flask-restful

Question

问题

我的应用程序的所有路由都是通过 flask-restful 资源定义的。如何找到正在处理当前请求的资源 object/class？

为什么我想要这个

我想记录处理请求时引发的所有异常。我连接到 flask.got_request_exception，如 http://flask.pocoo.org/docs/1.0/api/#signals 中所述，类似这样的东西运行良好：

from flask import got_request_exception, request

def log_exception(sender, exception, **extra):
    logger.info("URL: {}, Exception: {}".format(request.url, type(exception).__name__))

got_request_exception.connect(log_exception, app)

唯一的问题是我想记录一些请求数据，但不是所有数据 - 例如我想隐藏密码。我认为将日志记录数据逻辑与请求处理逻辑结合在一起是个好主意，如下所示：

from flask import request
import flask_restful

class SomeResource(flask_restful.Resource):
    def get(self):
        # ... GET processing
    def log_data(self):
        # log all body params
        return request.get_json()

class Login(flask_restful.Resource):
   def post(self):
       # ... authentication
   def log_data(self):
       # log selected body params
       return {'login': request.get_json()['login'], 'password': 'HIDDEN!'}

然后在我的 log_exception:

中使用它

from flask import got_request_exception, request

def log_exception(sender, exception, **extra):
    resource_class = # THIS IS THE THING I'M MISSING
    logger.info("URL: {}, Exception: {}, Data: {}".format(request.url, type(exception).__name__), 
                resource_class.log_data())

got_request_exception.connect(log_exception, app)

但也许这应该以其他方式完成？

Answer 1

一旦信号调用了日志记录方法，我就找不到访问对象的好方法。

如果您愿意处理所有可能的情况，您可以创建自己的自定义异常，在 class.

上调用 log_data 方法

相反，我选择处理自己在基地 class 中的记录。给你的简单例子，我只是使用了打印函数，但你可以调用 app.logging.info 来代替。

from flask import Flask, request
import flask_restful

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

class MyCustomResource(flask_restful.Resource):
    def get(self):
        try:
            self.my_get()
        except Exception as exception:
            # Will catch all errors in your subclass my_get method
            print("exception caught")
            print(request.url)
            print(type(exception).__name__)
            print(self.log_data())

            # Re-raise if you want (or not)
            raise exception

    def my_get(self):
        # Python equivalent of virtual method
        raise NotImplementedError()

    def log_data(self):
        # Python equivalent of virtual method
        raise NotImplementedError()


class SomeResource(MyCustomResource):
    def my_get(self):
        # Example unknown error occurs here
        raise Exception("error!")
        return "some data"

    def log_data(self):
        # Called by parent
        return "some logging data for the object"

api.add_resource(SomeResource, "/")

如果您想更深入地研究 flask restful 源代码，您还可以猴子修补（或构建您自己的包）任何方法正在调用 get/post。

Answer 2

您不想从 flask_restful.Resource 继承，而是想从自定义资源

继承全部

class MyResource(flask_restful.Resource):
    def dispatch_request(self, *args, **kwargs):
        try:
            return super(MyResource,self).dispatch_request(*args, **kwargs)
        except Exception as ex:
            setattr(ex, "_raised_by", self)
            raise ex

然后你可以使用异常处理器

def log_exception(sender, exception, **extra):
    _raised_by = getattr(exception, "_raised_by", None)
    if _raised_by:
        print(_raised_by)
    property("URL: {}, Exception: {}".format(request.url, type(exception).__name__))

这是我试过的完整代码

from flask import request, Flask
import flask_restful

app = Flask(__name__)

api = flask_restful.Api(app)


class MyResource(flask_restful.Resource):
    def dispatch_request(self, *args, **kwargs):
        try:
            return super(MyResource,self).dispatch_request(*args, **kwargs)
        except Exception as ex:
            setattr(ex, "_raised_by", self)
            raise ex

# MyResource = flask_restful.Resource

class SomeResource(MyResource):
    def get(self):
        raise Exception("Not implemented")

    def log_data(self):
        # log all body params
        return request.get_json()


class Login(MyResource):
    def post(self):
        raise Exception("Not implemented")

    def log_data(self):
        # log selected body params
        return {'login': request.get_json()['login'], 'password': 'HIDDEN!'}


from flask import got_request_exception, request

api.add_resource(Login, '/login')
api.add_resource(SomeResource, '/some')


def log_exception(sender, exception, **extra):
    _raised_by = getattr(exception, "_raised_by", None)
    if _raised_by:
        print(_raised_by)
    property("URL: {}, Exception: {}".format(request.url, type(exception).__name__))


got_request_exception.connect(log_exception, app)

if __name__ == '__main__':
    app.run(debug=True)

Edit-1：8 月 5 日

正如@jbet 所评论的，如果有人想要处理 class 总是一个更清晰的选择是使用 MyResource 如下

from flask import g

class MyResource(flask_restful.Resource):
    def dispatch_request(self, *args, **kwargs):
        g.processed_by = self
        return super(MyResource,self).dispatch_request(*args, **kwargs)

flask-restful - 当前请求的资源 class

flask-restful - resource class for current request

python

flask