简体   繁体   English

从feedparser媒体获取URL:内容

[英]Retrieving url from feedparser media:content

The following code works fine: 以下代码可以正常工作:

if 'media_content' in entry:
                mediaContent=entry.media_content
            else:
                mediaContent='No Media'

But this gives me an error: 但这给我一个错误:

 if 'media_content' in entry:
                mediaContent=entry.media_content['url']
            else:
                mediaContent='No Media'

This is the error: 这是错误:

list indices must be integers, not str

I've seen multiple uses of this syntax when searching on the web to pull out the url from the media_content element of feedparser, but it isn't working for me. 在网络上搜索从feedparser的media_content元素中提取URL时,我已经看到了这种语法的多种用法,但是它对我不起作用。 I'm new to python, so I'm sure it is just that I'm not accessing the dictionary properly. 我是python的新手,所以我确定这只是我没有正确访问字典。 I just want the raw url as a string. 我只想要原始网址作为字符串。

Thanks! 谢谢!

entry.media_content is a list of entries, each a dictionary: entry.media_content是条目列表 ,每个词典:

for item in entry.media_content:
    print item['url']

You could just grab the first one listed: 您可以仅获取列出的第一个:

if 'media_content' in entry:
    mediaContent = entry.media_content[0]['url']

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM