如何在Python中解析嵌套标记的XML

我有以下XML。如何在Python中解析嵌套标记的XML

<component name="QUESTIONS"> 
    <topic name="Chair"> 
     <state>active</state> 
     <subtopic name="Wooden"> 
      <links> 
       <link videoDuration="" youtubeId="" type="article"> 
        <label>Understanding Wooden Chair</label> 
        <url>http://abcd.xyz.com/1111?view=app</url> 
       </link> 
       <link videoDuration="" youtubeId="" type="article"> 
        <label>How To Assemble Wooden CHair</label> 
        <url>http://abcd.xyz.com/2222?view=app</url> 
       </link> 
       <link videoDuration="11:35" youtubeId="Qasefrt09_2" type="video"> 
        <label>Wooden Chair Tutorial</label> 
        <url>/</url> 
       </link> 
       <link videoDuration="1:06" youtubeId="MSDVN235879" type="video"> 
        <label>How To Access Wood</label> 
        <url>/</url> 
       </link> 
      </links> 
     </subtopic> 
    </topic> 
    <topic name="Table"> 
     <state>active</state> 
     <subtopic name=""> 
      <links> 
       <link videoDuration="" youtubeId="" type="article"> 
        <label>Understanding Tables</label> 
        <url>http://abcd.xyz.com/3333?view=app</url> 
       </link> 
       <link videoDuration="" youtubeId="" type="article"> 
        <label>Set-up Table</label> 
        <url>http://abcd.xyz.com/4444?view=app</url> 
       </link> 
       <link videoDuration="" youtubeId="" type="article"> 
        <label>How To Change table</label> 
        <url>http://abcd.xyz.com/5555?view=app</url> 
       </link> 
      </links> 
     </subtopic> 
    </topic> 
</component>

我试图解析这个XML Python和创建URL array其中将包含： 1.所有存在于XML 2.对于链接选项卡中的HTTP URL如果YouTube的存在，然后捕获和准备youtube网址并将其添加到URL array。

我有以下代码，但它没有给我的网址和链接。

from xml.etree import ElementTree 

with open('faq.xml', 'rt') as f: 
    tree = ElementTree.parse(f) 

for node in tree.iter(): 
    print node.tag, node.attrib.get('url') 

for node in tree.iter('outline'): 
    name = node.attrib.get('link') 
    url = node.attrib.get('url') 
    if name and url: 
     print ' %s :: %s' % (name, url) 
    else: 
     print name

我该如何做到这一点，以获得所有的网址？

根据下面的答案开发了以下代码：以下问题是，它只打印1个url并非全部。

from xml.etree import ElementTree 

def fetch_faq_urls(): 
    url_list = [] 
    with open('faq.xml', 'rt') as f: 
     tree = ElementTree.parse(f) 

    for link in tree.iter('link'): 
     youtube = link.get('youtubeId') 
     if youtube: 
      print "https://www.youtube.com/watch?v=" + youtube 
      video_url = "https://www.youtube.com/watch?v=" + youtube 
      url_list.append(video_url) 
      # print "youtubeId", link.find('label').text, '???' 
     else: 
      print link.find('url').text 
      article_url = link.find('url').text 
      url_list.append(article_url) 
      # print 'url', link.find('label').text, 
     return url_list 

faqs = fetch_faq_urls() 
print faqs

来源

2016-09-29 in_learning_software

你想要的信息是在<link>所以只是遍历这些。使用get()可以获取YouTube的id和find()以获取子对象<url>。

from xml.etree import ElementTree 

with open('faq.xml', 'rt') as f: 
    tree = ElementTree.parse(f) 

for link in tree.iter('link'): 
    youtube = link.get('youtubeId') 
    if youtube: 
     print "youtube", link.find('label').text, '???' 
    else: 
     print 'url', link.find('label').text, link.find('url').text

来源

2016-09-29 03:28:03 tdelaney

非常感谢。我明白了我应该做什么。谢谢tdelaney –

用开发的代码更新了我的问题？我不明白为什么它只向数组推送1个值？ –

@in_learning_software - 这只是一个小缩进问题。请注意，您的'return url_list'位于'for'块中，因此它会在循环的第一遍时执行。简单地向下一个更高层次倾斜。 – tdelaney

看看xmltodict。

>>> print(json.dumps(xmltodict.parse(""" 
... <mydocument has="an attribute"> 
... <and> 
...  <many>elements</many> 
...  <many>more elements</many> 
... </and> 
... <plus a="complex"> 
...  element as well 
... </plus> 
... </mydocument> 
... """), indent=4)) 
{ 
    "mydocument": { 
     "@has": "an attribute", 
     "and": { 
      "many": [ 
       "elements", 
       "more elements" 
      ] 
     }, 
     "plus": { 
      "@a": "complex", 
      "#text": "element as well" 
     } 
    } 
}

来源

2016-09-29 05:39:14 ge7600

如何在Python中解析嵌套标记的XML

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