Python feedparser返回第一個條目中第一個媒體項目的URL

Question

我正在第一次使用python，並且我有點卡住了。

使用feedparser解析一個RSS提要，我想進入0的第一媒體項目的URL，並將其加載到一個變量。下面

似乎工作，但我打兩次ENTER運行它，它返回的URL在入口0，在這裏我只希望第一（16：9）圖像URL的所有媒體項目。

>>> import feedparser 
>>> d = feedparser.parse(http://www.abc.net.au/news/feed/45910/rss) 
>>> for content in d.entries[0].media_content: print content['url'] 

-link to where i got the code above

RSS XML：

  <media:group> 
     <media:description>French fighter jets take off to drop bombs on the Islamic State stronghold of Raqqa in Syria. (Supplied)</media:description> 
     <media:content url="http://www.abc.net.au/news/image/6943630-16x9-2150x1210.jpg" medium="image" type="image/jpeg" width="2150" height="1210"/> 
      <media:content url="http://www.abc.net.au/news/image/6943630-4x3-940x705.jpg" medium="image" type="image/jpeg" width="940" height="705"/> 
      <media:content url="http://www.abc.net.au/news/image/6943630-3x2-940x627.jpg" medium="image" type="image/jpeg" width="940" height="627" isDefault="true"/> 
      <media:content url="http://www.abc.net.au/news/image/6943630-3x4-940x1253.jpg" medium="image" type="image/jpeg" width="940" height="1253"/> 
      <media:content url="http://www.abc.net.au/news/image/6943630-1x1-1400x1400.jpg" medium="image" type="image/jpeg" width="1400" height="1400"/> 
      <media:thumbnail url="http://www.abc.net.au/news/image/6943630-4x3-140x105.jpg" width="140" height="105"/> 
     </media:group> 

看起來這在Python運行時：

>>> for content in d.entries[0].media_content: print content['url'] 
... 
http://www.abc.net.au/news/image/6943630-16x9-2150x1210.jpg 
http://www.abc.net.au/news/image/6943630-4x3-940x705.jpg 
http://www.abc.net.au/news/image/6943630-3x2-940x627.jpg 
http://www.abc.net.au/news/image/6943630-3x4-940x1253.jpg 
http://www.abc.net.au/news/image/6943630-1x1-1400x1400.jpg 
>>> 

Answer 1

快速回答：

url = d.entries[0].media_content[0]['url'] 

d.entries[n].media_content是一個完整的類型的字典列表，這樣你可以得到的第一個項目在該列表中，並在「URL」的值存儲在一個變量。

這裏是它的外觀在Python外殼：

>>> import feedparser 
>>> d = feedparser.parse("http://www.abc.net.au/news/feed/45910/rss") 
>>> url = d.entries[0].media_content[0]['url'] 
>>> print url 
http://www.abc.net.au/news/image/6943798-16x9-2150x1210.jpg