flask-restful-当前请求的资源类

Question

问题

我所有的应用程序路由都是通过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？

我为什么要这个

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

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，但不是全部数据-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一起使用是一个好主意，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!'}

而不是在我的中使用它log_exception：

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

而不是继承自flask_restful.Resource您想要继承的所有自定义资源

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)

编辑 1：8 月 5 日

正如@jbet 所评论的，如果有人想要获得处理类，一个更干净的选择是使用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)