I want to use a memory logger in my project. It keeps track of the last n logging records. A minimal example main file looks like this:
import sys
import logging
from logging import StreamHandler
from test_module import do_stuff
logger = logging.getLogger(__name__)
class MemoryHandler(StreamHandler):
def __init__(self, n_logs: int):
StreamHandler.__init__(self)
self.n_logs = n_logs
self.my_records = []
def emit(self, record):
self.my_records.append(self.format(record))
self.my_records = self.my_records[-self.n_logs:]
def to_string(self):
return '\n'.join(self.my_records)
if __name__ == '__main__':
logging.basicConfig(stream=sys.stdout, level=logging.INFO)
mem_handler = MemoryHandler(n_logs=10)
logger.addHandler(mem_handler)
logger.info('hello')
do_stuff()
print(mem_handler.to_string())
The test module I am importing do_stuff from looks like this:
import logging
logger = logging.getLogger(__name__)
def do_stuff():
logger.info('doing stuff')
When I run the main function two log statements appear. The one from main and the one from doing stuff, but the memory logger only receives "hello" and not "doing stuff":
INFO:__main__:hello
INFO:test_module:doing stuff
hello
I assume that this is because mem_handler is not added to the test_module logger. I can fix this by adding the mem_handler explicitely:
logging.getLogger('test_module').addHandler(mem_handler)
But in general I don't want to list all modules and add the mem_handler manually. How can I add the mem_handler to all loggers in my project?
The Python logging system is federated. That means there is a tree like structure similar to the package structure. This structure works by logger name and the levels are separated by dots.
If you use the module's __name__ to get the logger it will be equivalant to the dotted name of the package. for example:
package.subpackage.module
In this federated system a message is send up the loggers structure (unless one of the loggers is explicitly configured with propagate=False).
So, the best way to add a handler is to add it to the root logger on the top of the structure and make sure all loggers below propagate.
You can get the root logger with logging.getLogger() (without any name) and then add handlers or other configuration as you like.
I am getting duplicate (double) logs when using the python logging. I have 3 files :
1. main.py
2. dependencies.py
3. resources.py
I am making only 1 call to the python logger constructor which is done inside the main.py
Following are my import statements in the 3 files
main.py
import xml.etree.ElementTree as et
from configparser import ConfigParser
from Craftlogger import Craftlogger
logger = Craftlogger().getLogger()
dependencies.py
import os,sys
from main import getJobDetails,postRequest,logger
from configparser import ConfigParser
resources.py
import os,sys
import xml.etree.ElementTree as et
And inside the main method in the main.py, I have the imports
def main():
from resources import getResourceDetails,setResources
from dependencies import setDependencies
..... Remaining code .....
My logging file looks like this
import logging
class Craftlogger:
def __init__(self):
self.logger = logging.getLogger(__name__)
handler = logging.StreamHandler()
formatter_string = '%(asctime)s | %(levelname)-8s | %(filename)s-%(funcName)s-%(lineno)04d | %(message)s'
formatter = logging.Formatter(formatter_string)
handler.setFormatter(formatter)
self.logger.addHandler(handler)
self.logger.setLevel(logging.DEBUG)
self.logger.propagate = False
def getLogger(self):
return self.logger
Note : I had to do the imports inside of main so as to be able to do circular imports.
My guess would be that two CraftLogger objects exist and both have the same self.logger member. logging.getLogger(__name__) probably returns the same object for another CraftLogger object, resulting in two addHandler calls on the same logger. This is just a guess, no guarantee.
Logging is a cross cutting concern. As such, I frown upon classes which set up logging on their own. The responsibility to configure logging (especially handlers) should be solely with the main executing function, e.g. your main function. No submodule / class / function should modify logging, except getting a logger via logging.getlogger(name).
This avoids most of these pitfalls and allows easy composition of modules.
