On Sun, 27 Dec 2009 11:46:50 -0800 (PST)
mdipierro <mdipie...@cs.depaul.edu> wrote:
> import xmlrpclib
> server=xmlrpclib.ServerProxy('http://127.0.0.1:8000/app/xmlrpctest/
> call/xmlrpc')
> print str(server.add(3,4)+server.sub(3,4))
>
I found my mistake.
The right format for calling is as below:
server=xmlrpclib.ServerProxy('http://localhost/test/xmlrpctest/call/xmlrpc')
print str(server.add(3,4))
Indeed it is also explained at the web2py tutorials but it seems it may cause a
confusion. It says,
server = ServerProxy('http://..../app/default/call/xmlrpc')
...: The domain name of the application, in my case it was localhost
app: application name
default: the name of your controller
call: the function name that is returning service() function
xmlrpc: use as it is
There is no need to mention the port number as 8000 also.
xmlrpc service calling is working with the same idea as a regular web2py
application. The idea is simple, just call the funtion of the controller with
the suitable parameters.
> (assuming the port is 8000).
--
Oguz Yarimtepe <oguzyarimt...@gmail.com>
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