使用POSTMAN将python代码转换为JSON格式

Question

我已经使用邮递员将数据转换为json格式。

我写了以下代码：

def customer_get_meals(request, restaurant_id):
    meals = MealSerializer(
        Meal.objects.filter(restaurant_id = restaurant_id).order_by("-id"),
        many = True,
        context = {"request": request}
    ).data

    return JsonResponse({"meals: meals"})

这是网址

url(r'^restaurant/account/$', views.restaurant_account, name = 'restaurant-account'),
url(r'^restaurant/meal/$', views.restaurant_meal, name = 'restaurant-meal'),
url(r'^restaurant/meal/add/$', views.restaurant_add_meal, name = 'restaurant-add-meal'),
url(r'^restaurant/meal/edit/(?P<meal_id>\d+)/$', views.restaurant_edit_meal, name = 'restaurant-edit-meal'),

来自rest_framework导入序列化器 从foodtaskerapp.models导入Restaurant，Meal

MealSerializer类（serializers.ModelSerializer）：     图片= serializers.SerializerMethodField（）

def get_image(self, meal):
    request = self.context.get('request')
    image_url = meal.image.url
    return request.build_absolute_uri(image_url)

class Meta:
    model = Meal
    fields = ("id", "name", "short_description", "image", "price")

毕竟，在这种情况下，我无法在Postman中获得JSON格式： 位于/ api / customer / meals / 2 /的TypeError 为了允许非字典对象被序列化，请将安全参数设置为False。