Question

我是 Django框架的新手，我试图生成一个从 Amazon S3 下载文件的链接。我尝试加载模板页面时收到此错误：

Reverse for 'myapp.views.handles3downloads' with arguments '(u'README.md',)'
and keyword arguments '{}' not found.

urls.py

urlpatterns = patterns('',
    url(r'^handles3downloads/(\d+)/$', handles3downloads),
)

views.py

def handles3downloads(request, fname):
    bucket_name = 'bucketname'
    key = s.get_bucket(bucket_name).get_key(fname)
    dfilename = key.get_contents_to_filename(fname)

    wrapper = HttpResponse(file(dfilename))
    response = HttpResponse(wrapper, content_type='text/plain')
    response['Content-Length'] = os.path.getsize(dfilename)
    return response

模板文件

<a href="{% url 'myapp.views.handles3downloads' sfile.linkUrl %}">{{sfile.linkUrl}}</a>

我查看了一些类似错误的解决方案，但它对我没有帮助。任何人都可以帮帮我。

提前谢谢

Answer 1

urls.py文件中的正则表达式似乎有误。请尝试使用此代码：

url(r'^handles3downloads/(\w+)/$', handles3downloads),

您将参数字符串传递给视图，regex匹配整数。

Answer 2

乔丹是对的，你的urls.py有问题。你可以通过错误来判断。您试图在“myapp.views.handles3downloads”上获得反转，但是已经识别出反向字符串了吗？试试这个。

urlpatterns = patterns('',
     url(r'^handles3downloads/([^/]+)/$', handles3downloads,
     name='myapp.views.handles3downloads'),
)

Answer 3

urls.py

url(r'^handles3downloads/', handles3downloads),

views.py

def handles3downloads(request):
  fname = request.GET['filename']
  bucket_name = 'bucketname'
  key = s.get_bucket(bucket_name).get_key(fname)
  key.get_contents_to_filename('/tmp/'+key.name)
  wrapper = FileWrapper(open('/tmp/'+fname, 'rb'))
  content_type = mimetypes.guess_type('/tmp/'+fname)[0]
  response = HttpResponse(wrapper,content_type=content_type)
  response['Content-Length'] = os.path.getsize('/tmp/'+fname)
  response['Content-Disposition'] = 'attachment; filename=%s' % smart_str(fname)

模板

<a href="/handles3downloads/?filename=file1.jpg" rel="external">Download</a>

使用django从amazon s3下载文件

3 个答案: