Question

我确信我在这里的某个地方犯了一个菜鸟错误。所以我有一个特定链接的详细信息页面。假设我的页面上有孵化器列表，如果我点击其中一个，我想显示其详细信息。我确信这可以通过使用主键来完成，但我不断收到错误。

models.py

class Incubators(models.Model):       
  # I have made all the required imports
    incubator_name = models.CharField(max_length=30)
    owner = models.CharField(max_length=30)
    city_location = models.CharField(max_length=30)
    description = models.TextField(max_length=100)
    logo = models.FileField()
    verify = models.BooleanField(default = False)

    def get_absolute_url(self):
        return reverse('main:details', kwargs={'pk': self.pk})

    def __str__(self):                  # Displays the following  stuff when a query is made
      return self.incubator_name + '-' + self.owner 

class Details(models.Model):
    incubator = models.ForeignKey(Incubators, on_delete = models.CASCADE)
    inc_name = models.CharField(max_length = 30)
    inc_img = models.FileField()
    inc_details = models.TextField(max_length= 2500)
    inc_address = models.TextField(max_length = 600, default = "Address")
    inc_doc = models.FileField()
    inc_policy = models.FileField()

    def __str__(self):
        return self.inc_name

views.py

def details(request, incubator_id):
    inc = get_object_or_404(Incubators, pk = incubator_id)    
    return render(request, 'main/details.html', {'inc': inc})

这是我的urls.py，但我确定此处没有错误：

url(r'^incubators/(?P<pk>[0-9]+)', views.details, name = 'details'),

你能解释一下为什么我会收到这个错误吗？

TypeError at /incubators/9

details() got an unexpected keyword argument 'pk'

Request Method:     GET
Request URL:    http://127.0.0.1:8000/incubators/9
Django Version:     1.11.3
Exception Type:     TypeError
Exception Value:    

details() got an unexpected keyword argument 'pk'

Answer 1

在您的网址格式中，您需要调用变量＆＃34; pk＆＃34;但在您看来，您将其称为incubator_id

要解决此问题，请从以下位置更改您的网址格式：

url(r'^incubators/(?P<pk>[0-9]+)', views.details, name = 'details'),

到

url(r'^incubators/(?P<incubator_id>[0-9]+)', views.details, name = 'details'),

我无法访问django中的主键？

1 个答案: