Re: Auto generated code in Python: How to specify controller name in Yaml file

2016-09-21 Thread Benjamin SOULAS
ame. > > If you find that’s not the case, please open a ticket on the codegen > project so that it can be addressed. > > > > > > *From: *<swagger-swaggersocket@googlegroups.com> on behalf of Benjamin > SOULAS <benjamin.soula...@gmail.com> > *Reply

Auto generated code in Python: How to specify controller name in Yaml file

2016-09-07 Thread Benjamin SOULAS
Hello everyone, Actually, I use the swagger editor to generate my API skeleton, but it generates only one Controller (named "default_controller.py"). I don't find any doc on "How specify a controller" for a route? The idea is to generate several controllers for the API in order to avoid a huge

Re: Swagger throws “Swagger Error Not a valid parameter definition”

2016-08-08 Thread Benjamin SOULAS
Thanks, I forgot to say I work on this editor. Le lundi 8 août 2016 10:19:37 UTC+2, tony tam a écrit : > > Change "paramType" to "in" for starters. I suggest checking syntax with > http://editor.swagger.io > > On Aug 8, 2016, at 9:54 AM, Benjamin SOULA

Swagger throws “Swagger Error Not a valid parameter definition”

2016-08-08 Thread Benjamin SOULAS
Hello people, I am trying to generate my own REST server in python Flask framework, actually I only have a GET method and a POST one. But the POST method IS the problem about paramters which seems not to correspond to the YAML Syntax in Swagger. The idea, I need the template ID which is into