Because you do not have the freedom to name files in upload folder any
way you like. web2py must name those files. replace your function with
this:
@service.xmlrpc
def checkflow(file,name):
import cStringIO
db.image.insert(file=db.image.file.store(cStringIO.StringIO(file.data),name))
return name
On Nov 25, 3:37 am, António Ramos <[email protected]> wrote:
> In this scenario where i transfer a file in uploads folder i have a problem.
>
> I cannot see or download the file
> viahttp://127.0.0.1:8000/CrossValidation1/appadmin/download/db/ico1.gif
>
> The ico1.gif is there but i get the error 404 Not found
>
> So i cannot create a link in my app to that file.
>
> What is the problem?
> best regards
> António
>
> 2010/11/25 mdipierro <[email protected]>
>
> > This reads the entire file in ram. Only ok for small files. May be a
> > vulnerability if anybody can access this service.
>
> > On Nov 24, 5:16 pm, António Ramos <[email protected]> wrote:
> > > i got xmlrpclib to transfer a file from python script to web2py app
>
> > > Here is my code
> > > *Python script*
> > > import xmlrpclib
> > > server = xmlrpclib.ServerProxy('
> >http://127.0.0.1:8000/CrossValidation1/default/call/xmlrpc')
> > > filepath='c:\\'
> > > filename='ico.gif'
> > > filebin = xmlrpclib.Binary(open(filepath+filename).read())
> > > server.checkflow(filebin,filename)
> > > *
> > > and in my default.py controller*
>
> > > import xmlrpclib
>
> > > import os
> > > @service.xmlrpc
> > > def checkflow(file,name):
> > > basedir= request.folder.replace('/','')
> > > f=open(basedir+'\\uploads\\'+name,'wb')
> > > db.image.insert(file=name)
> > > f.write(file.data)
> > > f.close()
> > > print "ok"
> > > return name
>
> > > def call():
> > > session.forget()
> > > return service()
> > > *
> > > my model*
> > > *
> > > *db.define_table('image',
> > > Field('file','upload'))
>
> > > Is this a good practice?
>
> > > 2010/11/24 mdipierro <[email protected]>
>
> > > > No
>
> > > > filename = request.vars.varwithFile.filename or 'filename.txt'
> > > > stream = request.vars.varwithFile.file or open(filename,'rb')
> > > > db.photo.insert(file=db.photo.file.store(stream),filename=filename)
>
> > > > On Nov 24, 3:37 pm, António Ramos <[email protected]> wrote:
> > > > > but if i want, using the shell, to add and image manually
> > > > > do i use
> > > > > db.photo.insert('file'=varwithFile)
> > > > > ?????
>
> > > > > *my test model is
> > > > > db.define_table('photo',
> > > > > Field('file','upload'))*
>
> > > > > Best regards
> > > > > António
>
>
