Question

我有这个HTML模板

<html><body>
<form method="POST">
   {% render_form submitSerializer %}
  <input type="submit" value="Save">
</form>
</body></html>

此序列化程序

class SubmitSerializer(serializers.ModelSerializer):
    submittedFile = serializers.FileField()
    class Meta:
        model = Submit
        fields = {'submittedFile', 'result'}
        read_only_fields = {'result'}

和这个观点

class SubmitView(APIView):
    renderer_classes = [renderers.TemplateHTMLRenderer]

    def get(self, request, *args, **kwargs):
        return Response({'submitSerializer':submitSerializer}, template_name='singlelesson.html')

    def post(self, request, *args, **kwargs):
        submit = Submit(lesson=lessonInstance, user=self.request.user, submittedFile = self.request.data['submittedFile'])
        serializer = SubmitSerializer(submit)

        #compare_files returns true if two files contain same text
        result = compare_files(correct_solution, self.request.POST.get('submittedFile'))

        serializer.save(result = result)
        return Response({'submitSerializer':serializer}, template_name='singlelesson.html')

问题是表单上传的不是文件而是其名称，比较两个文件时出现类型错误。如何让表单上传文件？

Answer 1

您应该从请求对象中的FILES QueryDict中提取文件，而不是POST。 e.g：

self.request.FILES['submittedFile']

或者django-rest-framework方式：

request.data['submittedFile']

有关详细说明，请查看： http://www.django-rest-framework.org/api-guide/parsers/#fileuploadparser

编辑：同时将表单更改为：

<html><body>
<form method="POST" enctype="multipart/form-data">
   {% render_form submitSerializer %}
  <input type="submit" value="Save">
</form>
</body></html>

并将SubmitView发送至：

class SubmitView(APIView):
    parser_classes = (MultiPartParser,)
...

django REST框架：HTML表单返回文件名而不是文件

1 个答案: