如何使用 django rest 框架创建 Django API？

Question

我处于需要创建一个django api的情况，它在从frontend调用时会给出从其他来源而不是从serializers器和models内部获取的响应。 目前我正在使用 django rest 框架来创建api如下所示

楷模

# Create your models here.
class submission(models.Model):
    # fixed fields to show on today panel/pending panel
    email_id = models.CharField(max_length=200, primary_key=True)
    email_sender = models.CharField(max_length=200, null=True, blank=True)


def __str__(self):
        return self.email_id

序列化器

class UserSerializer(serializers.ModelSerializer):
    class Meta:
        model = submission
        fields = ('__all__')

意见

class SAMPLE(viewsets.ModelViewSet):
    queryset = submission.objects.all()
    serializer_class = UserSerializer

网址

router = DefaultRouter()
# user table
router.register('user_table_all_data', SAMPLE, basename='user_table_all_data')

and it works as charm, but in order to work with django rest framework i always need to call like this 1. Create model or use existing model, 2. Create serializers,3. Create Views,4. Create URLs 2. Create serializers,3. Create Views,4. Create URLs 2. Create serializers,3. Create Views,4. Create URLs ，

但是如果我不想使用models例如我想显示来自其他来源的数据说其中包含示例json如下所示

sampleData = {"Name":"RAJNISH","Age":32,"ADDRESS":"India"}

所以如果我的API是说'sample/someid (not pk)'

返回响应为

{"Name":"RAJNISH","Age":32,"ADDRESS":"India"}

我怎么能在 django rest 框架中做到这一点？

Answer 1

使用基于 Function 的视图而不是基于 Class 的视图

def sample(request):
    return Response(data={"Name":"RAJNISH","Age":32,"ADDRESS":"India"})