Hi, this is how I stream the file:
def GET(self):
# assuming we have selected doc from the database
web.header("Content-Disposition", "attachment; filename=%s" %
doc.filename)
web.header("Content-Type", doc.filetype)
web.header('Transfer-Encoding','chunked')
f = open(doc.filename, 'rb')
while 1:
buf = f.read(1024 * 8)
if not buf:
break
yield buf
On Friday, October 12, 2012 4:48:02 AM UTC+4, Jason Macgowan wrote:
>
> Hello,
>
> I have a web app that accepts requests, and based on the request it builds
> a zip file. How can I get web.py to yield this file?
>
> I tried something along the lines of
>
> def GET(self):
> return open('/path/to/file')
>
> But that doesn't work, and is probably not even close to the right
> direction to go.
>
> Any advice?
>
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