4

我正在尝试使用 Python 解析 XML 文件以从 XML 提要中获取标题、作者、URL 和摘要。然后我确保我们收集数据的 XML 是这样的：

<?xml version="1.0" encoding="utf-8"?>
<feed xmlns="http://www.w3.org/2005/Atom"
  xmlns:grddl="http://www.w3.org/2003/g/data-view#"
  grddl:transformation="2turtle_xslt-1.0.xsl">

<title>Our Site RSS</title>
<link href="http://www.oursite.com" />
<updated>2013-08-14T20:05:08-04:00</updated>
<id>urn:uuid:c60d7202-9a58-46a6-9fca-f804s879f5ebc</id>
<rights>
    Original content available for non-commercial use under a Creative
    Commons license (Attribution-NonCommercial-NoDerivs 3.0 Unported),
    except where noted.
</rights>

<entry>
    <title>Headline #1</title>
    <author>
        <name>John Smith</name>
    </author>
    <link rel="alternate"
          href="http://www.oursite.com/our-slug/" />
    <id>1234</id>
    <updated>2013-08-13T23:45:43-04:00</updated>

    <summary type="html">
        Here is a summary of our story
    </summary>
</entry>
<entry>
    <title>Headline #2</title>
    <author>
        <name>John Smith</name>
    </author>
    <link rel="alternate"
          href="http://www.oursite.com/our-slug-2/" />
    <id>1235</id>
    <updated>2013-08-13T23:45:43-04:00</updated>

    <summary type="html">
        Here is a summary of our second story
    </summary>
</entry>

我的代码是：

import xml.etree.ElementTree as ET
tree = ET.parse('data.xml')
root = tree.getroot()

for child in root:
    print child.tag

当 Python 打印 child.tag 时，标签不是“条目”，而是“{ http://www.w3.org/2005/Atom }entry”。我曾尝试使用：

for entry in root.findall('entry'):

但这不起作用，因为 entry 标记包含作为根标记一部分的 w3 url。此外，获取 root 的孙辈将他们的标签显示为“{ http://www.w3.org/2005/Atom }author”

我无法更改 XML，但是如何修改它（将根设置为 ）并重新保存它或更改我的代码以便 root.findall('entry') 工作？

4

2 回答 2

5

This is standard ElementTree behavior. If the tags you're searching for are declared within a namespace, you have to specify that namespace when you search for those tags. However, you can do something like this:

import xml.etree.ElementTree as ET
tree = ET.parse('data.xml')
root = tree.getroot()

def prepend_ns(s):
    return '{http://www.w3.org/2005/Atom}' + s

for entry in root.findall(prepend_ns('entry')):
    print 'Entry:'
    print '    Title: '   + entry.find(prepend_ns('title')).text
    print '    Author: '  + entry.find(prepend_ns('author')).find(prepend_ns('name')).text
    print '    URL: '     + entry.find(prepend_ns('link')).attrib['href']
    print '    Summary: ' + entry.find(prepend_ns('summary')).text
于 2013-08-15T01:19:02.663 回答
1

试试 BeautifulSoup4，它不仅可以解析 XML，还可以解析 HTML 等。这是一个 to-go 代码，希望对您有所帮助。

from bs4 import BeautifulSoup

def main():
    input = """....""" 
    soup = BeautifulSoup(input)   
    for entry in soup.findAll("entry"):
        title = entry.find("title").text.strip()
        author = entry.find("author").text.strip()
        link  = entry.find("link").text.strip()
        summary = entry.find("summary").text.strip()
        print title, author, link, summary
if __name__ == '__main__':
    main()
于 2013-08-15T01:28:27.570 回答