Displaying view function in Django

Question

I am trying to display the result of my view function on a Django webpage but only get one line with no hyperlink.

code:

from django.http import HttpResponse
import feedparser

def index(content):
    YahooContent = feedparser.parse ("http://news.yahoo.com/rss/")
    for feed in YahooContent.entries:
            content = (feed.title + ": " + "\n"+feed.link + "\n" + feed.published + "\n")
            return HttpResponse(content)

Result on the webpage:

Man arrested in death of missing Ariz. girl: http://news.yahoo.com/arizona-girls-home-searched-body-found-154919366.html Thu, 04 Sep 2014 14:05:16 -0400

alecxe · Accepted Answer · 2014-09-04 18:53:49Z

4

You need to collect feeds in a list and only then, after the loop, return an HttpResponse instance:

content = []
for feed in YahooContent.entries:
    content.append(feed.title + ": " + "\n" + feed.link + "\n" + feed.published)

return HttpResponse('\n'.join(content))

Another option, that would follow Django philosophies about separation of concerns, would be to create and render a template and pass the data into the template context:

create a template, let's say index.html with the following content

<table>
    <tr>
        <th>Title</th>
        <th>Link</th>
        <th>Published</th>
    </tr>
    {% for entry in entries %}
        <tr>
            <td>{{ entry.title }}</td>
            <td>{{ entry.link }}</td>
            <td>{{ entry.published }}</td>
        </tr>
    {% endfor %}
</table>

put the template into templates directory of your app or project

render it in the view using, for example, render_to_response()

from django.shortcuts import render_to_response
import feedparser

def index(content):
    entries = feedparser.parse ("http://news.yahoo.com/rss/").entries
    return render_to_response('index.html', {'entries': entries})

edited Sep 4, 2014 at 18:53

answered Sep 4, 2014 at 18:42

alecxe

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2 Comments

Amir Over a year ago

that worked to display all content but still displays all as a single string. I want to be able to separate the news and also display the hyperlinks.

alecxe Over a year ago

@Amir okay, I've provided you with an alternative, better approach.

Vikas Mishra · Accepted Answer · 2014-09-04 18:47:08Z

0

your 'return' statement is inside the for loop due to which it is returning after the first iteration itself and hence giving just one feed instead of all, to be able to return all feeds you need to build a list of all the feeds and then return that.

answered Sep 4, 2014 at 18:47

Vikas Mishra

651 silver badge7 bronze badges

2 Comments

alecxe Over a year ago

I don't see the point to duplicate the answer after 5 minutes. Also, that would not really help without the actual code.

Vikas Mishra Over a year ago

I apologize but I dint see your answer before as had opened this question and then went to read my mails and then typed the answer.

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