我应该将图像上传到Django中的静态目录吗？

Question

我有这个包含图像字段的模型.

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


class Customer(models.Model):
    user = models.ForeignKey(User)
    name = models.CharField(max_length=127)
    logo = models.ImageField(upload_to='customer/logo', null=True, blank=True)

    def __str__(self):
        return self.name

在我看来,我从指定的URL下载图像并将其存储在图像字段中.为了测试,我使用测试用户作为外键.

import json
import urllib.request

from django.core.files.base import ContentFile
from django.http import HttpResponse
from django.contrib.auth.models import User

from customer.models import Customer


def create(request):
    values = json.loads(request.body.decode('utf-8'))
    values['user'] = User.objects.get(id=1)
    values['logo'] = ContentFile(urllib.request.urlopen(values['logo']).read(),
                                                                    'test.png')
    model = Customer.objects.create(**values)
    return HttpResponse('Created customer with ' + str(values))

图像customer/logo/test.png按预期上传.现在,我如何在前端显示这些图像？我可以将它们保存到静态文件目录中,但只有相关用户才能访问它.

(顺便说一句,Django管理界面显示有一个文件为该Customer对象上传.但它链接到http://localhost:8000/admin/customer/customer/20/customer/logo/test.png一个错误的位置,并导致一个未找到的页面.)

Answer 1

for FileField和的文件ImageField相对于上传,settings.MEDIA_ROOT并且应该可以通过附加到的相同的相对文件名访问settings.MEDIA_URL.这就是您的管理界面指向错误网址的原因.出于安全原因,这些应该与STATIC_ROOT和不同STATIC_URL,否则Django会提出一个ImproperlyConfiguredError.

这不会阻止用户访问他们不应该看到的文件,如果他们知道或可以猜到网址.为此,您需要通过Django而不是您选择的Web服务器来提供这些私有文件.基本上,您需要在Web根级别指定一个私有目录,如果用户有权查看该文件,则需要加载这些文件.例如:

from django.core.files.storage import FileSystemStorage

PRIVATE_DIR = os.path.join(ROOT_DIR, 'web-private')
fs = FileSystemStorage(location=PRIVATE_DIR)

class Customer(models.Model):
    logo = models.ImageField(upload_to='customer/logo', storage=fs, null=True, blank=True)

在您的视图中,您必须提供此文件.我当前项目中的一个自定义应用程序使用以下函数发送静态文件:

import mimetypes
from django.http import HttpResponse # StreamingHttpResponse
from django.core.servers.basehttp import FileWrapper

def send_file(file):
    """
    Send a file through Django without loading the whole file into
    memory at once. The FileWrapper will turn the file object into an
    iterator for chunks of 8KB.
    """
    filename = file.name
    if settings.PRIVATE_MEDIA_USE_XSENDFILE:
        # X-sendfile
        response = HttpResponse()
        response['X-Accel-Redirect'] = filename  # Nginx
        response['X-Sendfile'] = filename        # Apache 2, mod-xsendfile
        # Nginx doesn't overwrite headers, but does add the missing headers.
        del response['Content-Type']
    else:
        # Can use django.views.static.serve() method (which supports if-modified-since),
        # but this also does the job well, as it's mainly for debugging.
        mimetype, encoding = mimetypes.guess_type(filename)
        response = HttpResponse(FileWrapper(file), content_type=mimetype)
        response['Content-Length'] = os.path.getsize(filename)
    return response

然后send_file(customer.logo)在您的视图中使用.

Django> = 1.5应该使用new StreamingHttpResponse而不是HttpResponse.