CODEWITHSUNDEEP
pythonamazon-web-servicesaws-lambdatornadozappa
user9460641
user9460641

Reputation:

Error while deploying Tornado app to AWS Lambda

I am trying a simple "Hello World" to deploy a Python-Tornado app to AWS Lambda using Zappa.

The code for the same in app.py file is:

import tornado.ioloop
import tornado.web

class MainHandler(tornado.web.RequestHandler):
    def get(self):
        self.write("Hello, world")

def make_app():
    return tornado.web.Application([
        (r"/", MainHandler),
    ])
app = make_app()
app.listen(8891)
if __name__ == "__main__":

    tornado.ioloop.IOLoop.current().start()

The error I get after I run zappa deploy dev is 

Error: Warning! Status check on the deployed lambda failed. A GET request to '/' yielded a 500 response code.

The error displyed when I run zappa tail is 

__call__() takes 2 positional arguments but 3 were given

The zappa_settings.json file is:

{
    "dev": {
        "app_function": "app.app",
        "aws_region": "ap-south-1",
        "profile_name": "default",
        "project_name": "dmi-amort",
        "runtime": "python3.6",
        "s3_bucket": "zappa-mekp987ue",
        "manage_roles": false,
        "role_name": "lambda-role",
    }
}

How can I fix this issue?

Upvotes: 2

Views: 689

Answers (1)

Ben Darnell
Ben Darnell

Reputation: 22134

Zappa is based on WSGI; Tornado is not. The two are incompatible, so you'll have to replace one of them with an alternative. (I'm not aware of a simple way to combine Tornado with Lambda, so I'd suggest using Zappa with Flask)

In older versions of Tornado, you could use WGSIApplication to get partial support for Tornado in a WSGI environment, but this is no longer available in Tornado 6.0.

Upvotes: 2

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