Imagine you have to import two modules who both modify the logging system...fun
I'm having trouble getting child loggers named properly in python (2.7). I have the following file structure:
-mypackage
-__init__.py
-main.py
-log
-__init__.py
-logfile.log
-src
-__init__.py
-logger.py
-otherfile.py
The contents of main.py are:
import logging
import src.logger
from src.otherfile import Foo
logger = logging.getLogger(__name__)
logger.info('Logging from main')
foo = Foo()
The contents of otherfile.py are:
import logging
class Foo():
def __init__(self):
self.logger = logging.getLogger(__name__)
self.logger.info('Logging from class in otherfile')
The contents of logger.py are:
import os
import logging
logdir = os.path.dirname(__file__)
logfile = os.path.join(logdir, '../log/controller.log')
logger = logging.getLogger('__main__')
logger.setLevel(logging.DEBUG)
fh = logging.FileHandler(logfile)
fh.setLevel(logging.DEBUG)
formatter = logging.Formatter('%(asctime)s - $(name)s - %(levelname)s: %(message)s')
fh.setFormatter(formatter)
logger.addHandler(fh)
logger.info('logging from logger.py')
I used logging.getLogger(__name__) in each file based on the docs. The exception is logger.py, where I name the logger __main__ so that it would run from top down and not try to derive everything from a buried file.
When I run main.py, it logs correctly from logger.py and main.py, but the logger from otherfile.py isn't derived correctly from the main logger.
How do I get the logger in otherfile.py to derive from my main logger?
In logger.py you are configuring the "__main__" logger. I was tricked by the fact that in main.py you use __name__. Since you are invoking python main.py, __name__ evaluates to "__main__". Right.
This can become a problem since when imported (instead of executed), main.py's logger won't be "__main__" but "main". It can be fixed by making your package executable: rename main.py to __main__.py and running your package like this:
python -m mypackage
This way, logger names (actually module __name__'s) will remain consistent.
That said, in no way the logger that you configure in logger.py is a parent of the logger in otherfile.py. The real parent of that logger is called "mypackage" but you haven't configured it, so it's logs are invisible.
You have several choices, you can configure (set log level, handler and formatter):
the root logger : logger = logging.getLogger()
mypackage's logger : logger = logger.getLogger(__name__) in mypackage.__init__
... or go down to the level of granularity you wish.
You can easily create hierarchies of loggers by calling separating levels of loggers by a dot ("."). For example, the logger returned by calling logging.getLogger('__main__.' + __name__) inherits all properties from the logger returned by logging.getLogger('__main__'). This behaviour is described here: https://docs.python.org/2/library/logging.html#module-level-functions.
I have the following file structure in my Python project:
Main.py
classes
|--- __init__.py
|--- Common.py
|--- Logger.py
|--- Dictionary.py
I'm setting a static variable of the Common class from within my Main file:
from classes.Common import Common
# set path to the log file for log output
Common.set_log_file_path(C:\\logging\\output.txt"))
which gets set in the Common class:
class Common():
log_file_path = ''
#staticmethod
def set_log_file_path(log_file_path):
Common.log_file_path = log_file_path
Now I instantiate a Logger object from within my Main file:
from classes.Logger import Logger
# initiate logger
log = Logger()
The Logger object reads the log file path from the Common object which works fine:
from Common import Common
class Logger():
log_file_path = ''
def __init__(self, log_file_path=''):
# if not specified, take default log file path from Common class
if log_file_path == '':
self.log_file_path = Common.log_file_path
Now comes the problem: From my Main file I instantiate a Dictionary object:
from classes.Dictionary import Dictionary
# load dictionary
dictionary = Dictionary()
In the dictionary object I also want to have a logger, so I create one there:
from Logger import Logger
class Dictionary():
log = Logger()
def __init__(self):
Dictionary.log.info('Dictionary > __init__()')
But this one doesn't work. Somehow when the Logger from within the Dictionary tries to load the log file path from the Common class, it is empty.
Why is that so? Shouldn't that be the same Common class and therefore holding the same static information here? Am I missing something? Do I do the imports in a wrong way?
I'm working with Python 2.6.5, and my imports are as follows:
Main.py imports Dictionary, Logger, Common
Dictionary.py imports Logger
Logger.py imports Common
Common has no imports
Almost everything in a Python module is dynamically executed — including import and class statements.
When you import the Directory module the first time the module is executed which means the import statement which imports Logger is executed and after the Logger is imported the class Directory: is executed which instantiates a Logger object. At this point your Common.set_log_file_path() was not yet called so that Logger object takes the default value which is defined when class Common: was executed.
The ”solution” would be importing Common and set the default path before executing anything that actually uses that default path attribute:
from classes.Common import Common
Common.log_file_path = 'C:/logging/output.txt'
from classes.Directory import Directory
from classes.Logger import Logger
Solution in quotes because having imports depending on other code being executed before is something that can turn into little nightmares very easily. Therefore it is something very seldom seen in productive code.
I have a small python project that has the following structure -
Project
-- pkg01
-- test01.py
-- pkg02
-- test02.py
-- logging.conf
I plan to use the default logging module to print messages to stdout and a log file.
