Reputation: 1689
I am running a Django app where I can upload files. Now I want to download the files using requests. I was trying to create a view where the file is downloaded, so I can then make a call with requests. But it doesn't quite work
My model:
class FileCollection(models.Model):
name = models.CharField(max_length=120, null=True, blank=True)
store_file = models.FileField(storage=PrivateMediaStorage(), null=True, blank=True)
creation_date = models.DateTimeField(null=True, blank=True)
My views
def fileview(request, *args, **kwargs):
file = FileCollection.objects.first()
file_path = file.store_file
print(file_path)
FilePointer = open(file_path, "r")
response = HttpResponse(FilePointer, content_type='application/msexcel')
response['Content-Disposition'] = 'attachment; filename=NameOfFile'
return response
It tells me that TypeError: expected str, bytes or os.PathLike object, not FieldFile
If I pass in the url provided in the apiview/admin I get: FileNotFoundError: [Errno 2] No such file or directory
Also tried:
def fileview(request):
path = FileCollection.objects.first()
obj = path.store_file
o = str(obj)
file_path = os.path.join(settings.MEDIA_ROOT, o)
print(file_path)
if os.path.exists(file_path):
with open(file_path, 'rb') as fh:
response = HttpResponse(fh.read(),
content_type="application/vnd.ms-excel")
response[
'Content-Disposition'] = 'inline; filename=' + os.path.basename(
file_path)
return response
but this gives me ValueError: The view file_storage_api.api.v1.views.fileview didn't return an HttpResponse object. It returned None instead.
Is that the right way to go?
I'm very grateful for help or hints. Thanks so much
Upvotes: 0
Views: 733
Reputation: 1435
file_path = file.store_file
is not a file path but an instance of a FileField try using
file_path = file.store_file.name
and use the second snippet
EDIT: here is the code I use: l_targetFile is the path to the actual file
l_prjPath = os.path.realpath(os.path.dirname(__file__)).replace(<adapt the path here>)
l_userFileName= <file field>.name.replace('<upload to sub-dir>','')
l_targetFile = l_prjPath + '/media/' + l_fileObj.file_obj.name
#return the file
response = FileResponse(open(l_targetFile, 'rb'),\
(l_responseDisposition == 'attachment'))
#process the filename as stored on the local machine in case of download
try:
#check if it will throw
l_tmpUserName = l_userFileName.encode('ascii')
#no error use the non-encoded filename
l_fileExpr = 'filename="{0}"'.format(l_userFileName)
except UnicodeEncodeError:
# Handle a non-ASCII filename
l_fileExpr = "filename*=utf-8''{}".format(quote(l_userFileName))
response['Content-Disposition'] = '{0};{1};'.format(l_responseDisposition,l_fileExpr)
if '.pdf' in l_userFileName:
response['Content-Type'] = 'application/pdf'
elif l_dataSource == CMSArchiveEntry.SOURCE_TAG:
response['Content-Type'] = 'text/html; charset=utf-8'
return response
Upvotes: 1