Django下载csv后的帖子请求。

问题描述 投票:0回答:1

我想创建一个端点,在这个端点中可以发送POST请求,如果POST请求被验证,那么页面就会下载一个csv。

我创建了序列器表单,以便于对接收到的数据进行验证。

我的问题是,csv很容易下载到HttpResponse中,但我需要在后请求中做一个端点和数据验证。

我的主要问题是我不能返回 export 的下一个函数 validation

这是我的档案

#urls.py

from django.conf.urls import url
from . import views

urlpatterns = [
    url(r'^hello-view/', views.HelloApiView.as_view()),
]

序列器

#serializers.py
from rest_framework import serializers

class HelloSerializer(serializers.Serializer):
    """Serializes a name field """

    name = serializers.CharField(max_length=100)
    seller_id =  serializers.CharField(max_length=100)

    def validate_name(self, dob):
        UnitOfMeasureName = ["Each", "Grams", "Ounces", "Pounds", "Kilograms", "Metric Tons"]
        if dob in UnitOfMeasureName:
            return dob
        else:
            raise serializers.ValidationError('Wrong username')

和视图文件在这个文件中,我创建了导出功能,试图导出csv数据,但不工作。

import csv
from django.shortcuts import render
from django.http import HttpResponse
from rest_framework import viewsets
from rest_framework.views import APIView
from rest_framework.response import Response
from rest_framework import status

from . import serializers


class HelloApiView(APIView):

    def export(self, request):
        response = HttpResponse(content_type='text/csv')
        writer = csv.writer(response)
        writer.writerow(['First name', 'Last name', 'ID'])

        response['Content-Disposition'] = 'attachment; filename="one.csv"'
        return response

    serializer_class = serializers.HelloSerializer

    def get(self, request, format=None):

        an_apiview = [
            'Uses HTTP METHOD as function',
            'Its is similar to a traditional django view'
        ]

        return Response({'message': 'Hello', 'anapi': an_apiview})

    def post(self, request):
        serializer = serializers.HelloSerializer(data = request.data)

        if serializer.is_valid():
                in this place i want to return the file


        else:
            return Response(
                serializer.errors, status = status.HTTP_400_BAD_REQUEST)
django csv httpresponse django-serializer
1个回答
1
投票

在你的urls.py文件中添加一个新的端点

#urls.py

from django.conf.urls import url
from . import views

urlpatterns = [
    url(r'^hello-view/', views.HelloApiView.as_view()),
    url(r'^csv_download/$', views.csv_download, name="csv_download"),
]

然后在views.py文件中,添加一个叫做 csv_download 并将你所有的东西移到那里(这可能是不必要的,但它肯定会使代码更干净,更易读)。

# views.py

import csv
from django.shortcuts import render
from django.http import HttpResponse
from rest_framework import viewsets
from rest_framework.views import APIView
from rest_framework.response import Response
from rest_framework import status

from . import serializers


class HelloApiView(APIView):
    def get(self, request, format=None):
        an_apiview = [
            'Uses HTTP METHOD as function',
            'Its is similar to a traditional django view'
        ]
        return Response({'message': 'Hello', 'anapi': an_apiview})


def csv_download(request):


    if request.method == 'POST':
        serializer_class = serializers.HelloSerializer
        serializer = serializers.HelloSerializer(data = request.data)
        if serializer.is_valid():
            response = HttpResponse(content_type='text/csv')
            response['Content-Disposition'] = 'attachment; filename="one.csv"'
            writer = csv.writer(response,delimiter=',') # I always like to specify the delimeter
            writer.writerow(['First name', 'Last name', 'ID'])



            #Then you may actually want to write some data to the CSV file, currently, you've only defined the headers (first row). An example would be like:
            for value in list_of_objects:
                writer.writerow([
                value.first_name,
                value.last_name,
                value.id
                ])

            return response


        else:
            return Response(
                serializer.errors, status = status.HTTP_400_BAD_REQUEST)

    else:
        # redirect them or do something here if they hit this URL without a POST request
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