read the URL yourself:

from lxml import etree
from urllib.request import urlopen

parser = etree.HTMLParser()

with urlopen('https://pypi.python.org/simple') as f:
    tree = etree.parse(f, parser)

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