我正在尝试在django 1.3中创建两个用户类型。我正在使用以下models.py:
继承AUTH_PROFILE_MODULEclass Member(models.Model):
ROLE_CHOICES = (
(0, 'Guide'),
(1, 'Operator'),
)
user = models.ForeignKey(User, unique=True)
location = models.CharField(max_length=60)
role = models.IntegerField(choices=ROLE_CHOICES)
class Guide(Member):
bio = models.TextField(blank=True)
experience = models.TextField(blank=True)
image = models.ImageField(blank=True, upload_to='images')
fileupload = models.FileField(blank=True, upload_to='files')
def __unicode__(self):
return self.user.username
def get_absolute_url(self):
return '/profiles/guides/%s' % self.user.username
class Operator(Member):
bio = models.TextField(blank=True)
image = models.ImageField(blank=True, upload_to='images')
def __unicode__(self):
return self.user.username
def get_absolute_url(self):
return '/profiles/operators/%s' % self.user.username
我正在使用基于通用类的视图,并且可以使ListView适用于Guide和Operator模型我无法使DetailView工作。我的views.py如下:
class GuideDetailView(DetailView):
model = Guide
context_object_name = 'guide'
template_name = 'members/guide_detail.html'
class GuideListView(ListView):
model = Guide
context_object_name = 'guides'
template_name = 'members/guide_list.html'
知道可能缺少什么吗?
答案 0 :(得分:1)
提供查询集:
class GuideDetailView(DetailView):
queryset = Guide.objects.all()
或覆盖DetailView的get方法:
class GuideDetailView(DetailView):
def get(self):
return "Everything you want, maybe: Guide.object.get(id=1)"
在你的urls.py中给出这个:
url(r'^(?P<my_id>\d)/$', GuideDetailView.as_view(),),
您需要覆盖get,如下所示:
class GuideDetailView(DetailView):
def get(self, request, **kwargs):
# lookup Guide Id in your database and assign it object
self.object = Guide.objects.get(pk=kwargs.get('my_id'))
# add object to your context_data, so that you can access via your template
context = self.get_context_data(object=self.object)
return self.render_to_response(context)