I'm seeing a behavior that I have no way of explaining... Here's my simplified setup:
module x:
import logging
logger = logging.getLogger('x')
def test_debugging():
logger.debug('Debugging')
test for module x:
import logging
import unittest
from x import test_debugging
class TestX(unittest.TestCase):
def test_test_debugging(self):
test_debugging()
if __name__ == '__main__':
logger = logging.getLogger('x')
logger.setLevel(logging.DEBUG)
# logging.debug('another test')
unittest.main()
If I uncomment the logging.debug('another test') line I can also see the log from x. Note, it is not a typo, I'm calling debug on logging, not on the logger from module x. And if I call debug on logger, I don't see logs.
What is this, I can't even?..
In your setup, you didn't actually configure logging. Although the configuration can be pretty complex, it would suffice to set the log level in your example:
if __name__ == '__main__':
# note I configured logging, setting e.g. the level globally
logging.basicConfig(level=logging.DEBUG)
logger = logging.getLogger('x')
logger.setLevel(logging.DEBUG)
unittest.main()
This will create a simple StreamHandler with a predefined output format that prints all the log records to the stdout. I suggest you to quickly look over the relevant docs for more info.
Why did it work with the logging.debug call? Because the logging.{info,debug,warn,error} functions all call logging.basicConfig internally, so once you have called logging.debug, you configured logging implicitly.
Edit: let's take a quick look under the hood what is the actual meaning of the logging.{info,debug,error,warning} functions. Let's take the following snippet:
import logging
logger = logging.getLogger('mylogger')
logger.warning('hello world')
If you run the snippet, hello world will be not printed (and this is correct so!). Why not? It's because you didn't actually specify how the log records should be treated - should they be printed to stdout, or maybe printed to a file, or maybe sent to some server that will email them to the recipients? The logger mylogger will receive the log record hello world, but it doesn't know yet what to do with it. So, to actually print the record, let's do some configuration for the logger:
import logging
logger = logging.getLogger('mylogger')
formatter = logging.Formatter('Logger received message %(message)s at time %(asctime)s')
handler = logging.StreamHandler()
handler.setFormatter(formatter)
logger.addHandler(handler)
logger.warning('hello world')
We now attached a handler that handles the record by printing it to the stdout in the format specified by formatter. Now the record hello world will be printed to the stdout. We could attach more handlers and the record would be handled by each of the handler. Example: try to attach another StreamHandler and you will notice that the record is now printed twice.
So, what's with the logging functions now? If you have some simple program that has only one logger that should print the messages and that's all, you can replace the manual configuration by using convenience logging functions:
import logging
logging.warning('hello world')
This will configure the root logger to print the messages to stdout by adding a StreamHandler to it with some default formatter, so you don't have to configure it yourself. After that, it will tell the root logger to process the record hello world. Merely a convenience, nothing more. If you want to explicitly trigger this basic configuration of the root logger, issue
logging.basicConfig()
with or without the additional configuration parameters.
Now, let's go through my first code snippet once again:
logging.basicConfig(level=logging.DEBUG)
After this line, the root logger will print all log records with level DEBUG and higher to the command line.
logger = logging.getLogger('x')
logger.setLevel(logging.DEBUG)
We did not configure this logger explicitly, so why are the records still being printed? This is because by default, any logger will propagate the log records to the root logger. So the logger x does not print the records - it has not been configured for that, but it will pass the record further up to the root logger that knows how to print the records.
Related
I have put the following in my config.py:
import time
import logging
#logging.basicConfig(format='%(asctime)s %(message)s', datefmt='%Y-%m-%d %H:%M:%S', level=logging.INFO)
logFormatter = logging.Formatter('%(asctime)s %(message)s', datefmt='%Y-%m-%d %H:%M:%S')
rootLogger = logging.getLogger()
rootLogger.setLevel(logging.INFO)
fileHandler = logging.FileHandler("{0}.log".format(time.strftime('%Y%m%d%H%M%S')))
fileHandler.setFormatter(logFormatter)
rootLogger.addHandler(fileHandler)
consoleHandler = logging.StreamHandler()
consoleHandler.setFormatter(logFormatter)
rootLogger.addHandler(consoleHandler)
and then I am doing
from config import *
in all of my scripts and imported files.
Unfortunately, this causes multiple log files created.
How to fix this? I wan't centralized config.py with logging configured both to console and file.
Case 1: Independent Scripts / Programs
In case we are talking about multiple, independent scripts, that should have logging set up in the same way: I would say, each independent application should have its own log. If you definitively do not want this, you would have to
make sure that all applications have the same log file name (e.g. create a function in config.py, with a parameter "timestamp", which is provided by your script(s)
specify the append filemode for the fileHandler
make sure that config.py is not called twice somewhere, as you would add the log handlers twice, which would result in each log message being printed twice.
Case 2: One big application consisting of modules
In case we are talking about one big application, consisting of modules, you could adopt a structure like the following:
config.py:
def set_up_logging():
# your logging setup code
module example (some_module.py):
import logging
def some_function():
logger = logging.getLogger(__name__)
[...]
logger.info('sample log')
[...]
main example (main.py)
import logging
from config import set_up_logging
from some_module import some_function
def main():
set_up_logging()
logger = logging.getLogger(__name__)
logger.info('Executing some function')
some_function()
logger.info('Finished')
if __name__ == '__main__':
main()
Explanation:
With the call to set_up_logging() in main() you configure your applications root logger
each module is called from main(), and get its logger via logger = logging.getLogger(__name__). As the modules logger are in the hierarchy below the root logger, those loggings get "propagated up" to the root logger and handled by the handlers of the root logger.
For more information see Pythons logging module doc and/or the logging cookbook
I am having some difficulties using python's logging. I have two files, main.py and mymodule.py. Generally main.py is run, and it will import mymodule.py and use some functions from there. But sometimes, I will run mymodule.py directly.
I tried to make it so that logging is configured in only 1 location, but something seems wrong.
Here is the code.
# main.py
import logging
import mymodule
logger = logging.getLogger(__name__)
def setup_logging():
# only cofnigure logger if script is main module
# configuring logger in multiple places is bad
# only top-level module should configure logger
if not len(logger.handlers):
logger.setLevel(logging.DEBUG)
# create console handler with a higher log level
ch = logging.StreamHandler()
ch.setLevel(logging.DEBUG)
formatter = logging.Formatter('%(levelname)s: %(asctime)s %(funcName)s(%(lineno)d) -- %(message)s', datefmt = '%Y-%m-%d %H:%M:%S')
ch.setFormatter(formatter)
logger.addHandler(ch)
if __name__ == '__main__':
setup_logging()
logger.info('calling mymodule.myfunc()')
mymodule.myfunc()
and the imported module:
# mymodule.py
import logging
logger = logging.getLogger(__name__)
def myfunc():
msg = 'myfunc is being called'
logger.info(msg)
print('done with myfunc')
if __name__ == '__main__':
# only cofnigure logger if script is main module
# configuring logger in multiple places is bad
# only top-level module should configure logger
if not len(logger.handlers):
logger.setLevel(logging.DEBUG)
# create console handler with a higher log level
ch = logging.StreamHandler()
ch.setLevel(logging.DEBUG)
formatter = logging.Formatter('%(levelname)s: %(asctime)s %(funcName)s(%(lineno)d) -- %(message)s', datefmt = '%Y-%m-%d %H:%M:%S')
ch.setFormatter(formatter)
logger.addHandler(ch)
logger.info('myfunc was executed directly')
myfunc()
When I run the code, I see this output:
$>python main.py
INFO: 2016-07-14 18:13:04 <module>(22) -- calling mymodule.myfunc()
done with myfunc
But I expect to see this:
$>python main.py
INFO: 2016-07-14 18:13:04 <module>(22) -- calling mymodule.myfunc()
INFO: 2016-07-14 18:15:09 myfunc(8) -- myfunc is being called
done with myfunc
Anybody have any idea why the second logging.info call doesn't print to screen? thanks in advance!
