jwt is the emerging standard for apis and such, but if the usage is 
all-internal, why don't you just start simple and do


if request.vars.token not in ('aaaaxa', 'bbbbbbxb', 'cccccc'):
     raise HTTP(403)

 ?

On Wednesday, September 28, 2016 at 2:15:19 PM UTC+2, Marlysson Silva wrote:
>
> This is what you need:
>
> http://web2py.readthedocs.io/en/latest/tools.html#gluon.tools.Auth.jwt
>
> Json Web Token , communications to services "rest" .. I think that 
> facilitates your work
>
> Em quarta-feira, 28 de setembro de 2016 08:57:59 UTC-3, Lisandro escreveu:
>>
>> Hi there!
>> I have a web2py app that implements a JSON-RPC public webservice, over 
>> HTTPS.
>> It's public as it doesn't require user and password to authenticate.
>>
>> However, the webservice is only consumed by a set of other applications 
>> that I manage.
>> So I would like those apps to use a key in order to connect to the 
>> webservice.
>> I'm trying to do such thing with a custom decorator, but I can't make it 
>> work.
>>
>> This is what I have:
>>
>> from gluon.tools import Service
>>
>>
>> service = Service()
>>
>>
>> def validate_key():
>>     return True
>>
>>
>> @auth.requires(lambda: validate_key())
>> def call():
>>     session.forget()
>>     return service()
>>
>>
>> @service.jsonrpc
>> def test():
>>     return 'test ok'
>>
>>
>>
>> But when I try to connect to the webservice, I receive this error:
>>
>> ProtocolError: <ProtocolError for dev.medios/ws/call/jsonrpc: 303 SEE OTHER>
>>
>>
>> I'm not sure if what I'm tying to do is possible in that way. 
>> I know that I can apply the @auth.requires_login() decorator to the 
>> call() method, but that would force me to create users for every app using 
>> the webservice, and that's not wat I want. I would like to simply generate 
>> a key for every app that needs to use the webservice, and then ask the 
>> applications to use that key (either in the call to connect to the 
>> webservice, or in every call to any method of the webservice).
>>
>> What would be the correct approach?
>> Thanks in advance.
>> Regards, 
>> Lisandro.
>>
>

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