To use the logging module, some initialization is required -
import logging.config
logging.config.fileConfig('logging.conf')
logger = logging.getLogger('pyApp')
logger.info('testing')
At present, I perform this initialization in every module before I start logging messages. Is it possible to perform this initialization only once in one place such that the same settings are reused by logging all over the project?
Best practice is, in each module, to have a logger defined like this:
import logging
logger = logging.getLogger(__name__)
near the top of the module, and then in other code in the module do e.g.
logger.debug('My message with %s', 'variable data')
If you need to subdivide logging activity inside a module, use e.g.
loggerA = logging.getLogger(__name__ + '.A')
loggerB = logging.getLogger(__name__ + '.B')
and log to loggerA and loggerB as appropriate.
In your main program or programs, do e.g.:
def main():
"your program code"
if __name__ == '__main__':
import logging.config
logging.config.fileConfig('/path/to/logging.conf')
main()
or
def main():
import logging.config
logging.config.fileConfig('/path/to/logging.conf')
# your program code
if __name__ == '__main__':
main()
See here for logging from multiple modules, and here for logging configuration for code which will be used as a library module by other code.
Update: When calling fileConfig(), you may want to specify disable_existing_loggers=False if you're using Python 2.6 or later (see the docs for more information). The default value is True for backward compatibility, which causes all existing loggers to be disabled by fileConfig() unless they or their ancestor are explicitly named in the configuration. With the value set to False, existing loggers are left alone. If using Python 2.7/Python 3.2 or later, you may wish to consider the dictConfig() API which is better than fileConfig() as it gives more control over the configuration.
Actually every logger is a child of the parent's package logger (i.e. package.subpackage.module inherits configuration from package.subpackage), so all you need to do is just to configure the root logger. This can be achieved by logging.config.fileConfig (your own config for loggers) or logging.basicConfig (sets the root logger). Setup logging in your entry module (__main__.py or whatever you want to run, for example main_script.py. __init__.py works as well)
using basicConfig:
# package/__main__.py
import logging
import sys
logging.basicConfig(stream=sys.stdout, level=logging.INFO)
using fileConfig:
# package/__main__.py
import logging
import logging.config
logging.config.fileConfig('logging.conf')
and then create every logger using:
# package/submodule.py
# or
# package/subpackage/submodule.py
import logging
log = logging.getLogger(__name__)
log.info("Hello logging!")
For more information see Advanced Logging Tutorial.
A simple way of using one instance of logging library in multiple modules for me was following solution:
base_logger.py
import logging
logger = logging
logger.basicConfig(format='%(asctime)s - %(message)s', level=logging.INFO)
Other files
from base_logger import logger
if __name__ == '__main__':
logger.info("This is an info message")
I always do it as below.
Use a single python file to config my log as singleton pattern which named 'log_conf.py'
#-*-coding:utf-8-*-
import logging.config
def singleton(cls):
instances = {}
def get_instance():
if cls not in instances:
instances[cls] = cls()
return instances[cls]
return get_instance()
#singleton
class Logger():
def __init__(self):
logging.config.fileConfig('logging.conf')
self.logr = logging.getLogger('root')
In another module, just import the config.
from log_conf import Logger
Logger.logr.info("Hello World")
This is a singleton pattern to log, simply and efficiently.
Throwing in another solution.
In my module's init.py I have something like:
# mymodule/__init__.py
import logging
def get_module_logger(mod_name):
logger = logging.getLogger(mod_name)
handler = logging.StreamHandler()
formatter = logging.Formatter(
'%(asctime)s %(name)-12s %(levelname)-8s %(message)s')
handler.setFormatter(formatter)
logger.addHandler(handler)
logger.setLevel(logging.DEBUG)
return logger
Then in each module I need a logger, I do:
# mymodule/foo.py
from [modname] import get_module_logger
logger = get_module_logger(__name__)
When the logs are missed, you can differentiate their source by the module they came from.
Several of these answers suggest that at the top of a module you you do
import logging
logger = logging.getLogger(__name__)
It is my understanding that this is considered very bad practice. The reason is that the file config will disable all existing loggers by default. E.g.
#my_module
import logging
logger = logging.getLogger(__name__)
def foo():
logger.info('Hi, foo')
class Bar(object):
def bar(self):
logger.info('Hi, bar')
And in your main module :
#main
import logging
# load my module - this now configures the logger
import my_module
# This will now disable the logger in my module by default, [see the docs][1]
logging.config.fileConfig('logging.ini')
my_module.foo()
bar = my_module.Bar()
bar.bar()
Now the log specified in logging.ini will be empty, as the existing logger was disabled by fileconfig call.
While is is certainly possible to get around this (disable_existing_Loggers=False), realistically many clients of your library will not know about this behavior, and will not receive your logs. Make it easy for your clients by always calling logging.getLogger locally. Hat Tip : I learned about this behavior from Victor Lin's Website.
So good practice is instead to always call logging.getLogger locally. E.g.
#my_module
import logging
logger = logging.getLogger(__name__)
def foo():
logging.getLogger(__name__).info('Hi, foo')
class Bar(object):
def bar(self):
logging.getLogger(__name__).info('Hi, bar')
Also, if you use fileconfig in your main, set disable_existing_loggers=False, just in case your library designers use module level logger instances.
I would like to add my solution (which is based on logging cookbook and other articles and suggestions from this thread. However it took me quite a while to figure out, why it wasn't immediately working how I expected. So I created a little test project to learn how logging is working.
Since I have figured it out, I wanted to share my solution, maybe it can be of help to someone.
I know some of my code might not be best practice, but I am still learning. I left the print() functions in there, as I used them, while logging was not working as expected. Those are removed in my other application. Also I welcome any feedback on any parts of the code or structure.
my_log_test project structure (cloned/simplified from another project I work on)
my_log_test
├── __init__.py
├── __main__.py
├── daemon.py
├── common
│ ├── my_logger.py
├── pkg1
│ ├── __init__.py
│ └── mod1.py
└── pkg2
├── __init__.py
└── mod2.py
Requirements
A few things different or that I have not seen explicitly mentioned in the combination I use:
the main module is daemon.pywhich is called by __main__.py
I want to be able to call the modules mod1.py and mod2.py seperately while in development/testing
At this point I did not want to use basicConfig() or FileConfig() but keep it like in the logging cookbook
So basically, that means, I need to initialize the root logger in daemon.py (always) and in the modules mod1.py and mod2.py (only when calling them directly).
To make this init in several modules easier, I created my_logger.py which does, what is described in the cookbook.
My mistakes
Beforehand, my mistake in that module was to init the logger with logger = logging.getLogger(__name__) (module logger) instead of using logger = logging.getLogger() (to get the root logger).
The first problem was, that when called from daemon.py the logger's namespace was set to my_log_test.common.my_logger. The module logger in mod1.py with an "unmatching" namespace my_log_test.pkg1.mod1 could hence not attach to the other logger and I would see no log output from mod1.
The second "problem" was, that my main program is in daemon.py and not in __main__.py. But after all not a real problem for me, but it added to the namespace confusion.
Working solution
This is from the cookbook but in a separate module. I also added a logger_cleanup function that I can call from daemon, to remove logs older than x days.
## my_logger.py
from datetime import datetime
import time
import os
## Init logging start
import logging
import logging.handlers
def logger_init():
print("print in my_logger.logger_init()")
print("print my_logger.py __name__: " +__name__)
path = "log/"
filename = "my_log_test.log"
## get logger
#logger = logging.getLogger(__name__) ## this was my mistake, to init a module logger here
logger = logging.getLogger() ## root logger
logger.setLevel(logging.INFO)
# File handler
logfilename = datetime.now().strftime("%Y%m%d_%H%M%S") + f"_{filename}"
file = logging.handlers.TimedRotatingFileHandler(f"{path}{logfilename}", when="midnight", interval=1)
#fileformat = logging.Formatter("%(asctime)s [%(levelname)s] %(message)s")
fileformat = logging.Formatter("%(asctime)s [%(levelname)s]: %(name)s: %(message)s")
file.setLevel(logging.INFO)
file.setFormatter(fileformat)
# Stream handler
stream = logging.StreamHandler()
#streamformat = logging.Formatter("%(asctime)s [%(levelname)s:%(module)s] %(message)s")
streamformat = logging.Formatter("%(asctime)s [%(levelname)s]: %(name)s: %(message)s")
stream.setLevel(logging.INFO)
stream.setFormatter(streamformat)
# Adding all handlers to the logs
logger.addHandler(file)