Loggers exist in a hierarchy, with a root logger (retrieved with logging.getLogger(), no arguments) at the top. Each logger inherits configuration from its parent, with any configuration on the logger itself overriding the inherited configuration. In this case, you are never configuring the root logger, only the module-specific logger in main.py. As a result, the module-specific logger in mymodule.py is never configured.
The simplest fix is probably to use logging.basicConfig in main.py to set options you want shared by all loggers.
Chepner is correct. I got absorbed into this problem. The problem is simply in your main script
16 log = logging.getLogger() # use this form to initialize the root logger
17 #log = logging.getLogger(__name__) # never use this one
If you use line 17, then your imported python modules will not log any messages
In you submodule.py
import logging
logger = logging.getLogger()
logger.debug("You will not see this message if you use line 17 in main")
Hope this posting can help someone who got stuck on this problem.
While the logging-package is conceptually arranged in a namespace hierarchy using dots as separators, all loggers implicitly inherit from the root logger (like every class in Python 3 silently inherits from object). Each logger passes log messages on to its parent.
In your case, your loggers are incorrectly chained. Try adding print(logger.name) in your both modules and you'll realize, that your instantiation of logger in main.py is equivalent to
logger = logging.getLogger('__main__')
while in mymodule.py, you effectively produce
logger = logging.getLogger('mymodule')
The call to log INFO-message from myfunc() passes directly the root logger (as the logger in main.py is not among its ancestors), which has no handler set up (in this case the default message dispatch will be triggered, see here)
I have been working on this almost all day couldn't figure what I am missing. I am trying to add a custom handler to emit all log data into a GUI session. It works but the handler doesn't extend to the submodules and just emits records from the main module. Here is a small snippet I tried
I have two files
# main.py
import logging
import logging_two
def myapp():
logger = logging.getLogger('myapp')
logging.basicConfig()
logger.info('Using myapp')
ch = logging.StreamHandler()
logger.addHandler(ch)
logging_two.testme()
print logger.handlers
myapp()
Second module
#logging_two
import logging
def testme():
logger = logging.getLogger('testme')
logger.info('IN test me')
print logger.handlers
I would expect the logger in logging_two.testme to have the handler I have added in the main module. I looked at the docs to me it seems this should work but I am not sure if I got it wrong?
the result I get is
[]
[<logging.StreamHandler object at 0x00000000024ED240>]
In myapp() you are adding the handler to the logger named 'myapp'. Since testme() is getting the logger named 'testme' it does not have the handler since it is a different part of the logging hierarchy.
If you just have logger = logger.getLogger() in myapp() then it would work since you are adding the handler to the root of the hierarchy.
Check out the python logging docs.
I want to use the logging module instead of printing for debug information and documentation.
The goal is to print on the console with DEBUG level and log to a file with INFO level.
I read through a lot of documentation, the cookbook and other tutorials on the logging module but couldn't figure out, how I can use it the way I want it. (I'm on python25)
I want to have the names of the modules in which the logs are written in my logfile.
The documentation says I should use logger = logging.getLogger(__name__) but how do I declare the loggers used in classes in other modules / packages, so they use the same handlers like the main logger? To recognize the 'parent' I can use logger = logging.getLogger(parent.child) but where do I know, who has called the class/method?`
The example below shows my problem, if I run this, the output will only have the __main__ logs in and ignore the logs in Class
This is my Mainfile:
# main.py
import logging
from module import Class
logger = logging.getLogger(__name__)
logger.setLevel(logging.DEBUG)
# create file handler which logs info messages
fh = logging.FileHandler('foo.log', 'w', 'utf-8')
fh.setLevel(logging.INFO)
# create console handler with a debug log level
ch = logging.StreamHandler()
ch.setLevel(logging.DEBUG)
# creating a formatter
formatter = logging.Formatter('- %(name)s - %(levelname)-8s: %(message)s')
# setting handler format
fh.setFormatter(formatter)
ch.setFormatter(formatter)
# add the handlers to the logger
logger.addHandler(fh)
logger.addHandler(ch)
if __name__ == '__main__':
logger.info('Script starts')
logger.info('calling class Class')
c = Class()
logger.info('calling c.do_something()')
c.do_something()
logger.info('calling c.try_something()')
c.try_something()
Module:
# module.py
imnport logging
class Class:
def __init__(self):
self.logger = logging.getLogger(__name__) # What do I have to enter here?
