Python Django 1.8文件上载示例

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/片段。因此，如果您能在这里发布或向我推荐一个最小但完整的（模型、视图、模板）示例代码，我将不胜感激

我在网上看到了许多其他的例子，但我注意到我的版本在默认情况下没有MEDIA\u ROOT或MEDIA\u 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URL.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)),
)

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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。没有媒体根目录或媒体URL，我如何实现这一点？

欢迎使用Stackoverlfow！如果您包括您的输入、您尝试过的内容、您的预期输出与实际输出以及您收到的任何错误的完整堆栈跟踪，您将大大增加获得问题答案的机会。您还可以阅读您希望创建一个包含特定上传位置的字段，当您从视图中引用表单时，添加“request.FILES or None，…”；在模板中，使用enctype=“multipart/form data”设置表单。已回答此问题。请参阅此欢迎使用Stackoverlfow！如果您包括您的输入、您尝试过的内容、您的预期输出与实际输出以及您收到的任何错误的完整堆栈跟踪，您将大大增加获得问题答案的机会。您还可以阅读您希望创建一个包含特定上传位置的字段，当您从视图中引用表单时添加'request.FILES or None，…'；在模板中，使用enctype=“multipart/form data”设置表单。已回答此问题，请参见此