There is any authentication on that? If yes, you should try something like
this:
server = ServerProxy('http://'+ ws_user + ':' + ws_pass + '
@127.0.0.1:8000/app22/default/call/jsonrpc
<http://127.0.0.1:8000/app22/default/call/jsonrpc>')
And If you created by yourself the WS through web2py, make sure that you
put the decorator on your service function "load". Besides, try (this time
is a guess, ok?) return your "result" variable instead of locals().
On Tuesday, July 8, 2014 7:41:44 AM UTC-3, lyn2py wrote:
>
> I'm wanting to get JSON RPC to work within web2py controller functions,
> without using pyjamas (which is the example provided in the book)
>
> This is my code for the JSON RPC call:
> def test_call():
> from gluon.contrib.simplejsonrpc import ServerProxy
> service = ServerProxy('
> http://127.0.0.1:8000/app22/default/call/jsonrpc')
> result = service.load(1000)
> return locals()
>
> (1) Running it within web2py controller functions = It returns an error:
> ValueError: No JSON object could be decoded
>
> (2) Running it in the browser =
> http://127.0.0.1:8000/app22/default/call/jsonrpc/load/1000 it runs into
> the same error as (1).
>
> (3) Running it in command prompt = it works like a charm.
>
> How do I get it to work within web2py controller functions? Thanks!
>
>
>
--
Resources:
- http://web2py.com
- http://web2py.com/book (Documentation)
- http://github.com/web2py/web2py (Source code)
- https://code.google.com/p/web2py/issues/list (Report Issues)
---
You received this message because you are subscribed to the Google Groups
"web2py-users" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
For more options, visit https://groups.google.com/d/optout.