self.logger.info('creating an instance of Class')
self.dict = {'a':'A'}
def do_something(self):
self.logger.debug('doing something')
a = 1 + 1
self.logger.debug('done doing something')
def try_something(self):
try:
logging.debug(self.dict['b'])
except KeyError, e:
logging.exception(e)
Output in console:
- __main__ - INFO : Script starts
- __main__ - INFO : calling class Class
- __main__ - INFO : calling c.do_something()
- __main__ - INFO : calling c.try_something()
No handlers could be found for logger "module"
Besides: Is there a way to get the module names were the logs ocurred in my logfile, without declaring a new logger in each class like above? Also like this way I have to go for self.logger.info() each time I want to log something. I would prefer to use logging.info() or logger.info() in my whole code.
Is a global logger perhaps the right answer for this? But then I won't get the modules where the errors occur in the logs...
And my last question: Is this pythonic? Or is there a better recommendation to do such things right.
In your main module, you're configuring the logger of name '__main__' (or whatever __name__ equates to in your case) while in module.py you're using a different logger. You either need to configure loggers per module, or you can configure the root logger (by configuring logging.getLogger()) in your main module which will apply by default to all loggers in your project.
I recommend using configuration files for configuring loggers. This link should give you a good idea of good practices: http://victorlin.me/posts/2012/08/26/good-logging-practice-in-python
EDIT: use %(module) in your formatter to include the module name in the log message.
The generally recommended logging setup is having at most 1 logger per module.
If your project is properly packaged, __name__ will have the value of "mypackage.mymodule", except in your main file, where it has the value "__main__"
If you want more context about the code that is logging messages, note that you can set your formatter with a formatter string like %(funcName)s, which will add the function name to all messages.
If you really want per-class loggers, you can do something like:
class MyClass:
def __init__(self):
self.logger = logging.getLogger(__name__+"."+self.__class__.__name__)
I do have the following logger class (as logger.py):
import logging, logging.handlers
import config
log = logging.getLogger('myLog')
def start():
"Function sets up the logging environment."
log.setLevel(logging.DEBUG)
formatter = logging.Formatter(fmt='%(asctime)s [%(levelname)s] %(message)s', datefmt='%d-%m-%y %H:%M:%S')
if config.logfile_enable:
filehandler = logging.handlers.RotatingFileHandler(config.logfile_name, maxBytes=config.logfile_maxsize,backupCount=config.logfile_backupCount)
filehandler.setLevel(logging.DEBUG)
filehandler.setFormatter(formatter)
log.addHandler(filehandler)
console = logging.StreamHandler()
console.setLevel(logging.DEBUG)
console.setFormatter(logging.Formatter('[%(levelname)s] %(message)s')) # nicer format for console
log.addHandler(console)
# Levels are: debug, info, warning, error, critical.
log.debug("Started logging to %s [maxBytes: %d, backupCount: %d]" % (config.logfile_name, config.logfile_maxsize, config.logfile_backupCount))
def stop():
"Function closes and cleans up the logging environment."
logging.shutdown()
For logging, I launch logger.start() once, and then import from logger import log at any project file. Then I just use log.debug() and log.error() when needed.
It works fine from everywhere on the script (different classes, functions and files) but it won't work on different processes lanuched through the multiprocessing class.
I get the following error: No handlers could be found for logger "myLog".
What can I do?
from the python docs: logging to a single file from multiple processes is not supported, because there is no standard way to serialize access to a single file across multiple processes in Python.
See: http://docs.python.org/howto/logging-cookbook.html#logging-to-a-single-file-from-multiple-processes
BTW: What I do in this situation, is use Scribe which is a distributed logging aggregator, that I log to via TCP. This allows me to log all servers I have to the same place, not just all processes.
See this project: http://pypi.python.org/pypi/ScribeHandler