TypeError使用BOTO Library将图像文件上传到Django中的Amazon S3

Question

我是编程和Django的初学者,所以我很感激帮助初学者可以得到他的头脑!

我正在按照一个教程来展示如何使用Boto库将图像上传到Amazon S3帐户,但我认为它适用于较旧版本的Django(我在1.1.2和Python 2.65)并且有些变化.我收到一个错误:异常类型:TypeError异常值:'InMemoryUploadedFile'对象是unsubscriptable

我的代码是:

Models.py:

from django.db import models
from django.contrib.auth.models import User
from django import forms
from datetime import datetime

class PhotoUrl(models.Model):
    url = models.CharField(max_length=128)
    uploaded = models.DateTimeField()
    def save(self):
        self.uploaded = datetime.now()
        models.Model.save(self)

views.py:

import mimetypes
from django.http import HttpResponseRedirect
from django.shortcuts import render_to_response
from django.core.urlresolvers import reverse
from django import forms
from django.conf import settings
from boto.s3.connection import S3Connection
from boto.s3.key import Key

def awsdemo(request):   
def store_in_s3(filename, content):
    conn = S3Connection(settings.AWS_ACCESS_KEY_ID, settings.AWS_SECRET_ACCESS_KEY)
    b = conn.create_bucket('almacmillan-hark')
    mime = mimetypes.guess_type(filename)[0]
    k = Key(b)
    k.key = filename
    k.set_metadata("Content-Type", mime)
    k.set_contents_from_strong(content)
    k.set_acl("public-read")

photos = PhotoUrl.objects.all().order_by("-uploaded")
if not request.method == "POST":
    f = UploadForm()
    return render_to_response('awsdemo.html', {'form':f, 'photos':photos})

f = UploadForm(request.POST, request.FILES)
if not f.is_valid():
    return render_to_response('awsdemo.html', {'form':f, 'photos':photos})

file = request.FILES['file']
filename = file.name
content = file['content']
store_in_s3(filename, content)
p = PhotoUrl(url="http://almacmillan-hark.s3.amazonaws.com/" + filename)
p.save()
photos = PhotoUrl.objects.all().order_by("-uploaded")   
return render_to_response('awsdemo.html', {'form':f, 'photos':photos})

urls.py:

(r'^awsdemo/$', 'harkproject.s3app.views.awsdemo'), 

awsdemo.html:

<div class="form">
    <strong>{{form.file.label}}</strong>
    <form method="POST" action ="." enctype="multipart/form-data">
        {{form.file}}<br/>
        <input type="submit" value="Upload">
    </form> 
</div>  

我真的很感激帮助.我希望我提供了足够的代码.

亲切的问候AL

Answer 1

我认为你的问题是这一行:

content = file['content']

来自Django文档:

FILES中的每个值都是一个UploadedFile对象,包含以下属性:

• read(num_bytes = None) - 从文件中读取一些字节.
• name - 上载文件的名称.
• size - 上载文件的大小(以字节为单位).
• chunk(chunk_size = None) - 生成顺序数据块的生成器.

试试这个:

content = file.read()