Re: [web2py] Process a file received via xmlrpc

Fri, 16 Sep 2011 09:06:19 -0700 

Its not excel. it will be a txt file

my problem is how do i get the file.

you say open("/uploads/%s"%fname)
I will try

Thank you

2011/9/16 Bruno Rocha <[email protected]>

> open("/uploads/%s"%fname)

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