On Thu, Feb 12, 2009 at 8:02 AM, huw_at1 <huwdjo...@gmail.com> wrote:

>
> Hi all,
>
> I am trying to write a simple app that allows a user to select a file
> (via the browse button) from which data can be read in and parsed. I
> have read several examples but none seem to do exactly what I want. I
> have built a form and so am trying to use forms.FileField(). I can add
> this field no problem to my form. However I am not sure I understand
> how I am supposed to pull the data contained within the file from
> within my views. I tried:
>
> ....
> file = forms.FileField()
>
> ....
>
> if request.FILES:
>   filename = request.FILES['file']
>
>   ...
>
> However I cant seem to get this to work. Does anyone have any good
> tutorials or can offer any suggestions?
>

Can't get it to work means what, exactly?  Your get an error message?  Your
filename variable winds up being set to to something you aren't expecting?
Your computer catches fire?

This page:

http://docs.djangoproject.com/en/dev/topics/http/file-uploads/

discusses handling uploaded files in some detail.  As described there,
request.FILES['file'] will be an UploadedFile object, so your assignment of
it to a variable named 'filename' rather sounds like you have misunderstood
something there, since request.FILES['file'] will be more than the file's
name, it will be an object with methods and attributes that (as detailed on
that page) that let you access the uploaded file's name, size, and data
(either in chunks or in one piece).  There's an example handle_uploaded_file
function that iterates through the data chunks and does something with
them.  Obviously it's not going to do exactly what you want to do with your
file data, but that page lays out a whole working structure and would seem
to be a good starting point for you to use in developing your specific
file-handling code.

Karen

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