[Zope] extraction of complete filepath of a <input type=file> object

[Loren Stafford]

Here's the external method I wrote to do what you want:

import string, os

def getMimeType(self,file):
  __allow_access_to_unprotected_subobjects__=1
  contentHeader=file.headers.headers[1]
  mimeType=string.split(contentHeader, ': ')[1]
  return mimeType


def uploadFileName(self,client_name):
  """ Convert the file path supplied by the browser into a UNIX file name.
      We don't use any of the path except the basename, i.e. the file name
itself.
  """
  # if client_name has any \ in it, it's probably a Windows-style name
  our_name=string.replace(client_name, "\\", "/")
  base_name=os.path.basename(our_name)
  t=string.maketrans(' \'"','___')
  base_name=string.translate(base_name,t)
  return base_name

-- Loren

> -----Original Message-----
> From: Laurie Nason [mailto:laurien@tiger.3dem.bioch.bcm.tmc.edu]
> Sent: Friday, June 29, 2001 14:00
> To: Chris Withers; Loren Stafford; Andrei Belitski; zope@zope.org
> Subject: RE: [Zope] extraction of complete filepath of a <input
> type=file> object
>
>
> I have the opposite problem - I would like windoze to give me
> only the file
> name - am I going to have to parse the complete path to strip out the \'s?
>
> Laurie
>
> -----Original Message-----
> From: zope-admin@zope.org [mailto:zope-admin@zope.org]On Behalf Of Chris
> Withers
> Sent: Friday, June 01, 2001 2:06 PM
> To: Loren Stafford; Andrei Belitski; zope@zope.org
> Subject: Re: [Zope] extraction of complete filepath of a <input
> type=file> object
>
>
> > I thought REQUEST['my_file'].filename gave you the entire path. Are you
> sure
> > it doesn't.
>
> Only on Windoze, on Linux you only get the filename, which is more secure
> IMHO.
>
> Why do you want the complete file path anyway?
>
> cheers,
>
> Chris
>
>
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