如何在Python中解析嵌套标记的XML

Question

我有以下XML。在

<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>

我试图用python解析这个xml并创建一个URL array，它将包含： 1xml中存在的所有http URL 2对于链接选项卡（如果有youtube），则捕获并准备youtube url并将其添加到URL array。在

我有以下代码，但它没有给我的网址和链接。在

^{pr2}$

我怎样才能得到所有的网址？在

根据以下答案制定了以下代码： 下面的问题是，它只打印1个url而不是全部。在

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

Answer 1

看看xmltodict。在

>>> 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"
        }
    }
}

Answer 2

您需要的信息在<link>下，所以只需重复这些信息。使用get()获取youtube id，find()获取子对象{}。在

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