I hate to give the question this heading but I actually don't know whats happening so here it goes.
I was doing another project in which I wanted to use logging module. The code is distributed among few files & instead of creating separate logger objects for seperate files, I thought of creating a logs.py with contents
import sys, logging
class Logger:
def __init__(self):
formatter = logging.Formatter('%(filename)s:%(lineno)s %(levelname)s:%(message)s')
stdout_handler = logging.StreamHandler(sys.stdout)
stdout_handler.setFormatter(formatter)
self.logger=logging.getLogger('')
self.logger.addHandler(stdout_handler)
self.logger.setLevel(logging.DEBUG)
def debug(self, message):
self.logger.debug(message)
and use this class like (in different files.)
import logs
b = logs.Logger()
b.debug("Hi from a.py")
I stripped down the whole problem to ask the question here. Now, I have 3 files, a.py, b.py & main.py. All 3 files instantiate the logs.Logger class and prints a debug message.
a.py & b.py imports "logs" and prints their debug message.
main.py imports logs, a & b; and prints it own debug message.
The file contents are like this: http://i.imgur.com/XoKVf.png
Why is debug message from b.py printed 2 times & from main.py 3 times?
Specify a name for the logger, otherwise you always use root logger.
import sys, logging
class Logger:
def __init__(self, name):
formatter = logging.Formatter('%(filename)s:%(lineno)s %(levelname)s:%(message)s')
stdout_handler = logging.StreamHandler(sys.stdout)
stdout_handler.setFormatter(formatter)
self.logger=logging.getLogger(name)
self.logger.addHandler(stdout_handler)
self.logger.setLevel(logging.DEBUG)
def debug(self, message):
self.logger.debug(message)
http://docs.python.org/howto/logging.html#advanced-logging-tutorial :
A good convention to use when naming loggers is to use a module-level
logger, in each module which uses logging, named as follows:
logger = logging.getLogger(__name__)
logging.getLogger('') will return exactly the same object each time you call it. So each time you instantiate a Logger (why use old-style classes here?) you are attaching one more handler resulting in printing to one more target. As all your targets pointing to the same thing, the last call to .debug() will print to each of the three StreamHandler objects pointing to sys.stdout resulting in three lines being printed.
First. Don't create your own class of Logger.
Just configure the existing logger classes with exising logging configuration tools.
Second. Each time you create your own class of Logger you also create new handlers and then attach the new (duplicating) handler to the root logger. This leads to duplication of messages.
If you have several modules that must (1) run stand-alone and (2) also run as part of a larger, composite, application, you need to do this. This will assure that logging configuration is done only once.
import logging
logger= logging.getLogger( __file__ ) # Unique logger for a, b or main
if __name__ == "__main__":
logging.basicConfig( stream=sys.stdout, level=logging.DEBUG, format='%(filename)s:%(lineno)s %(levelname)s:%(message)s' )
# From this point forward, you can use the `logger` object.
logger.info( "Hi" )
Related
I have a root logging class that I created, which I'd like to use for each micro-service function that I'm deploying.
Example output log: [2023-01-01 13:46:26] - INFO - [utils.logger.<module>:5] - testaaaaa
The logger is defined in utils.logger so that's why it's showing that in the log, hence %(name)s.
Instead of using the same root logger name which is set with logger = logging.getLogger(__name__), how can I get the same structure in dot notation where the logger is being instantiated and called?
Even if I have to modify my logger class to accept a name parameter when initializing the object, that is fine. But I like the dot notation because I will have files like routes.users.functiona, routes.users.functionb, routes.database.functiona and so on.
So I want to show which module the logging came from. Can't seem to follow how logging is capturing the path when using __name__.
Also if you have any suggestions about making the following more robust :)
Here is my class:
import typing
import logging
class GlobalLogger:
MINIMUM_GLOBAL_LEVEL = logging.DEBUG
GLOBAL_HANDLER = logging.StreamHandler()
LOG_FORMAT = "[%(asctime)s] - %(levelname)s - [%(name)s.%(funcName)s:%(lineno)d] - %(message)s"
LOG_DATETIME_FORMAT = "%Y-%m-%d %H:%M:%S"
def __init__(self, level: typing.Union[int, str] = MINIMUM_GLOBAL_LEVEL):
self.level = level
self.logger = self._get_logger()
self.log_format = self._log_formatter()
self.GLOBAL_HANDLER.setFormatter(self.log_format)
def _get_logger(self):
logger = logging.getLogger(__name__)
logger.setLevel(self.level)
logger.addHandler(self.GLOBAL_HANDLER)
return logger
def _log_formatter(self):
return logging.Formatter(fmt=self.LOG_FORMAT, datefmt=self.LOG_DATETIME_FORMAT)
I think you misunderstood the logging concept.
The variable __name__ holds the name of Python package where this variable is used. I think in your case it will have the value where class GlobalLogger is defined. And it does not matter from where you will create instance of your class. => Check it with print command.
At the top of every of your Python module (*.py), initialize module logger simply by logger = logging.getLogger(__name__). In your access the logging methods via module variable logger.
The logging data are automatically propagated to the parent logger based on the dot notation. If you have module A.B.C, the data are propagated C -> B -> A.
The output handlers, formatting can be defined for every logger or just for root logger e.g.: A or just (empty) "".
I usually have function InitLogging, where I get handle to the root logger and setup the output handlers. You do not need setup logging handlers in every module.
Conclusion:
If you still want use your class, give name of your logger as named parameter into the __init__. When you create instance of your class use variable __name__ during initialization.
I think there is not needed to create so many specific logger classes. You can define the logger properties later.
I've developed a module that I would like to be imported into any script and automatically have the module logger hiearchy below the logger established in the main script, no matter what the main logger is called (root, main, etc.)
i.e.
a.py
import module
log = logging.getLogger()
log.info("Test Main")
test()
b.py
import module
log = logging.getLogger('main')
log.info("Test Main")
test()
module.py
mod_log = logging.getLogger(__name__)
def test():
mod_log.info("Test Mod")
If the script ran successfully, I would expect the following output. I just can't seem to get it to run?
a.py
ROOT - Test Main
ROOT.module - Test Mod
b.py
main - Test Main
main.module - Test Mod
This is not possible because there is no way to tell where a logger was defined and which one the "main" logger is supposed to be. The main script could define any number of loggers, including zero. The only logger that is guaranteed to exist is the root logger.
If you are fine with just taking a guess you could make your logger a child of whatever is the first logger that was created.
module.py
import logging
logger_name = __name__
if logging.root.manager.loggerDict:
parent = next(iter(logging.root.manager.loggerDict))
logger_name = parent + '.' + logger_name
mod_log = logging.getLogger(logger_name)
Implement a Singleton Design pattern of the Logger class - whose basically job is to add a formatter, setting logging level and filtering and finally returning an instance of the logger.
Then use this to create an object:
logger = Logger.__call__().get_logger()
print(Logger())
Following the above method will create one and only on instance of logger and no matter where you create the object of the Logger class, it will be the same old instance if it was created before.
I have python project with multiple modules with logging. I perform initialization (reading log configuration file and creating root logger and enable/disable logging) in every module before start of logging the messages. Is it possible to perform this initialization only once in one place (like in one class may be called as Log) such that the same settings are reused by logging all over the project?
I am looking for a proper solution to have only once to read the configuration file and to only once get and configure a logger, in a class constructor, or perhaps in the initializer (__init__.py). I don't want to do this at client side (in __main__ ). I want to do this configuration only once in separate class and call this class in other modules when logging is required.
setup using #singleton pattern
#log.py
import logging.config
import yaml
from singleton_decorator import singleton
#singleton
class Log:
def __init__(self):
configFile = 'path_to_my_lof_config_file'/logging.yaml
with open(configFile) as f:
config_dict = yaml.load(f)
logging.config.dictConfig(config_dict)
self.logger = logging.getLogger('root')
def info(self, message):
self.logger.info(message)
#module1.py
from Log import Log
myLog = Log()
myLog.info('Message logged successfully)
#module2.py
from Log import Log
myLog = Log() #config read only once and only one object is created
myLog.info('Message logged successfully)
From the documentation,
Note that Loggers should NEVER be instantiated directly, but always through the module-level function logging.getLogger(name). Multiple calls to getLogger() with the same name will always return a reference to the same Logger object.
You can initialize and configure logging in your main entry point. See Logging from multiple modules in this Howto (Python 2.7).
I had the same problem and I don't have any classes or anything, so I solved it with just using global variable
utils.py:
existing_loggers = {}
def get_logger(name='my_logger', level=logging.INFO):
if name in existing_loggers:
return existing_loggers[name]
# Do the rest of initialization, handlers, formatters etc...
I was wondering what would be the best way for me to structure my logs in a special situation.
I have a series of python services that use the same python files for communicating (ex. com.py) with the HW. I have logging implemented in this modules and i would like for it to be dependent(associated) with the main service that is calling the modules.
How should i structure the logger logic so that if i have:
main_service_1->module_for_comunication
The logging goes to file main_serv_1.log
main_service_2->module_for_comunication
The logging goes to file main_serv_2.log
What would be the best practice in this case without harcoding anything?
Is there a way to know the file which is importing the com.py, so that i am able inside of the com.py, to use this information to adapt the logging to the caller?
In my experience, for a situation like this, the cleanest and easiest to implement strategy is to pass the logger to the code that does the logging.
So, create a logger for each service you want to have log to a different file, and pass that logger in to the code from your communications module. You can use __name__ to get the name of the current module (the actual module name, without the .py extension).
In the example below I implemented a fallback for the case when no logger is passed in as well.
com.py
from log import setup_logger
class Communicator(object):
def __init__(self, logger=None):
if logger is None:
logger = setup_logger(__name__)
self.log = logger
def send(self, data):
self.log.info('Sending %s bytes of data' % len(data))
svc_foo.py
from com import Communicator
from log import setup_logger
logger = setup_logger(__name__)
def foo():
c = Communicator(logger)
c.send('foo')
svc_bar.py
from com import Communicator
from log import setup_logger
logger = setup_logger(__name__)
def bar():
c = Communicator(logger)
c.send('bar')
log.py
from logging import FileHandler
import logging
def setup_logger(name):
logger = logging.getLogger(name)
handler = FileHandler('%s.log' % name)
logger.addHandler(handler)
return logger
main.py
from svc_bar import bar
from svc_foo import foo
import logging
# Add a StreamHandler for the root logger, so we get some console output in
# addition to file logging (for easy of testing). Also set the level for
# the root level to INFO so our messages don't get filtered.
logging.basicConfig(level=logging.INFO)
foo()
bar()
So, when you execute python main.py, this is what you'll get:
On the console:
INFO:svc_foo:Sending 3 bytes of data
INFO:svc_bar:Sending 3 bytes of data
And svc_foo.log and svc_bar.log each will have one line
Sending 3 bytes of data
If a client of the Communicator class uses it without passing in a logger, the log output will end up in com.log (fallback).
I see several options:
Option 1
Use __file__. __file__ is the pathname of the file from which the module was loaded (doc). depending of your structure, you should be able to identify the module by performing an os.path.split() like so:
If the folder structure is
+- module1
| +- __init__.py
| +- main.py
+- module2
+- __init__.py
+- main.py
you should be able to obtain the module name with a code placed in main.py:
def get_name():
module_name = os.path.split(__file__)[-2]
return module_name
This is not exactly DRY because you need the same code in both main.py. Reference here.
Option 2
A bit cleaner is to open 2 terminal windows and use an environment variable. E.g. you can define MOD_LOG_NAME as MOD_LOG_NAME="main_service_1" in one terminal and MOD_LOG_NAME="main_service_2" in the other one. Then, in your python code you can use something like:
import os
LOG_PATH_NAME os.environ['MOD_LOG_NAME']
This follows separation of concerns.
