阅读RSS feed并在Django模板中显示它| feedparser [英] Reading RSS feed and displaying it in Django Template | feedparser

问题描述

请参阅此博客： http://johnsmallman.wordpress.com/author/johnsmallman/feed /

我想为我的应用程序获取RSS feed。上面的博客是wordpress博客。

I want to fetch the RSS feed for my application. The above blog is a wordpress blog.

我正在使用 feedparser

import feedparser
feeds = feedparser.parse('http://johnsmallman.wordpress.com/author/johnsmallman/feed/')

现在 feeds ['feed'] ['title'] 输出 u Johnsmallman的博客\xbb John Smallman

我的问题是我在应用程序中的显示方式如何。可以说这个博客包含100篇文章。所以我想循环访问并获取所有数据。

My question is How exactly i present this in my app. Lets say this blog contains 100s of articles. So i want to loop over and fetch all data.

没有任何直接的方法吗？有任何预定义的库或方法吗？

Isn't there any direct way of doing this? Any pre-defined library or method?

我经常在Google上搜索但遇到了困难。

I have ofcouse googled but had hard time.

我基本上是希望将其呈现给Django模板。

I am basically looking to render it to Django Template. So would really be looking something towards it.

需要指导人员：）

推荐答案

如果将 Feed 添加到模板上下文中，则应该能够在模板中对其进行循环：

If you add feeds to your template context, you should be able to loop through it in your template:

<ul>
{% for entry in feeds.entries %}
    <li><a href="{{entry.link}}">{{entry.title}}</a></li>

{% endfor %}
</ul>

这篇关于阅读RSS feed并在Django模板中显示它| feedparser的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！