Logging Tastypie requests (access log)

Question

I'm having problems with creating an access logger for my tastypie restfull app. I'd like to log several HTTP headers in requests to server and pass them to logger/handler defined in django Settings file. The idea is to log every HTTP request into a access log file.

I've encountered several logging modules (apps) but they all use database, I want something simpler for a basic access log.

Lovro Pandžić · Accepted Answer · 2014-09-04T16:08:28.500

3

I ended up making my own Middleware class in middleware.py inside application root.

Also I placed 'appname.middleware.RequestLoggerMiddleware', inside Settings.py Middleware section.

Here is the code for my access logging middleware class:

import logging

logger = logging.getLogger('access')

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

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

For more info about Middleware components see Django Middleware documentation.

edited Sep 04 '14 at 16:08

answered Nov 20 '12 at 13:00

Lovro Pandžić

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The link is helpful, but the code isn't good anymore..... – wogsland Oct 11 '21 at 11:17

score 1 · Answer 2 · answered Sep 04 '14 at 16:59

Another possibility is to override of the ModelResource.dispatch() method, in a custom ModelResource object:

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

It could be a proxy thing, but I get a KeyError on `request.META['USER']` — mlissner, Mar 14 '16 at 18:21

Michael Samoylov · Answer 3 · 2012-11-19T20:10:30.763

-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)

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

edited Nov 19 '12 at 20:10

answered Nov 19 '12 at 20:04

Michael Samoylov

2,933
3
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But I don't have any views... I'm just using Resources from Tastypie. Is extending existing classes the only solution? – Lovro Pandžić Nov 19 '12 at 20:23
What I think all I need to do is to implement my own middleware (https://docs.djangoproject.com/en/dev/topics/http/middleware/?from=olddocs) with process_request returning None. – Lovro Pandžić Nov 19 '12 at 21:01

Logging Tastypie requests (access log)

3 Answers3