如何撤销django feed url？

I've been searching for hours to try and figure this out, and it seems like no one has ever put an example online - I've just created a Django 1.2 rss feed view object and attached it to a url. When I visit the url, everything works great, so I know my implementation of the feed class is OK.

我一直在寻找几个小时试图弄明白这一点,似乎没有人在网上举例 - 我刚刚创建了一个Django 1.2 rss feed视图对象并将其附加到一个url。当我访问网址时,一切都很好,所以我知道我的Feed类的实现是可以的。

The hitch is, I can't figure out how to link to the url in my template. I could just hard code it, but I would much rather use {% url %}

问题是,我无法弄清楚如何链接到我的模板中的网址。我可以硬编码,但我宁愿使用{%url%}

I've tried passing the full path like so:

我试过像这样传递完整路径:

{% url app_name.lib.feeds.LatestPosts blog_name=name %}

And I get nothing. I've been searching and it seems like everyone else has a solution so obvious it's not worth posting online. Have I just been up too long?

我一无所获。我一直在寻找,似乎其他人都有一个解决方案,显而易见,它不值得在网上发布。我刚刚起得太久了吗?

Here is the relevent url pattern:

这是相关的url模式:

from app.lib.feeds import LatestPosts

urlpatterns = patterns('app.blog.views',
    (r'^rss/(?P<blog_name>[A-Za-z0-9]+)/$', LatestPosts()),
    #snip...
)

Thanks for your help.

谢谢你的帮助。

1 个解决方案

#1

You can name your url pattern, which requires the use of the url helper function:

您可以命名您的网址格式,这需要使用网址辅助功能:

from django.conf.urls.defaults import url, patterns

urlpatterns = patterns('app.blog.views',
    url(r'^rss/(?P<blog_name>[A-Za-z0-9]+)/$', LatestPosts(), name='latest-posts'),
    #snip...
)

Then, you can simply use {% url latest-posts blog_name="myblog" %} in your template.

然后,您只需在模板中使用{%url latest-posts blog_name =“myblog”%}即可。

#1