在Django中使用feedparser时，“QuerySet”对象没有属性“url” [英] 'QuerySet' object has no attribute 'url' when using feedparser in Django

问题描述

这是从这里的问题的后续 Django / feedparser中的bozo_exception

This is follow up to the question from here bozo_exception in Django / feedparser

我想通过模型/数据库中的许多Feed进行迭代，并将其中的每一个都显示在html模板中。虽然我明白我需要在html模板中迭代思考x.feed.entries，我认为通过每个rss源的迭代需要在视图函数中正确执行？

I would like to iterate through many feeds from models/DB and have each of them displayed in the html template. While I do understand that I need to iterate thought x.feed.entries in the html template, I assume that iteration through each rss source needs to happen in the view function correct?

def feed5(request):
    source = Feed.objects.all()
    for item in source.url:
        rss = feedparser.parse(item)
    context = {'rss': rss,}
    return render(request, 'feedreader/feed5.html', context)

给我这个错误：'QuerySet'对象没有属性'url'。不确定我该怎么办？

gives me this error: 'QuerySet' object has no attribute 'url'. Not sure how should I go about it?

谢谢

推荐答案

那么它实际上并不是 - Python不是对你说谎。请参阅 source 是一个 QuerySet ，一个类似列表的结果结果，而不是一个结果。如果是您的 Feed 模型，应该具有一个url属性，然后查找它，而不是查询集：

Well, it actually doesn't - Python isn't lying to you. See, source is a QuerySet, a list-like structure of results, not a single result. If it's your Feed model that should have a url attribute, then look it up on it and not the query set:

for item in source:
    rss = feedparser.parse(item.url)

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