如何在 big Flask 中访问 api.url_for-Restful API

Question

我正在使用 Flask-Restful 开发 Restful API，资源比我想保留在 app.py 中的资源多。所以我应用了资源中建议的 project structure. Now I'd like to access api.url_for() 来生成一些链接，但似乎我必须 from app import api 才能做到这一点。

为了避免循环导入，我目前的解决方案是进行惰性导入。但是有更好的方法，对吧？

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（分别为bar.py）：

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

您可以将导入向下移动到 api = Api(app) 行下方：

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>')

现在 api 名称已定义，您可以安全地在 resources 模块中导入 from app import api：

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

请参阅 Flask 文档中的 Larger Applications pattern。

如何在 big Flask 中访问 api.url_for-Restful API

How to access api.url_for in big Flask-Restful API

python

flask

python-import

flask-restful