Django 1.8文件上传示例

我是Django开发平台的新手 . 我一直在努力在Django 1.8.2（最新版本）中创建一个基本的文件上传应用程序 . 我找不到任何最新的示例/代码段 . 所以，如果你能在这里发帖或者向我推荐一个最小但完整的（模型，视图，模板）示例代码，我感谢你 .

我在线观察了很多其他的例子，但我注意到我的版本默认没有MEDIA_ROOT或MEDIA_URL属性 . 例如：

MEDIA_ROOT = '/path/to/myproject/media/'
MEDIA_URL = '/media/'

这是我的Models.py页面

from django.db import models

# Create your models here.

class Test(models.Model):
   first_name = models.SlugField()
   last_name = models.SlugField()
   email = models.EmailField()
   timestamp = models.DateTimeField(auto_now_add = True, auto_now = False)
   updated = models.DateTimeField(auto_now_add = False, auto_now = True)

def __unicode__(self):
    return self.email

这是我的Forms.py页面

from django import forms
from .models import Test

class EmailForm(forms.Form):
    first_name = forms.CharField(required=False)
    last_name = forms.CharField(required=False)
    email = forms.EmailField()

class TestForm(forms.ModelForm):
    class Meta:
    model = Test

这是我的Views.py页面

from django.shortcuts import render
from django.shortcuts import render_to_response
from django.template import RequestContext
from django.http import HttpResponseRedirect
from django.core.urlresolvers import reverse

# Create your views here.
from .forms import EmailForm, TestForm
from .models import Test

def home(request):
   # print request.POST["email"], request.POST["email2"]

   form = EmailForm(request.POST or None)
   if form.is_valid():
      first_name = form.cleaned_data['first_name']
      last_name = form.cleaned_data['last_name']
      email = form.cleaned_data['email']
      new_test, created = Test.objects.get_or_create(email = email,        first_name = first_name, last_name = last_name)

    # # This is using Model Forms
    # form = TestForm(request.POST or None)
    # if form.is_valid():
    #     new_test = form.save(commit = False)
    #     #We Might Do Something Here
    #     email = form.cleaned_data['email']
    #     new_test_old, created = Test.objects.get_or_create(email = email)
    #     # new_test.save()

    context = {"form": form}
    template = "home.html"
    return render(request, template, context)

这是我的URLs.py页面

from django.conf.urls import patterns, include, url
from django.conf import settings
from django.conf.urls.static import static

from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('',
# Examples:
url(r'^$', 'tests.views.home', name='home'),
# url(r'^home2/$', 'lwc.views.home2', name='home'),
# url(r'^blog/', include('blog.urls')),

#This admin can be changed to anything but the
#link must also be changed to access admin area
url(r'^admin/', include(admin.site.urls)),
)

我的应用程序的目标是允许用户上传文件并提交到相关的SQLite数据库 . 如何在没有MEDIA_ROOT或MEDIA_URL的情况下完成此操作？