如何在Flask-Restful API中访问api.url_for

Question

I'm working on a Restful API using Flask-Restful with more resources than I'd like to keep in my app.py. So I've applied the suggested project structure . Now I'd like to access api.url_for() from a resource to generate some links, but it seems I'd have to from app import api to do that.

To avoid circular imports, my current solution is to do a lazy import. But there has be a better way, right?

app.py :

from flask import Flask
from flask_restful import Api
from myapi.resources.foo import Foo
from myapi.resources.bar import Bar
from myapi.resources.baz import Baz

app = Flask(__name__)
api = Api(app)

api.add_resource(Foo, '/Foo', '/Foo/<str:id>')
api.add_resource(Bar, '/Bar', '/Bar/<str:id>')
api.add_resource(Baz, '/Baz', '/Baz/<str:id>')

resource/foo.py ( bar.py respectively):

from flask_restful import Resource
from bar import Bar

class Foo(Resource):
    def get(self):
        from app import api
        related = api.url_for(Bar, foo=self.id)
        return {'Foo':self.id, 'related_bar':related}, 200

    def post(self):
        pass

Answer 1

You can move your imports down , to below the api = Api(app) line:

from flask import Flask
from flask_restful import Api

app = Flask(__name__)
api = Api(app)

from myapi.resources.foo import Foo
from myapi.resources.bar import Bar
from myapi.resources.baz import Baz

api.add_resource(Foo, '/Foo', '/Foo/<str:id>')
api.add_resource(Bar, '/Bar', '/Bar/<str:id>')
api.add_resource(Baz, '/Baz', '/Baz/<str:id>')

Now the api name has been defined and you can safely import from app import api in your resources modules:

from flask_restful import Resource
from app import api
from bar import Bar

class Foo(Resource):
    def get(self):
        related = api.url_for(Bar, foo=self.id)
        return {'Foo':self.id, 'related_bar':related}, 200

    def post(self):
        pass

See the Larger Applications pattern in the Flask documentation.