我正在尝试在用户的计算机上创建用于下载上传文件的脚本。问题是下载根本不起作用(它要么下载一个空文件,要么给我一些错误)。
最后一个错误是: 强制转换为Unicode:需要字符串或缓冲区,找到FieldFile
def download_course(request, id):
course = Courses.objects.get(pk = id).course
path_to_file = 'root/cFolder'
filename = course # Select your file here.
wrapper = FileWrapper(file(course))
content_type = mimetypes.guess_type(filename)[0]
response = HttpResponse(wrapper, content_type = content_type)
response['Content-Length'] = os.path.getsize(filename)
response['Content-Disposition'] = 'attachment; filename=%s/' % smart_str(course)
return response
如何正确声明文件名,以便每次知道要下载的文件时: 文件名实际上是“课程”,如上所述
谢谢!
答案 0 :(得分:5)
<强>编辑强>
我认为您需要从FileField对象中提取path
值:
def download_course(request, id):
course = Courses.objects.get(pk = id).course
path = course.path # Get file path
wrapper = FileWrapper( open( path, "r" ) )
content_type = mimetypes.guess_type( path )[0]
response = HttpResponse(wrapper, content_type = content_type)
response['Content-Length'] = os.path.getsize( path ) # not FileField instance
response['Content-Disposition'] = 'attachment; filename=%s/' % \
smart_str( os.path.basename( path ) ) # same here
return response
为什么:
假设我有(好吧,我实际上)模型:
class DanePracodawcy( DaneAdresowe, DaneKontaktowe ):
# other fields
logo = ImageWithThumbnailsField( upload_to = 'upload/logos/',
thumbnail = {'size': (180, 90)},
blank = True )
ImageWithThumbnailsField
是FileField的子类,因此它的行为方式相同。现在,当我做SELECT时:
mysql> select logo from accounts_danepracodawcy;
+-----------------------------+
| logo |
+-----------------------------+
| upload/logos/Lighthouse.jpg |
+-----------------------------+
1 row in set (0.00 sec)
它显示(相对于MEDIA_ROOT)存储文件的路径。但是当我访问logo
模型属性时:
[D:projekty/pracus]|1> from accounts.models import DanePracodawcy
[D:projekty/pracus]|4> DanePracodawcy.objects.get().logo
<4> <ImageWithThumbnailsFieldFile: upload/logos/Lighthouse.jpg>
[D:projekty/pracus]|5> type( _ )
<5> <class 'sorl.thumbnail.fields.ImageWithThumbnailsFieldFile'>
我得到一些对象的实例。如果我尝试将该实例传递给os.path.getsize
:
[D:projekty/pracus]|8> import os.path
[D:projekty/pracus]|9> os.path.getsize( DanePracodawcy.objects.get().logo )
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
D:\projekty\pracus\<ipython console> in <module>()
C:\Python26\lib\genericpath.pyc in getsize(filename)
47 def getsize(filename):
48 """Return the size of a file, reported by os.stat()."""
---> 49 return os.stat(filename).st_size
50
51
TypeError: coercing to Unicode: need string or buffer, ImageWithThumbnailsFieldFile found
我和你一样得到TypeError。所以我需要文件路径作为字符串,可以使用path
属性获取:
[D:projekty/pracus]|13> os.path.getsize( DanePracodawcy.objects.get().logo.path )
<13> 561276L
或者,我可以使用MEDIA_ROOT设置获取name
属性和os.path.join
:
[D:projekty/pracus]|11> from django.conf import settings
[D:projekty/pracus]|12> os.path.getsize( os.path.join( settings.MEDIA_ROOT, DanePracodawcy.objects.get().logo.name ) )
<12> 561276L
但这是不必要的打字。
最后要注意:因为path
是绝对路径,我需要提取文件名以将其传递给Content-Disposition标头:
[D:projekty/pracus]|16> DanePracodawcy.objects.get().logo.path
<16> u'd:\\projekty\\pracus\\site_media\\upload\\logos\\lighthouse.jpg'
[D:projekty/pracus]|17> os.path.basename( DanePracodawcy.objects.get().logo.path )
<17> u'lighthouse.jpg'
答案 1 :(得分:2)
除非您让用户下载动态生成的文件,否则我不明白您为什么要这样做。
您可以让此视图重定向到相应的路径,并且相应的标头由提供静态文件的服务器设置;通常是apache或nginx
我的观点如下:
from django.conf import settings
def download_course(request,id):
course = get_object_or_404(Course,id=id)
filename = course.course
return redirect('%s/%s'%(settings.MEDIA_URL,filename))
享受:)