Now I have in controller:
def upload():
form = FORM("Upload
file:",INPUT(_type='file',_name='myfile'),INPUT(_type='submit',_name='submit',_value='Submit'))
if request.vars:
if form.accepts(request,session):
my_file = request.vars.myfile.file
my_filename = request.vars.myfile.filename
filepath = os.path.join(request.folder, 'uploads') // this
path work - it is absolute path in hard drive
fp =open(filepath.my_filename,'wb')
fp.write(my_file)
fp.close()
return dict(form=form)
request.vars.myfile -> make return on stored object
request.vars.myfile.file -> make return address of stored object
I understood what you told me, and logical check on conditions, but
how to put everything in one function.
the pseudo algorithm in my head is next:
make def func():
make form = FORM(...)
check if condition existing
make file = request.vars.myfile.file
make filename = request.vars.myfile.filename
make filepath = os.path.join(...)
make write file to filepath filename
go to page and show empty form - wait for new file
else:
just show empty form without submission
return form
but I just loss my mind after 24 ours of trying this or similar
uploads.
Miroslav
On Nov 19, 6:12 pm, Anthony <[email protected]> wrote:
> You can tell if the function is being called with a form submission by
> checking for request.vars:
>
> if request.vars:
> print 'this is a form submission'
>
>
>
>
>
>
>
> On Saturday, November 19, 2011 12:06:40 PM UTC-5, miroslavgojic wrote:
>
> > The error is caused when file is not selected.
> > By default on first run form is empty (file is not selected), and form
> > must wait for selecting and submitting.
>
> > How access to file before calling form? What that mean?
>
> > Miroslav
>
> > On Nov 19, 5:52 pm, Anthony <[email protected]> wrote:
> > > You might need to access the file before calling form.accepts (first
> > you'll
> > > have to check that form.vars.myfile exists). You can also access it via
> > > request.vars.myfile (which won't change, even after form.accepts).
>
> > > Anthony
