python - 如何使用 feedparser 解析 ""？

Question

rss文件如下所示，我想获取media:group部分的内容。我查看了feedparser的文档，但似乎没有提到这一点。怎么做？任何帮助表示赞赏。

<?xml version="1.0" encoding="UTF-8"?>
<rss xmlns:ymusic="http://music.yahoo.com/rss/1.0/ymusic/" xmlns:media="http://search.yahoo.com/mrss/" xmlns:content="http://purl.org/rss/1.0/modules/content/" xmlns:cf="http://www.microsoft.com/schemas/rss/core/2005" xmlns:dc="http://purl.org/dc/elements/1.1/" version="2.0"><channel>
        <title>XYZ InfoX:  Special hello  </title>
        <link>http://www1.XYZInfoX.com/learninghello/home</link>
        <description>hello</description>
        <language>en</language>         <copyright />
        <pubDate>Wed, 17 Mar 2010 08:50:06 GMT</pubDate>
        <dc:creator />
        <dc:date>2010-03-17T08:50:06Z</dc:date>
        <dc:language>en</dc:language> <dc:rights />
        <image>
            <title>Voice of America</title>
            <link>http://www1.XYZInfoX.com/learninghello</link>
            <url>http://media.XYZInfoX.com/designimages/XYZRSSIcon.gif</url>
        </image>

        <item>
                <title>Who Were the Deadliest Gunmen of the Wild West?</title>
                <link>http://www1.XYZInfoX.com/learninghello/home/Deadliest-Gunmen-of-the-Wild-West-87826807.html</link>
                <description> The story of two of them: "Killin'" Jim Miller was an outlaw, "Texas" John Slaughter was a lawman | EXPLORATIONS  </description>
                <pubDate>Wed, 17 Mar 2010 00:38:48 GMT</pubDate>
                <guid isPermaLink="false">87826807</guid>
                <dc:creator></dc:creator>
                <dc:date>2010-03-17T00:38:48Z</dc:date>                                                                                                                                     
                <media:group>
                    <media:content url="http://media.XYZInfoX.com/images/archives_peace_comm_480_16mar_se.jpg" medium="image" isDefault="true" height="300" width="480" />
                    <media:content url="http://media.XYZInfoX.com/images/archives_peace_comm_230_16mar_se_edited-1.jpg" medium="image" isDefault="false" height="230" width="230" />
                    <media:content url="http://media.XYZInfoX.com/images/tex_trans_lawmans_230_16mar10_se.jpg" medium="image" isDefault="false" height="230" width="230" />
                    <media:content url="http://www.XYZInfoX.com/MediaAssets2/learninghello/dalet/se-exp-outlaws-part2-17mar2010.Mp3" type="audio/mpeg" medium="audio" isDefault="false" />
                </media:group>
     </item>

Answer 1

PyPi 提供的 feedparser 4.1 有这个错误。

我的解决方案是从存储库中获取最新的 feedparser.py(4.2 pre)。

svn checkout http://feedparser.googlecode.com/svn/trunk/ feedparser-readonly
cd feedparser-readonly
python setup.py install

现在您可以访问所有 mrss 项目

>>> import feedparser  # the new version!
>>> d = feedparser.parse(MY_XML_URL)
>>> for content in d.entries[0].media_content: print content['url']

应该为你做这份工作