記錄Tastypie請求（訪問日誌）

Question

我在爲我的tastypie restfull應用程序創建訪問記錄器時遇到問題。 我想在服務器的請求中記錄多個HTTP頭，並將它們傳遞給在django設置文件中定義的記錄器/處理程序。這個想法是將每個HTTP請求記錄到訪問日誌文件中。

我遇到了幾個日誌模塊（應用程序），但他們都使用數據庫，我希望更簡單的基本訪問日誌。

Answer 1

最後我使我自己的中間件類中的應用程序根middleware.py。

另外我在Settings.py中間件部分放置了'appname.middleware.RequestLoggerMiddleware',。

這裏是我的訪問日誌記錄中間件類的代碼：

import logging 

logger = logging.getLogger('access') 

class RequestLoggerMiddleware(object): 
    def process_request(self, request): 
     ... logging logic here... 

     logger.info('logging message')) 
     return None 

欲瞭解更多信息有關中間件組件看到Django Middleware documentation。

Answer 2

import logging 


logger = logging.getLogger('project.app.view') 


def my_view(request): 
    entry = '%s %s for %s' % (request.method, request.get_full_path(), request.META['REMOTE_ADDR']) 
    logger.info(entry) 

見https://docs.djangoproject.com/en/1.4/topics/logging/

Answer 3

另一種可能性是覆蓋ModelResource.dispatch()方法，在自定義ModelResource對象：

class CustomModelResource(ModelResource): 
    def dispatch(self, request_type, request, **kwargs): 
     """ 
     Override for systematic logging. 
     """ 
     log_user = request.META['USER'] 
     log_request_type = request_type 
     log_resource_name = kwargs['resource_name'] 
     log_api_name = kwargs['api_name'] 
     log_response = {} 

     try: 
      response = super(CustomModelResource, self).dispatch(request_type, request, **kwargs) 
      log_response['response_code'] = response.status_code 
     # Also log what could go wrong 
     except Exception, e: 
      log_response['error_type'] = e.__class__.__name__ 
      log_response['error_message'] = e.message 
      log_response['response_code'] = http.HttpBadRequest.status_code 
      raise 
     finally: 
      # Log all the things 
      logger.debug('%s asked for %s on %s through api %s: \n%s' % (
       log_user, 
       log_request_type, 
       log_resource_name, 
       log_api_name, 
       log_response, 
      )) 

     return response 

class Meta: 
    # Other custom stuff