On Friday, April 13, 2018 at 1:15:36 PM UTC-7, Multiplery wrote:
>
> Hey folks,
>
> I'm attempting to upload a file directly without the use of a SQLForm but
> rather the ordinary FORM.
> From previous threads regarding the topic i figured my controller should
> look something like this:
>
> [code]
> def index():
>
> import shutil
> form = FORM(
> INPUT( _name="upload", _type="file"),
> INPUT( _type="Submit", _value="Upload" ))
> if form.accepts(request,session):
> if form.vars:
>
> filename=request.vars.upload.filename
> file=request.vars.upload.file
> shutil.copyfileobj(file,open('/path'+filename,'wb'))
> return dict (form = form)
> [/code]
>
> yet no file appears in the folder of my path. I'm not sure if the problem
> is with my code or maybe elsewhere, for example permission based. I'm using
> a raspberry with a raspiban os. i Already tried to start python as root
> user.
>
> Can someone spot a mistake in the code? has someone tried to run this on a
> debian based os and knows if its a permission problem? I'm aware of the
> potential dangers of the shutil.copyfileobj, so no worries about that.
>
> Sincerely,
>
>
Shouldn't you be passing the [relevant] request.post_vars to form.accepts() ?
See the top-level paragraph for *FORM* in chapter 7, and also
<URL:http://web2py.com/books/default/chapter/29/07/forms-and-validators#The-process-and-validate-methods>
/dps
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