Question

我在我的＆＃34; test＆＃34;中为我的博客应用创建了ImageField模型。使用sqllite在我的本地服务器上的django项目。

我有我的settings.py

MEDIA_ROOT = '/Users/me/Sites/python/djangotut/media/' MEDIA_ROOT_URL = 'http://127.0.0.1:8000/media/images/photos/'

和我的博客/ models.py

photo = models.ImageField(upload_to='images/photos/')

但问题是我的blog.urls.py我不知道如何添加网址来处理文档https://docs.djangoproject.com/en/1.6/howto/static-files/#serving-files-uploaded-by-a-user-during-development

中的模式

from django.conf.urls import url
from django.conf.urls.static import static

from .views import index, post


urlpatterns = [
    url(
        regex=r'^$',
        view=index,
        name='blog-index'
    ),
    url(
        regex=r'^(?P<slug>[\w\-]+)/$',
        view=post,
        name='blog-detail'
    ),

]

此外，我已经阅读了有关为生产环境设置网址的信息。当您分发应用程序时。在这种情况下，我的网址需要看起来像什么？

Answer 1

更新。看一下：https://docs.djangoproject.com/en/dev/ref/models/fields/。

models.py

UPLOAD_PATH = "photos"
photo = models.ImageField(upload_to=UPLOAD_PATH)

settings.py

from path import path
PROJECT_DIR = path(__file__).dirname().parent
MEDIA_ROOT = PROJECT_DIR / 'media'
MEDIA_URL = '/images/'

urls.py

from django.conf import settings
urlpatterns += static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

Answer 2

我发现我所寻找的与你的答案非常相似。特别是设置中的MEDIA_ROOT和MEDIA_URL不是确切的URL。

settings.py

MEDIA_ROOT = os.path.join(BASE_DIR, 'media')
MEDIA_URL = '/media/'

urls.py

from django.conf.urls import include, url
from django.conf.urls.static import static
from django.conf import settings
from django.contrib import admin
admin.autodiscover()

urlpatterns = [
    url(r'^polls/', include('polls.urls')),
    url(r'^newsletter/', include('newsletter.urls')),
    url(r'^blog/', include('blog.urls')),
    url(r'^admin/', include(admin.site.urls))
] + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

models.py保持不变。

Django应用程序网址为媒体

2 个答案: