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

Question

我正在使用 Flask-Restful 开发 Restful API，其资源比我想在 app.py 中保留的资源多。所以我应用了建议的项目结构。现在我想从资源访问api.url_for()from app import api以生成一些链接，但似乎我必须这样做。

为了避免循环导入，我当前的解决方案是进行惰性导入。但还有更好的方法，对吗？

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分别）：

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该线下方api = Api(app)：

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名称已经定义，您可以安全地导入from app import api您的resources模块：

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 文档中的大型应用程序模式。