Question

我使用Django Rest框架（DRF）构建了以下基于泛型类的视图

class ExampleDetail(generics.RetrieveUpdateDestroyAPIView):
    queryset = Example.objects.all()
    serializer_class = ExampleSerializer
    renderer_classes = (JSONRenderer, TemplateHTMLRenderer)

    def get(self, request, *args, **kwargs):

        response = self.retrieve(request, *args, **kwargs)
        if request.accepted_renderer.format == 'html':
            form = ExampleForm(data=response.data)
            return Response({'data': response.data, 'form': form}, template_name='example.html')

        return response

此视图允许我通过指定format = json或html从同一端点获取JSON数据或HTML表单。

我想以编程方式调用该视图以从另一个视图中获取呈现的HTML表单，以便将此表单包含在另一个包含更多内容的页面中。

Answer 1

我在文档中找到了解决方案...... https://docs.djangoproject.com/en/1.7/topics/class-based-views/mixins/

提示来自他们的例子：

class AuthorDetail(View):

    def get(self, request, *args, **kwargs):
        view = AuthorDisplay.as_view()
        return view(request, *args, **kwargs)

    def post(self, request, *args, **kwargs):
        view = AuthorInterest.as_view()
        return view(request, *args, **kwargs)

Answer 2

Sub Mail_Outlook()
    Dim OutApp As Object
    Dim OutMail As Object, signature As String
    Dim irow, i As Integer

    Dim objItem As Object
    irow = InputBox("How many Invites do you want to send?")

    Set OutApp = CreateObject("Outlook.Application")

    LastRow = ActiveSheet.Range("E" & Rows.Count).End(xlUp).Row

    For i = 0 To irow - 1

      Set OutMail = OutApp.CreateItem(o)

      If Cells(LastRow - i, 1).Value <> "" Then
          Mailto = Cells(LastRow - i, 5).Value
          If Mailto = "Colleague1" Or _
               Mailto = " Colleague2" Or _
               Mailto = " Colleague3" Then
             Mailto = Mailto & "@hotmail.com"
          End If
      End If

      MailSubject = Cells(LastRow - i, 4).Value & " – Important Message"
      MailBody = "Hi " & Cells(LastRow - i, 5).Value & "," & vbNewLine & vbNewLine & _
        " I’m inviting you for the conference being held today." & _
        vbNewLine

      With OutMail
        .Display
        signature = .body
        .Subject = MailSubject
        .To = Mailto
        .body = MailBody & vbNewLine & signature
      End With

    Next i

    Set OutMail = Nothing
    Set OutApp = Nothing

End Sub

OR

html_from_view = ExampleDetail.as_view({'get': 'list'})(request).content

Answer 3

从Django 2.2和DRF 3.9.2开始，我可以使用以下代码获得响应。

response = UserItemsApiView.as_view()(request=request._request).data

以上示例解决了以下问题：

request参数必须是django.http.HttpRequest的实例，而不是rest_framework.request.Request
使用content属性代替了data，从该视图中获得了结果。

Answer 4

我没有尝试过，但是我认为以下应该可行。

服务器端：

调用DRF并通过url.open或请求进行轮询

as this answer suggest

示例：

import requests
from django.shortcuts import render


def home(request):
   # get the list of todos
   response = 
    requests.get('https://jsonplaceholder.typicode.com/todos/')
   # transfor the response to json objects
   todos = response.json()
   return render(request, "main_app/home.html", {"todos": todos})

Complete example

客户端：

您可以调用DRF并通过ajax请求从任何其他视图指向。

Answer 5

如果我理解正确，你需要从视图B中获取结果，同时在视图A中。

使用请求 / urllib2 和 json 库可以解决您的问题（如this answer中所述）。

要获取URL，您可以结合使用 request.get_absolute_uri（）和/或 request.get_host（）和 django.core.urlresolvers .reverse

如何以编程方式在另一个视图中调用Django Rest Framework视图？

5 个答案: