Django Rest Framework URL调度程序 [英] Django Rest Framework url dispatcher

问题描述

我正在尝试与SWAPI API进行交互.

我想要一个视图，该视图可提取所有电影( films/)，而一个视图可提取一部电影( film/id ).

当我点击 http://127.0.0.1:8000/api/films/?id=4 时，它将进入 get_films 函数.

我正在使用Django == 1.11和Django Rest Framework == 3.9.0.

我的urs.py:

  urlpatterns = [url(r'films/'，views.get_films，name ="get-films")，url(r'films/(?P&id; id [0-9])/'，views.get_film，name ="get-film")，] 

我的views.py:

  MAX_RETRIES = 5API_URL =" https://swapi.co/api/@api_view(['GET'，'POST'])def get_films(request):request_url = API_URL +电影"打印(request_url)如果request.method =="GET":try_num = 0#跟踪我们重试了多少次而try_num<MAX_RETRIES:r = request.get(request_url，timeout = 10)如果r.status_code == 200:数据= r.json()返回响应(数据，状态=状态.HTTP_200_OK)别的:try_num + = 1#您可能可以使用记录器在此处记录错误time.sleep(5)#等待5秒钟，然后重试返回响应({错误":请求失败"}，状态= r.status_code)别的:返回响应({错误":不允许使用方法"}，状态=状态.HTTP_400_BAD_REQUEST)@api_view(['GET'，'POST'])def get_film(自己，要求):打印(输入)request_url = API_URL +电影/" + request.query_params.get('id')如果request.method =="GET":try_num = 0#跟踪我们重试了多少次而try_num<MAX_RETRIES:r = request.get(request_url，timeout = 10)如果r.status_code == 200:数据= r.json()返回响应(数据，状态=状态.HTTP_200_OK)别的:try_num + = 1#您可能可以使用记录器在此处记录错误time.sleep(5)#等待5秒钟，然后重试返回响应({错误":请求失败"}，状态= r.status_code)别的:返回响应({错误":不允许使用方法"}，状态=状态.HTTP_400_BAD_REQUEST) 

解决方案

您的实现存在两点错误.首先， get_film .该方法中的参数是错误的.您需要这样定义它:

  @api_view(['GET'，'POST'])def get_film(request，id):#<-第一个参数是用于`HttpRequest`对象的request.第二个是`id`，用于将其与url的id映射.打印(输入)request_url = API_URL +电影/" + str(id)#其余代码 

第二，在调用此 api 时，您需要这样做: http://127.0.0.1:8000/api/films/4/

I am trying to interact with the SWAPI API.

I want a view that fetches all the movies(films/) and one that fetches one (film/id).

When I hit http://127.0.0.1:8000/api/films/?id=4 it enters the get_films function.

I am using Django==1.11 and Django Rest Framework==3.9.0 .

my urs.py:

urlpatterns = [
    url(r'films/',views.get_films,name="get-films"),
    url(r'films/(?P<id>[0-9])/',views.get_film,name="get-film"),
]

my views.py:

MAX_RETRIES = 5 
API_URL= "https://swapi.co/api/" 

@api_view(['GET', 'POST'])
def get_films(request):
    request_url = API_URL + "films"
    print(request_url)
    if request.method == "GET":
        attempt_num = 0  # keep track of how many times we've retried
        while attempt_num < MAX_RETRIES:
            r = requests.get(request_url, timeout=10)
            if r.status_code == 200:
                data = r.json()
                return Response(data, status=status.HTTP_200_OK)
            else:
                attempt_num += 1
                # You can probably use a logger to log the error here
                time.sleep(5)  # Wait for 5 seconds before re-trying
        return Response({"error": "Request failed"}, status=r.status_code)
    else:
        return Response({"error": "Method not allowed"}, status=status.HTTP_400_BAD_REQUEST)

@api_view(['GET', 'POST'])
def get_film(self, request):
    print('entered')
    request_url = API_URL + "films/" + request.query_params.get('id')
    if request.method == "GET":
        attempt_num = 0  # keep track of how many times we've retried
        while attempt_num < MAX_RETRIES:
            r = requests.get(request_url, timeout=10)
            if r.status_code == 200:
                data = r.json()
                return Response(data, status=status.HTTP_200_OK)
            else:
                attempt_num += 1
                # You can probably use a logger to log the error here
                time.sleep(5)  # Wait for 5 seconds before re-trying
        return Response({"error": "Request failed"}, status=r.status_code)
    else:
        return Response({"error": "Method not allowed"}, status=status.HTTP_400_BAD_REQUEST)

解决方案

There are 2 things wrong with your implementation. First, get_film. the arguments are wrong in this method. you need to define it like this:

@api_view(['GET', 'POST'])
def get_film(request, id):  # <-- First argument is request which is for `HttpRequest` object. Second is `id` which is for mapping it with the url's id. 
    print('entered')
    request_url = API_URL + "films/" + str(id) 
    # rest of the code

Second, when calling this api, you need to do it like this: http://127.0.0.1:8000/api/films/4/

这篇关于Django Rest Framework URL调度程序的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！