Return File

[Jason Macgowan]

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?

[Andrey Kuzmin]

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