Python XML 解析、lxml、urllib.request

Question

我在尝试解析从 url 检索到的 XML 文件时遇到了一点困难，我的目标是将此 xml 文件转换为结构良好的对象，以便轻松检索其数据。我当前的代码导致以下错误:

>>> tree = etree.parse(data)
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:72421)
  File "parser.pxi", line 1791, in lxml.etree._parseDocument (src/lxml/lxml.etree.c:105883)
  File "parser.pxi", line 1817, in lxml.etree._parseDocumentFromURL (src/lxml/lxml.etree.c:106182)
  File "parser.pxi", line 1721, in lxml.etree._parseDocFromFile (src/lxml/lxml.etree.c:105181)
  File "parser.pxi", line 1122, in lxml.etree._BaseParser._parseDocFromFile (src/lxml/lxml.etree.c:100131)
  File "parser.pxi", line 580, in lxml.etree._ParserContext._handleParseResultDoc (src/lxml/lxml.etree.c:94254)
  File "parser.pxi", line 690, in lxml.etree._handleParseResult (src/lxml/lxml.etree.c:95690)
  File "parser.pxi", line 618, in lxml.etree._raiseParseError (src/lxml/lxml.etree.c:94722)
OSError: Error reading file '<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
    xmlns:content="http://purl.org/rss/1.0/modules/content/"
    xmlns:wfw="http://wellformedweb.org/CommentAPI/"
    xmlns:dc="http://purl.org/dc/elements/1.1/"
    xmlns:atom="http://www.w3.org/2005/Atom"

代码:

(scraper) gmf:scr gmf$ python3
Python 3.4.2 (default, Jan  2 2015, 20:14:16) 
[GCC 4.2.1 Compatible Apple LLVM 6.0 (clang-600.0.54)] on darwin
Type "help", "copyright", "credits" or "license" for more information.

>>> import urllib.request
>>> from lxml import etree
>>>
>>> opener = urllib.request.build_opener()
>>> f = opener.open('https://nordfront.se/feed')
data = f.read()
f.close()
>>> tree = etree.parse(data)

非常感谢你的帮助

Answer 1

根据文档字符串(参见 help(ET.parse))，ET.parse 需要第一个参数成为

• 文件名/路径

  import lxml.etree as ET    
  tree = ET.parse(filename)
  

• 文件对象

  with open('data.xml') as f:
      tree = ET.parse(f)
  

• 类文件对象

  import io
  tree = ET.parse(io.BytesIO(data))
  

• 使用 HTTP 或 FTP 协议(protocol)的 URL

  import urllib.request
  opener = urllib.request.build_opener()
  tree = ET.parse(opener.open(url))
  

这是最后一个选项，它将 opener.open(url) 直接传递给 ET.parse 而不是定义 data = f.read() 可能是您想要使用的选项。

或者，当您已经在字符串 data 中拥有 XML 时，您可以使用 ET.fromstring:

root = ET.fromstring(data)

但是请注意，parse 返回一个 ElementTree，而 fromstring 返回一个 Element。