python - django s3 boto文件上传TypeError : invalid file:

转载 作者：行者123 更新时间：2023-11-28 17:30:00 26 4

我正在尝试将文件从 postman 上传到 s3 并在k.set_contents_from_filename(文件)类型错误:无效文件:你能看看吗？非常感谢。

序列化器.py

from rest_framework import serializers
class ResourceSerializer(serializers.Serializer):
    file = serializers.FileField(required=True, max_length=None, use_url=True)
    name = serializers.CharField(required=True, max_length=500)

View .py

import logging
from boto.s3.key import Key

from rest_framework.views import APIView
from rest_framework.response import Response
from rest_framework import status

from .serializers import ResourceSerializer
from .utils import create_boto_connection
from django.conf import settings


class Resource(APIView):
    def post(self, request, format=None):
        serializer = ResourceSerializer(data=request.data)
        if serializer.is_valid():
            context = {}
            file = serializer.validated_data['file']
            name = serializer.validated_data['name']
            ext = file.name.split('.')[-1]
            new_file_name = '{file}.{ext}'.format(file=name, ext=ext)
            file_name_with_dir = 'profile_photos/{}'.format(new_file_name)
            # Create s3boto connection
            conn = create_boto_connection()
            try:
                bucket = conn.get_bucket(settings.AWS_STORAGE_BUCKET_NAME)
                k = Key(bucket)
                k.key = file_name_with_dir
                k.set_contents_from_filename(file)
                k.make_public()
                context['file'] = new_file_name
            except Exception as e:
                context['message'] = 'Failed to process request'
                # Logging Exceptions
                logging.exception(e)
                logging.debug("Could not upload to S3")
            return Response(context, status=status.HTTP_201_CREATED)
        return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)

实用程序.py

from boto.s3.connection import S3Connection

from django.conf import settings


def create_boto_connection():
    conn = S3Connection(settings.AWS_ACCESS_KEY_ID, settings.AWS_SECRET_ACCESS_KEY)
    # conn = boto.connect_s3()
    return conn

网址.py

from django.conf.urls import url

from s3boto import views

urlpatterns = [
    # v1
    url(r'^v1/s3boto/upload-resource/$', views.Resource.as_view(), name="upload-resource"),
]

postman :

最佳答案

您正在将 Django InMemoryUploadedFile 传递给方法 set_content_from_filename，它需要一个字符串。

来自boto documentation :

set_contents_from_filename(filename, headers=None, replace=True, cb=None, num_cb=10, policy=None, md5=None, reduced_redundancy=False, encrypt_key=False)

Store an object in S3 using the name of the Key object as the key in S3 and the contents of the file named by ‘filename’. See set_contents_from_file method for details about the parameters.

要么使用set_content_from_file或者将文件保存到本地临时文件并将该文件名传递给 set_content_from_filename。

关于python - django s3 boto文件上传TypeError : invalid file: <InMemoryUploadedFile:，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/35142529/

26 4 0

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