Question

我正在尝试使用DRF api上传并获取对图像的响应。图像上传到指定的位置，但我想要的是获取图像的链接，以便我可以点击它，并在我得到响应时可以看到图像。如果有人能帮助我纠正这个问题我会非常感激

models.py文件

class employees(models.Model):
name_of_employee = models.CharField(max_length=20, unique=True)
name_of_department = models.CharField(max_length=20, unique=True)
emp_id = models.IntegerField()
image = models.ImageField(upload_to='Images/', default='Images/None/No-img.jpg') 

def __str__(self):
    return '{} {}' .format(self.name_of_employee, self.name_of_department)

views.py文件

class employeeList(APIView):

def get(self, request):
    employees1 = employees.objects.all()
    serializer = employeesSerializer(employees1, many=True)
    return Response(serializer.data)

def post(self, request):
    serializer = employeesSerializer(data=request.data)
    if serializer.is_valid():
        serializer.save()
        return Response(serializer.data, status=status.HTTP_201_CREATED)
    return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)

def delete(self, request, pk):
    employees1 = self.get_object(pk)
    employees1.delete()
    return Response(status=status.HTTP_204_NO_CONTENT)

class employeeDetail(APIView):

def get_object(self,pk):
    try:
        return employees.objects.get(pk=pk)
    except employees.DoesNotExist:
        raise Http404

def get(self, request,pk):
    employees1 = self.get_object(pk)
    employees1 = employeesSerializer(employees1)
    return Response(employees1.data)

def put(self, request, pk):
    employees1 = self.get_object(pk)
    serializer = employeesSerializer(employees1, data=request.data)
    if serializer.is_valid():
        serializer.save()
        return Response(serializer.data)
    return Response(serializer.errors, status=status.HTTP_404_BAD_REQUEST)

def delete(self, request, pk):
    employees1 = self.get_object(pk)
    employees1.delete()
    return Response(status=status.HTTP_204_NO_CONTENT)

serializers.py文件

class employeesSerializer(serializers.ModelSerializer):

image = serializers.ImageField(max_length=None, use_url=True)

class Meta:
    model = employees
    fields = '__all__'

url.py文件

urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'^employees/', views.employeeList.as_view()),
url(r'^employees/(?P<pk>[0-9]+)/$', views.employeeDetail.as_view()),
]+static(settings.MEDIA_URL,document_root=settings.MEDIA_ROOT)

settings.py文件

STATIC_URL = '/static/'

STATICFILES_DIRS = (
os.path.join(BASE_DIR, "static"),
'/static/',
)
MEDIA_ROOT = os.path.join(BASE_DIR, 'media')

MEDIA_URL = '/media/'

在运行服务器时我的回复是

HTTP 200 OK
Allow: GET, POST, DELETE, HEAD, OPTIONS
Content-Type: application/json
Vary: Accept

[
 {
    "id": 1,
    "image": "/media/Images/share-1.jpg",
    "name_of_employee": "Abhinav",
    "name_of_department": "electrical",
    "emp_id": 1
 },

现在我想获取图像字段上的链接，以便我可以点击它并查看存储的图像

提前致谢

Answer 1

我认为你想要返回绝对而不是相对路径。我使用以下序列化程序的自定义图像实现来返回绝对路径：

class PhotoSerializer(serializers.ModelSerializer):
    image_upload = serializers.ImageField(write_only=True)
    image = serializers.SerializerMethodField('get_image_url')

    class Meta:
        model = Photo
        fields = ('id', 'image', 'image_upload')

    def get_image_url(self, obj):
        '''
        Return an absolute url for the image
        '''
        return self.context['request'].build_absolute_uri(str(obj.image))

这可能不是您想要的，因为它不会使用相同的字段进行上传/下载。

如何获取get（）方法

1 个答案: