Question

问题

我所有应用的路由都是通过flask-restful资源定义的。如何找到正在处理当前请求的资源对象/类？

我为什么要这么做

我想记录处理请求时引发的所有异常。如http://flask.pocoo.org/docs/1.0/api/#signals所述，我连接到flask.got_request_exception，类似这样的方法很有效：

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，但不是全部数据-例如我想隐藏密码。我认为最好将Logging-data-logic和请求处理逻辑一起使用，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)

但是也许应该以其他方式完成？

Answer 1

一旦信号调用了日志记录方法，我就找不到找到访问对象的好方法。

如果您愿意处理所有可能的情况，则可以创建自己的自定义异常，以在类上调用log_data方法。

相反，我选择处理自己在基类中的记录。为您提供的简单示例，我只使用了打印功能，但是您可以改为调用app.logging.info。

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烧瓶中的宁静源，那么无论调用get / post的哪种方法，您都可以猴子补丁（或构建自己的包）。

Answer 2

您想从自定义资源中继承全部内容，而不是从flask_restful.Resource继承

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

然后您可以使用异常处理程序

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__))

这是我尝试过的完整代码

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)

flask-restful-当前请求的资源类

2 个答案: