If uploadseparate=True then you have to reconstruct the full path with
a routine like this:
def get_path(filename,folder='uploads'):
"""Get full path to file stored in uploads"""
f=filename.split('.')
path=os.path.join(request.folder,folder,
'%s.%s'%(f[0],f[1]),
'%c%c'%(f[2][0],f[2][1]),
filename)
return path
where filename passed to get_path is db.mytable[row].file_fieldname
On Dec 12, 1:07 pm, "G. Clifford Williams" <[email protected]>
wrote:
> Is ther a prescribed mechanism for getting the full path to uploaded files?
>
> I've got an onaccept function that contains:
> os.symlink(os.path.join(request.folder,'uploads',
> form.vars.file_newfilename), link_path)
>
> which worked until I'd changed the field to use 'uploadseparate=True'
>
> Thanks in advance
> --Cliff
