我需要将网页转换为
XML(使用
Python 3.4.3).如果我将URL的内容写入文件,那么我可以完美地阅读和解析它,但如果我尝试直接从网页上读取,我的终端中会出现以下错误:
File “./AnimeXML.py”, line 22, in
xml = ElementTree.parse (xmlData)
File “/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/xml/etree/ElementTree.py”, line 1187, in parse
tree.parse(source, parser)
File “/Library/Frameworks/Python.framework/Versions/3.4/lib/python3.4/xml/etree/ElementTree.py”, line 587, in parse
source = open(source, “rb”)
OSError: [Errno 36] File name too long:
我的python代码:
# AnimeXML.py
#! /usr/bin/Python
# Import xml parser.
import xml.etree.ElementTree as ElementTree
# XML to parse.
sampleUrl = "http://cdn.animenewsnetwork.com/encyclopedia/api.xml?anime=16989"
# Read the xml as a file.
content = urlopen (sampleUrl)
# XML content is stored here to start working on it.
xmlData = content.readall().decode('utf-8')
# Close the file.
content.close()
# Start parsing XML.
xml = ElementTree.parse (xmlData)
# Get root of the XML file.
root = xml.getroot()
for info in root.iter("info"):
print (info.attrib)
有什么方法可以修复我的代码,以便我可以直接将网页读入python而不会出现此错误吗?
最佳答案 正如ElementTree文档的
Parsing XML部分所述:
We can import this data by reading from a file:
import xml.etree.ElementTree as ET
tree = ET.parse('country_data.xml')
root = tree.getroot()
Or directly from a string:
root = ET.fromstring(country_data_as_string)
您将整个XML内容作为巨型路径名传递.您的XML文件可能大于2K,或者您的平台的最大路径名大小,因此错误.如果不是,那么你只会得到一个不同的错误,即没有名为[所有内容都在你的XML文件中的第一个/所有内容].
只需使用fromstring
而不是解析.
或者,请注意parse
可以获取文件对象,而不仅仅是文件名.而urlopen
返回的东西是一个文件对象.
另请注意该部分的下一行:
fromstring()
parses XML from a string directly into anElement
, which is the root element of the parsed tree. Other parsing functions may create anElementTree
.
所以,你不希望root = tree.getroot().
所以:
# ...
content.close()
root = ElementTree.fromstring(xmlData)