如何编写没有模型的 Django REST Api 以使用请求发送文件 [英] How to write Django REST Api without model to send a file using Requests

问题描述

我想编写一个没有模型的rest方法，以便我可以使用python请求模块发送一个csv文件.应从服务器远程访问此 csv 文件.

I want to write a rest method without model so that I can send a csv file using python requests module. This csv file should be remotely accessed from the server.

例如——我已使用请求登录到我的项目并获取 cookie 和标头，以便我可以将其传递给以下请求方法..

For example - I have logged in to my project using requests and get the cookies and headers so that I can pass it to the following requests method..

files = {'file': open('test.csv', 'rb')}
response = requests.post(url, files=files, headers=api_headers,
                      cookies=api_cookies)

所以这个 url 应该是:调用那个 rest 方法.

So this url should be : call for that rest method.

views.py 文件:

views.py file :

class FileUploadView(APIView):
    parser_classes = (FileUploadParser,)

    def post(self, request, format=None):
        csvfile = request.data['file']
        #reader = csv.DictReader(csvfile)
        #for r in reader:
            #print(r)
        return Response(status=204)

请注意 - 我正在使用请求模块发送一个 csv 文件.

Just to note - I am sending a csv file using requests module.

谁能帮我写下这个rest方法?

Can anyone please help me on how to write this rest method?

推荐答案

普通 django 视图

Normal django view

def myview(request):
    f = request.FILES['file']
    with open('some/folder/name.txt', 'wb+') as destination: 
        #f.name or f.filename (dont know which one)will get filename.So you can replace it name.txt
        for chunk in f.chunks():
            destination.write(chunk)
    return JsonResponse({"message": "Uploaded!"})

更新

# views.py
class FileUploadView(views.APIView):
    parser_classes = (FileUploadParser,)

    def post(self, request, filename, format=None):
        file_obj = request.data['file']
        # ...
        # do some stuff with uploaded file
        # ...
        return Response(status=200)

# urls.py
urlpatterns = [
    # ...
    url(r'^upload/(?P<filename>[^/]+)$', FileUploadView.as_view())
]

然后

url = 'http://127.0.0.1:8000/upload/test.csv' #filename should be in url
files = {'file': open('test.csv', 'rb')}
response = requests.post(url, files=files, headers=api_headers,
                      cookies=api_cookies)

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