尝试运行 flask 应用程序时出现错误“功能”object 没有属性“as_view”

Question

I started writing flask app after a long time more than a year, guess I have forgot something. This below code results in an error:

from flask import Flask
from flask import jsonify

from flask_restplus import Resource, Api

from home_iot.config import reader
from download_audio.ydla import download


app = Flask(__name__)

_api = Api(app, catch_all_404s=True, version=0.1,
          title="REST HTTP API's Gateway",
          descrition="REST API gateway")


api_ns = _api.namespace("iot", description="API.")


@api_ns.route("/tcpserver", methods=["GET"])
def advertise_tcpserver():
    ip = reader.get_server_ip()
    return jsonify({"tcpserver": ip})


if __name__ == "__main__":
    app.run(host='127.0.0.1')

Error is:

$ python app.py

Traceback (most recent call last):
  File "app.py", line 29, in <module>
    @api_ns.route("/tcpserver", methods=["GET"])
  File "/Users/ciasto/pyenvs/flaskrestplusiot/lib/python2.7/site-packages/flask_restplus/namespace.py", line 98, in wrapper
    self.add_resource(cls, *urls, **kwargs)
  File "/Users/ciasto/pyenvs/flaskrestplusiot/lib/python2.7/site-packages/flask_restplus/namespace.py", line 87, in add_resource
    api.register_resource(self, resource, *ns_urls, **kwargs)
  File "/Users/ciasto/pyenvs/flaskrestplusiot/lib/python2.7/site-packages/flask_restplus/api.py", line 264, in register_resource
    self._register_view(self.app, resource, namespace, *urls, **kwargs)
  File "/Users/ciasto/pyenvs/flaskrestplusiot/lib/python2.7/site-packages/flask_restplus/api.py", line 287, in _register_view
    resource_func = self.output(resource.as_view(endpoint, self, *resource_class_args,
AttributeError: 'function' object has no attribute 'as_view'

Answer 1

Don't think that's the correct way to define the namespace with flask_restplus . Have a look at scaling docs .

You're probably looking for something like:

iot.py

from flask_restplus import Namespace

api_ns = Namespace("iot", description="API.")

@api_ns.route("/tcpserver", methods=["GET"])
def advertise_tcpserver():
    ip = reader.get_server_ip()
    return jsonify({"tcpserver": ip})

Then in your main app.py :

# other imports 

from .iot import api_ns

app = Flask(__name__)

_api = Api(app, catch_all_404s=True, version=0.1,
          title="REST HTTP API's Gateway",
          descrition="REST API gateway")

_api.add_namespace(api_ns, path='/some/prefix')

Also you appear to be using Python 2.7 which has been discontinued. I'd suggest upgrading to the latest version, using either a virutal environment or docker so as not to mess with your system's python.

Answer 2

Hope this can helps those who have this same error and have not found the solution

To complete the answer given by @v25 you must provide ressources to your namespace by inheriting from Ressource class in flask_restplus.

The following example works for me

Environment:

• ubuntu 18.04
• python 3.7.1

python requirements:

• flask==1.1.2
• flask-restplus==0.13.0
• werkzeug==0.16.1

Source code: iot.py

from flask_restplus import Namespace,Resource

api_ns = Namespace("iot", description="API.")

@api_ns.route("/tcpserver")
class AdvertiseTcpserver(Resource):
    def get(self):
        #TODO return the correct ip value
        return {"tcpserver": "ip"}

app.py

from .iot import api_ns
from flask import Flask
from flask_restplus import Api

app = Flask(__name__)

_api = Api(app, catch_all_404s=True, version=0.1,
      title="REST HTTP API's Gateway",
      descrition="REST API gateway")

_api.add_namespace(api_ns, path='/some/prefix')

app.run()

Test command:

#!/bin/sh
wget localhost:5000/some/prefix/tcpserver

Please let me know if this helped.