我想创建一个显示两个独立 Django 模型的页面:
class Client(models.Model):
name = models.CharField(max_length=100)
slug = AutoSlugField(populate_from='name', blank=True, unique=True)
order = models.IntegerField(editable=False, default=0)
class Meta:
ordering = ('order',)
def __unicode__(self):
return self.name
class Press(models.Model):
title = models.CharField(max_length=50)
article = models.ImageField(upload_to = 'images')
def image_thumb(self):
if self.article:
return u'<img src="%s" height="125"/>' %self.article.url
else:
return "no image"
image_thumb.short_description = "article"
image_thumb.allow_tags = True
class Meta:
verbose_name_plural = "press"
我不确定如何在 Views.py 中编写我的查询集。我试过这样的东西......
class ClientView(generic.ListView):
template_name = 'clients.html'
context_object_name = 'client'
def queryset(request):
client_page = {'press': Press.objects.all(), 'client': Clients.objects.all()}
return client_page
然后在我的 urls.py 中...
url(r'^clients/', views.ClientView.as_view(), name = 'client_model'),
我在堆栈答案中读到,我可以通过使用“get_extra_context”来做到这一点,但有人可以告诉我它是如何使用的吗?