如何使用cql引擎将图像存储为cassandra数据库中的Bytes字段？

Question

我想将我的网站图像存储在cassandra数据库中!我必须读取并存储图像的字节.你有一个很好的代码吗？我正在使用python2.7,django框架和cql-engine!

这是我的代码:

1-我的模特:

from cqlengine import columns
from cqlengine.models import Model

class UserImage(Model):
    Email = columns.Text(primary_key=True)
    image=columns.Bytes(required=False)

2-我的表格:

class UserImage(forms.Form):
    image=forms.ImageField()

3-我的观点:

from MainAPP.models import UserImage as UserImageModel
from MainAPP.forms import UsersForms

from django.http import HttpResponse
from cqlengine import connection
from PIL import Image

def UploadImage(request):
    if request.method == 'POST':
        form = UsersForms.UserImage(request.POST, request.FILES)
        if form.is_valid():
            try:
                image_data=Image.open(request.FILES['image'])
            except IOError:
                return HttpResponse("cannot upload %s"% request.FILES['image'].name)
            connection.setup(['127.0.0.1:9160'])
            UserImageModel.create(Email='ebrahimi.omid69@gmail.com', image=image_data)
            return HttpResponse('Stored Successfully!')
    else:
        form= UsersForms.UserImage()
        return render_to_response('Users/uploadImage.html', {'form': form}, context_instance=RequestContext(request))

我的模板:

{% block content %}
    <form enctype="multipart/form-data" method="post" action="">
    {% csrf_token %}
    {{ form.as_p }}
    <button type="submit">Upload</button>
    </form>
{% endblock %}

我在呈现模板中的错误:

AttributeError at /uploadImage
encode
Request Method: POST
Request URL:    http://127.0.0.1:8000/uploadImage
Django Version: 1.5.1
Exception Type: AttributeError
Exception Value:    encode

你有什么想法？请指导我......

Answer 1

我的编辑视图： 我们可以在不使用 PIL 的情况下将图像存储在 cassandra 中：

def UploadImage(request):
    if request.method == 'POST':
        form = UsersForms.UserImage(request.POST, request.FILES)
        if form.is_valid():
            try:
                image_data=request.FILES['image'].read()
            except IOError:
                return HttpResponse("cannot convert %s"% request.FILES['image'].name)
            connection.setup(['127.0.0.1:9160'])
            UserImageModel.create(Email='ebrahimi.omid69@gmail.com', image=image_data)
            return HttpResponse(request.FILES['image'].name)
    else:
        form= UsersForms.UserImage()
        return render_to_response('Users/uploadImage.html', {'form': form}, context_instance=RequestContext(request))