Question

是否可以在URL中获取传递给视图的变量。例如。我有一个包含许多位置的位置表。如果我单击某个位置，它将转到该位置的详细信息页面。位置ID将添加到URL。然后我有一个指向另一个页面/视图的链接，但想在下一个视图中引用位置ID。

urls.py：

url(r'^locations/get/(?P<location_id>\d+)/$', 'assessments.views.location'),
url(r'^locations/get/(?P<location_id>\d+)/next_page/$', 'assessments.views.next_page'),

view.py：

def location(request, location_id=1):
    return render_to_response('dashboard/location.html', {'location': Location.objects.get(id=location_id) })

def next_page(request):
    loc = Location.objects.get(id='id of that location')

Answer 1

只需在视图功能中使用其名称

访问它

def next_page(request,location_id):
    loc = Location.objects.get(id=location_id)

Answer 2

好的，我还不确定你究竟是什么意思，但我认为以下两个片段中的一个可以解决你的问题：

网址包含location_id，因此只需在next_page()：

中使用即可

def location(request, location_id):
    return render_to_response('dashboard/location.html', {'location': Location.objects.get(id=location_id) })

def next_page(request, location_id):
    loc = Location.objects.get(id=location_id)

在request.session中传递location_id：

def location(request, location_id):
    request.session['selected_location'] = location_id
    return render_to_response('dashboard/location.html', {'location': Location.objects.get(id=location_id) })

def next_page(request):
    try: 
        selected_location = request.session['selected_location']
    except KeyError:
        # Not sure if 404 is the best status code here but you get the idea
        raise Http404
    loc = Location.objects.get(id=selected_location)

Django从URL变为视图变量

2 个答案: