python - 为什么 Python 的 feedparser 看不到 feed 中的所有元素

Question

我使用以下代码:

import feedparser as fp

if __name__ == '__main__':
    url = 'http://www.careerbuilder.de/RTQ/rss20.aspx?rssid=RSS_PD&num=25&geoip=false&ddcompany=false&ddtitle=false&cat=JN038'  
    d = fp.parse(url)
    for entry in d.entries:
        print entry
        print '----------------------'

结果我得到:

{'guidislink': 0, 'published': u'Wed, 23 Apr 2014 04:00:00 Z', 'published_parsed': time.struct_time(tm_year=2014, tm_mon=4, tm_mday=23, tm_hour=4, tm_min=0, tm_sec=0, tm_wday=2, tm_yday=113, tm_isdst=0), 'title': u'Bankkaufmann (m/w)'}
----------------------
{'guidislink': 0, 'published': u'Wed, 23 Apr 2014 04:00:00 Z', 'published_parsed': time.struct_time(tm_year=2014, tm_mon=4, tm_mday=23, tm_hour=4, tm_min=0, tm_sec=0, tm_wday=2, tm_yday=113, tm_isdst=0), 'title': u'Anlagenbuchhalter (m/w)'}
----------------------
{'guidislink': 0, 'published': u'Wed, 23 Apr 2014 04:00:00 Z', 'published_parsed': time.struct_time(tm_year=2014, tm_mon=4, tm_mday=23, tm_hour=4, tm_min=0, tm_sec=0, tm_wday=2, tm_yday=113, tm_isdst=0), 'title': u'Bankkaufleute (m/w)'}
----------------------

提要中的条目似乎没有“摘要”和“链接”元素。作为确认，如果我尝试使用 entry.summary 或 entry.description，我会收到一条错误消息。这对我来说很奇怪，因为如果我在浏览器中打开 feed，我确实会在 xml 中看到 link 和 description 元素。

有人知道我做错了什么吗？

Answer 1

来自revision history of feedparser :

Universal Feed Parser 3.0b18 was released on February 17, 2004.

• always map description to summary_detail (Andrei)
• use libxml2 (if available)

来自 here

Some RSS feeds use guid when they mean link. guid can also be used as an opaque identifier that has nothing to do with links. If an RSS feed uses guid as the entry link and no link is present, Universal Feed Parser detects this and makes the guid available in d.entries[i].link.

也许这就是为什么我可以访问 entry.link 和 entry.description 而没有任何错误，尽管 print entry.keys() 给出

['summary_detail', 'published_parsed', 'links', 'title', 'summary', 'guidislink', 'title_detail', 'link', 'published', 'id']