Django Rest API自定义类实现

问题描述

我创建了一个类，用于从farsecret API检索有关特定食品的数据。 在课堂上，我为以下项目创建了3个函数：

*获得授权

*获取我们正在寻找的商品的ID

*项目的下载数据

class IngredientImport(APIView):

def get_authorization_token(self):
    request_token_url = "https://oauth.fatsecret.com/connect/token"
    consumer_key = os.environ.get('NTS_Client_ID')
    consumer_secret = os.environ.get('NTS_Client_Secret')
    data = {'grant_type':'client_credentials',"scope":"basic"}

    access_token_response = requests.post(
        request_token_url,data=data,verify=False,allow_redirects=False,auth=(consumer_key,consumer_secret)
    )
    return access_token_response.json()["access_token"]


def get_list_by_name(self,name,access_token):
    api_url = "https://platform.fatsecret.com/rest/server.api"
    params={
        "method":"foods.search","search_expression":name,"page_number":1,"max_results":1,"format":"json"
    }
    header = {"Authorization":access_token}
    api_call_headers = {"Authorization": "Bearer " + access_token}

    response = requests.get(
        api_url,params=params,headers=api_call_headers
    )
    items = response.json()["foods"]
    try:
        return response.json()["foods"]["food"]["food_id"]
    except KeyError:
        return None


def import_item(self,item_id,api_token):
    if item_id == None:
        return None

    api_url = "https://platform.fatsecret.com/rest/server.api"
    params={"method":"food.get","food_id":item_id,"format":"json"}
    api_call_headers = {"Authorization": "Bearer " + access_token}  

    response = requests.get(
        api_url,headers=api_call_headers
    )
    item = response.json()["food"]["servings"]["serving"]
    item_name = response.json()["food"]["food_name"]
    if type(item) == list:
        item = item[0]
    try:
        portion_size = float(item["metric_serving_amount"])
        carbs = round(float(item["carbohydrate"]) / portion_size * 100,2)
        fats = round(float(item["fat"]) / portion_size * 100,2)
        proteins = round(float(item["protein"]) / portion_size * 100,2)            
    except KeyError:
        return None

如何在我的应用中实现此类，以避免在urls.py中为每个函数创建3个不同的路径。有可能还是应该将其分解为基于函数的视图？

解决方法

class IngredientImport(APIView):

    def get_authorization_token(self):
       ...
 
    def get_list_by_name(self,name,access_token):
       ...

    def import_item(self,item_id,api_token):
       ...

    def get(self,request):
       # get name from query param
       name = self.request.GET.get('name')

       token = self.get_authorization_token()
       food_list = self.get_list_by_name(name,token)

       for food_id in food_list:
           self.import_item(food_id,token)

       return Response({'imported_foods': food_list})

然后，在您的urls.py上

urlpatterns = [
   path('import_foods',IngredientImport.as_view())
]