Question

为了避免为我的模型编写api方法视图。我希望创建一种更优雅的继承方式。

class ModelCrudAPI(MethodView):

    def __init__(self, model):
      self.model = model

    def get(self):
        store_id = request.args.get("store_id", None, int)
        waiters = self.model.query.filter_by(store_id=store_id).all()

        if waiters is None or store_id is None:
            return jsonify({'items': [], 'store_id': store_id}), 204

        # TODO: change personnel_data to list
        else:
            items = []
            for waiter in waiters:
                items.append(row2dict(waiter))
            return jsonify({"items": items, 'store_id': store_id})

我之前在这个问题中设置过类似的继承。 previous question about inheritance and decorators in methodviews

注意我想将sqlalcdhemy模型传递给我的班级。

第一个场景

class Test(ModelCrudAPI):

    def __init__(self, model):
        super(Test, self).__init__(model)

api.add_url_rule("/test", view_func=Test(MYMODEL).as_view("test"), methods=['GET','POST','PUT','DELETE'])

ERROR：

flask debuger :TypeError: __init__() takes exactly 2 arguments (1 given)

第二种情景）

class Test(ModelCrudAPI):

    def __init__(self, MYMODEL):
        super(Test, self).__init__(MYMODEL)

api.add_url_rule("/test", view_func=Test.as_view("test"), methods=['GET','POST','PUT','DELETE'])

ERROR：

flask debuger :TypeError: __init__() takes exactly 2 arguments (1 given)

我真的会帮助你。这将清理我的代码。

编辑此作品

class Test(ModelCrudAPI):
    def __init__(self):
        ModelCrudAPI.__init__(self, Personnel)


api.add_url_rule("/test", view_func=Test.as_view("test"), methods=['GET','POST','PUT','DELETE'])

Answer 1

as_view is a classmethod, and it forwards arguments provided to it to the class' constructor. So all you have to do is pass your model as the second argument to as_view:

class Test(ModelCrudAPI):
    def __init__(self, model):
        super(Test, self).__init__(model)

api.add_url_rule("/test", view_func=Test.as_view("test", MYMODEL), methods=['GET','POST','PUT','DELETE'])

如何在Flask中设置一个Inherited MethodView来对sqlalchemy模型进行CRUD操作

1 个答案: