[web2py] Re: jsonrpc call failure
Can you describe what happens in the uncessful test? Do you see a new
ticket file being generate in the errors/ folder? What does it say?
On May 11, 7:46 pm, Miguel Lopes mig.e.lo...@gmail.com wrote:
I've managed to successfully call on a jsonrpc service using localhost.
However I cannot replicate this with a remote server.
For the client I'm using jsonrpc at the Python prompt:
sv = ServiceProxy('http://127.0.0.1:8000/testapp/default/call/jsonrpc')
sv.getRequests()
['json test']
The very standard controller is:
@service.jsonrpc
def getRequests():
return [json test]
def call():
session.forget()
return service()
Notably I'm running web2py 1.74.4 on OSX on the localhost - the successful
test
And web2py 1.65.5 on Debian on the remote server - the unsuccessful test
I believe 1.65.5 supports jsonrpc and that this is served by default on port
8000. Am I wrong? What am I missing?
Miguel
Re: [web2py] Re: jsonrpc call failure
On Wed, May 12, 2010 at 1:48 AM, mdipierro mdipie...@cs.depaul.edu wrote: Can you describe what happens in the uncessful test? Do you see a new ticket file being generate in the errors/ folder? What does it say? Upps! Sorry false alarm. I was calling the remote server with the port and this was causing the connection to be refused. Unfortunately I've been doing variations on this for about 20 minutes or so!!! Sorry for this and Txs for the help! Miguel
[web2py] Re: jsonrpc call failure [FALSE ALARM]
:-) On May 11, 7:55 pm, Miguel Lopes mig.e.lo...@gmail.com wrote: On Wed, May 12, 2010 at 1:48 AM, mdipierro mdipie...@cs.depaul.edu wrote: Can you describe what happens in the uncessful test? Do you see a new ticket file being generate in the errors/ folder? What does it say? Upps! Sorry false alarm. I was calling the remote server with the port and this was causing the connection to be refused. Unfortunately I've been doing variations on this for about 20 minutes or so!!! Sorry for this and Txs for the help! Miguel
