功能这是当前项目stucture如何调用内部view.py
R6Scorextractor
R6Scoreex
migrations
templates
R6Scoreex
header.html
home.html
__Init__.py
settings.py
urls.py
views.py
models.py
apps.py
admin.py
tests.py
R6Scorextractor
__Init__.py
settings.py
urls.py
manage.py
R6Scorextractor/R6scoreex/urls.py
from django.conf.urls import url
from . import views
from django.conf.urls import include
urlpatterns = [
url(r'^$', views.index, name='index'),
]
R6Scorextractor/R6scoreex/views.py
# -*- coding: utf-8 -*-
from __future__ import unicode_literals
from django.shortcuts import render
from django.shortcuts import render
from django.conf import settings
from django.core.files.storage import FileSystemStorage
# Create your views here.
from django.http import HttpResponse
import pdb;
def index(request):
return render(request, 'R6scoreex/home.html')
def simple_upload(request):
print "Entered simple_upload"
if request.method == 'POST' and request.FILES['myfile']:
myfile = request.FILES['myfile']
fs = FileSystemStorage()
filename = fs.save(myfile.name, myfile)
uploaded_file_url = fs.url(filename)
return render(request, 'R6scoreex/home.html', {
'uploaded_file_url': uploaded_file_url
})
return render(request, 'R6scoreex/home.html')
R6Scorextractor/R6Scorextractor/url.py
from django.conf.urls import url
from django.contrib import admin
from django.conf.urls import include
urlpatterns = [
url(r'^admin/', admin.site.urls),
url(r'', include('R6scoreex.urls')),
]
我只是想知道如何称呼里面R6scoreex module.How的views.py simple_upload写URL它,服务器给了我404错误,当我与去以下
url(r'^/simple_upload/$', views.simple_upload, name='simple_upload'),
那么,什么是我得到404错误的原因,甚至加入上面的代码我在做什么错在这里
后
嗨!你写了这个url(r'^/simple_upload $ /',views.simple_upload,name ='simple_upload'),其中的urls.py?正确的是R6Scorextractor/R6scoreex/urls.py尝试删除正则表达式部分的$和/以检查它是否有效。你也不需要导入包括那里。 – Tico
是的,我写了R6scoreex/urls.py我写了这个\t url(r'^ simple_upload/$',views.simple_upload,name ='simple_upload'),它给了我404。超过24小时。 –