我正在制作一个博客应用程序,我想创建一个显示特定用户的所有博客的视图。为此,我需要将用户实例传递给我的看法是如何将用户实例传递给Django中的视图
def blogs(request,author=None,slug=None):
# If the author context has been passed then display the blogs of that author
if author:
# Displays a particular blog
if slug:
this_blog = get_object_or_404(Entry, creator = author, slug = slug)
return render_to_response('blog/blog_view.html', {'blog': this_blog,},context_instance=RequestContext(request))
# Displays all the blogs of a particular user
else:
blog_list = Entry.objects.filter(creator = author)
return render_to_response('blog/all_blogs_user.html', {'Blogs': blog_list},context_instance=RequestContext(request))
虽然语法,这是正确的,但现在我不如何真正在我的网址通过这个用户上下文。早些时候,我尝试传递只是用户标识,但没有奏效。有没有其他的选择做这件事。当我在内部构建url或者重定向到这个特定的视图时,这很好,但url如何看起来像是在外部。我的urls.py是
from django.conf.urls.defaults import patterns, include, url
from django.contrib.staticfiles.urls import staticfiles_urlpatterns
from django.views.generic.simple import direct_to_template
urlpatterns = patterns('',
url(r'^$', 'blog.views.blogs', name='all_blogs'),
url(r'^(?P<author>\d+)/(?P<slug>[-\w]+)/$', 'blog.views.blogs', name='view_blog'),
url(r'^(?P<author>\d+)/(?P<slug>[-\w]+)/edit/$', 'blog.views.post_form', name='edit_blog'),
url(r'^new/$', 'blog.views.post_form', name='new_blog'),
url(r'^(?P<author>\d+)/(?P<slug>[-\w]+)/delete/$', 'blog.views.delete_blog', name='delete_blog'),
)
urlpatterns += staticfiles_urlpatterns()
你是什么意思“刚刚路过的用户ID ..不起作用“?怎么了? –
'creator'需要是'User'实例不是整数。首先通过'pk'获取'User',然后在'Entry'查询中使用它。 'author = User.objects.get(pk = author)' –
@isbadawi其实问题是我能够传递用户ID,但我直接在过滤器中提到creator.id = author_id,这是错误的,因为rebus有指出我需要首先获取该对象,然后使用它 – Sachin