Update (since the question evolved a bit)
Once you've established the distinct name, all you have to do is to configure the logger:
import logging
logging.basicConfig(filename=LOG_PATH_NAME,level=logging.DEBUG)
(or get_name())and run the program.
I was wondering what the standard set up is for performing logging from within a Python app.
I am using the Logging class, and I've written my own logger class that instantiates the Logging class. My main then instantiates my logger wrapper class. However, my main instantiates other classes and I want those other classes to also be able to write to he log file via the logger object in the main.
How do I make that logger object such that it can be called by other classes? It's almost like we need some sort of static logger object to get this to work.
I guess the long and short of the question is: how do you implement logging within your code structure such that all classes instantiated from within main can write to the same log file? Do I just have to create a new logging object in each of the classes that points to the same file?
I don't know what you mean by the Logging class - there's no such class in Python's built-in logging. You don't really need wrappers: here's an example of how to do logging from arbitrary classes that you write:
import logging
# This class could be imported from a utility module
class LogMixin(object):
#property
def logger(self):
name = '.'.join([__name__, self.__class__.__name__])
return logging.getLogger(name)
# This class is just there to show that you can use a mixin like LogMixin
class Base(object):
pass
# This could be in a module separate from B
class A(Base, LogMixin):
def __init__(self):
# Example of logging from a method in one of your classes
self.logger.debug('Hello from A')
# This could be in a module separate from A
class B(Base, LogMixin):
def __init__(self):
# Another example of logging from a method in one of your classes
self.logger.debug('Hello from B')
def main():
# Do some work to exercise logging
a = A()
b = B()
with open('myapp.log') as f:
print('Log file contents:')
print(f.read())
if __name__ == '__main__':
# Configure only in your main program clause
logging.basicConfig(level=logging.DEBUG,
filename='myapp.log', filemode='w',
format='%(name)s %(levelname)s %(message)s')
main()
Generally it's not necessary to have loggers at class level: in Python, unlike say Java, the unit of program (de)composition is the module. However, nothing stops you from doing it, as I've shown above. The script, when run, displays:
Log file contents:
__main__.A DEBUG Hello from A
__main__.B DEBUG Hello from B
Note that code from both classes logged to the same file, myapp.log. This would have worked even with A and B in different modules.
Try using logging.getLogger() to get your logging object instance:
http://docs.python.org/3/library/logging.html#logging.getLogger
All calls to this function with a given name return the same logger instance. This means that logger instances never need to be passed between different parts of an application.
UPDATE:
The recommended way to do this is to use the getLogger() function and configure it (setting a handler, formatter, etc...):
# main.py
import logging
import lib
def main():
logger = logging.getLogger('custom_logger')
logger.setLevel(logging.INFO)
logger.addHandler(logging.FileHandler('test.log'))
logger.info('logged from main module')
lib.log()
if __name__ == '__main__':
main()
# lib.py
import logging
def log():
logger = logging.getLogger('custom_logger')
logger.info('logged from lib module')
If you really need to extend the logger class take a look at logging.setLoggerClass(klass)
UPDATE 2:
Example on how to add a custom logging level without changing the Logging class:
# main.py
import logging
import lib
# Extend Logger class
CUSTOM_LEVEL_NUM = 9
logging.addLevelName(CUSTOM_LEVEL_NUM, 'CUSTOM')
def custom(self, msg, *args, **kwargs):
self._log(CUSTOM_LEVEL_NUM, msg, args, **kwargs)
logging.Logger.custom = custom
# Do global logger instance setup
logger = logging.getLogger('custom_logger')
logger.setLevel(logging.INFO)
logger.addHandler(logging.FileHandler('test.log'))
def main():
logger = logging.getLogger('custom_logger')
logger.custom('logged from main module')
lib.log()
if __name__ == '__main__':
main()
Note that adding custom level is not recommended: http://docs.python.org/2/howto/logging.html#custom-levels
Defining a custom handler and maybe using more than one logger may do the trick for your other requirement: optional output to stderr.