[Django]-How can I get Django to print the debug information to the console?

The best way to do this is to set up a logging handler for django.requests. You do this by adding the LOGGING setting to your settings.py file. An example is included below:

LOGGING = {
    'version': 1,
    'disable_existing_loggers': True,
    'formatters': {
        'verbose': {
            'format': '%(levelname)s %(asctime)s %(module)s %(process)d %(thread)d %(message)s'
        },
    },
    'handlers': {
        'console': {
            'level': 'NOTSET',
            'class': 'logging.StreamHandler',
            'formatter': 'verbose'
        }
    },
    'loggers': {
        '': {
            'handlers': ['console'],
            'level': 'NOTSET',
        },
        'django.request': {
            'handlers': ['console'],
            'propagate': False,
            'level': 'ERROR'
        }
    }
}

With this in place, you’ll see a nice stack trace in your terminal every time a 500 response is returned.

For more information on Django’s implementation of the Python logger, see the documentation about logging.

If you’re running the development server locally, you can just use print.

Also, you can use the logging module and redirect output to stdout. In settings.py put something like:

import logging.config

logging.config.fileConfig('/path/to/logging.conf')

Then make a logging.conf file:

[loggers]
keys=root

[handlers]
keys=consoleHandler

[formatters]
keys=simpleFormatter

[logger_root]
level=DEBUG
handlers=consoleHandler

[handler_consoleHandler]
class=StreamHandler
level=DEBUG
formatter=simpleFormatter
args=(sys.stdout,)

[formatter_simpleFormatter]
format=format=%(asctime)s - %(name)s - %(levelname)s - %(message)s

Check out the python documentation for more details and more examples.

I was getting 500 Internal Server Error and was not able to debug it as the code was not accessed through a browser and terminal didn’t show the traceback.
This bit of code in a middleware made all the difference:

import traceback
import sys

class ProcessExceptionMiddleware(object):
    def process_response(self, request, response):
        if response.status_code != 200:                
            print '\n'.join(traceback.format_exception(*sys.exc_info()))
        return response

Instead of processing the exception I, am processing the response object and printing the traceback when the status code not 200 and voila I can now see what is causing the error. Which in this particular case was an improperly indented block.

Yuji’s answer above and the middleware documentation helped me reach this solution.

Ok guys, I think this an important thing to note, as it’s year 2016 and many people are using Django Rest Framework instead of standard django forms, I’ve wasted 2 hours on figuring out why my exceptions are being handled by a renderer and it never gets to the middleware, the solution to the problem is simple.

http://www.django-rest-framework.org/api-guide/exceptions/#custom-exception-handling

So like this:

import traceback

from rest_framework.views import exception_handler

def custom_exception_handler(exc, context):
    # Call REST framework's default exception handler first,
    # to get the standard error response.
    response = exception_handler(exc, context)
    traceback.print_exc()
    return response

And then in settings define the exception handler as such:

REST_FRAMEWORK = {
    'EXCEPTION_HANDLER': 'utils.custom_exception_handler.custom_exception_handler'
}

The answer for this is to create a django middleware to catch all exeptions.

When you catch them, you can redirect them to whatever you want.

Documentation is here:

http://docs.djangoproject.com/en/dev/topics/http/middleware/#process-exception