Python feedparser返回第一个条目中第一个媒体项的URL

Question

I'm working with python for the first time and I am a bit stuck.

Using feedparser to parse a RSS feed, I want to get the URL of the first media item of entry 0 and load it into a variable.

The below seems to work, but I have to hit enter twice to run it and it returns the URLs for ALL media items in the entry 0, where I only want the first (16x9) image URL.

>>> 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>

Looks like this when run in python:

>>> 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

Quick answer:

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

d.entries[n].media_content is a list full of dicts, so you can just get the first item in that list and store the value at "url" in a variable.

Here's how it looks in the Python shell:

>>> 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