logger.addHandler(stream)
def logger_cleanup(path, days_to_keep):
lclogger = logging.getLogger(__name__)
logpath = f"{path}"
now = time.time()
for filename in os.listdir(logpath):
filestamp = os.stat(os.path.join(logpath, filename)).st_mtime
filecompare = now - days_to_keep * 86400
if filestamp < filecompare:
lclogger.info("Delete old log " + filename)
try:
os.remove(os.path.join(logpath, filename))
except Exception as e:
lclogger.exception(e)
continue
to run deamon.py (through __main__.py) use python3 -m my_log_test
## __main__.py
from my_log_test import daemon
if __name__ == '__main__':
print("print in __main__.py")
daemon.run()
to run deamon.py (directly) use python3 -m my_log_test.daemon
## daemon.py
from datetime import datetime
import time
import logging
import my_log_test.pkg1.mod1 as mod1
import my_log_test.pkg2.mod2 as mod2
## init ROOT logger from my_logger.logger_init()
from my_log_test.common.my_logger import logger_init
logger_init() ## init root logger
logger = logging.getLogger(__name__) ## module logger
def run():
print("print in daemon.run()")
print("print daemon.py __name__: " +__name__)
logger.info("Start daemon")
loop_count = 1
while True:
logger.info(f"loop_count: {loop_count}")
logger.info("do stuff from pkg1")
mod1.do1()
logger.info("finished stuff from pkg1")
logger.info("do stuff from pkg2")
mod2.do2()
logger.info("finished stuff from pkg2")
logger.info("Waiting a bit...")
time.sleep(30)
if __name__ == '__main__':
try:
print("print in daemon.py if __name__ == '__main__'")
logger.info("running daemon.py as main")
run()
except KeyboardInterrupt as e:
logger.info("Program aborted by user")
except Exception as e:
logger.info(e)
To run mod1.py (directly) use python3 -m my_log_test.pkg1.mod1
## mod1.py
import logging
# mod1_logger = logging.getLogger(__name__)
mod1_logger = logging.getLogger("my_log_test.daemon.pkg1.mod1") ## for testing, namespace set manually
def do1():
print("print in mod1.do1()")
print("print mod1.py __name__: " +__name__)
mod1_logger.info("Doing someting in pkg1.do1()")
if __name__ == '__main__':
## Also enable this pkg to be run directly while in development with
## python3 -m my_log_test.pkg1.mod1
## init root logger
from my_log_test.common.my_logger import logger_init
logger_init() ## init root logger
print("print in mod1.py if __name__ == '__main__'")
mod1_logger.info("Running mod1.py as main")
do1()
To run mod2.py (directly) use python3 -m my_log_test.pkg2.mod2
## mod2.py
import logging
logger = logging.getLogger(__name__)
def do2():
print("print in pkg2.do2()")
print("print mod2.py __name__: " +__name__) # setting namespace through __name__
logger.info("Doing someting in pkg2.do2()")
if __name__ == '__main__':
## Also enable this pkg to be run directly while in development with
## python3 -m my_log_test.pkg2.mod2
## init root logger
from my_log_test.common.my_logger import logger_init
logger_init() ## init root logger
print("print in mod2.py if __name__ == '__main__'")
logger.info("Running mod2.py as main")
do2()
Happy if it helps. Happy to receive feedback as well!
You could also come up with something like this!
def get_logger(name=None):
default = "__app__"
formatter = logging.Formatter('%(levelname)s: %(asctime)s %(funcName)s(%(lineno)d) -- %(message)s',
datefmt='%Y-%m-%d %H:%M:%S')
log_map = {"__app__": "app.log", "__basic_log__": "file1.log", "__advance_log__": "file2.log"}
if name:
logger = logging.getLogger(name)
else:
logger = logging.getLogger(default)
fh = logging.FileHandler(log_map[name])
fh.setFormatter(formatter)
logger.addHandler(fh)
logger.setLevel(logging.DEBUG)
return logger
Now you could use multiple loggers in same module and across whole project if the above is defined in a separate module and imported in other modules were logging is required.
a=get_logger("__app___")
b=get_logger("__basic_log__")
a.info("Starting logging!")
b.debug("Debug Mode")
#Yarkee's solution seemed better. I would like to add somemore to it -
class Singleton(type):
_instances = {}
def __call__(cls, *args, **kwargs):
if cls not in cls._instances.keys():
cls._instances[cls] = super(Singleton, cls).__call__(*args, **kwargs)
return cls._instances[cls]
class LoggerManager(object):
__metaclass__ = Singleton
_loggers = {}
def __init__(self, *args, **kwargs):
pass
#staticmethod
def getLogger(name=None):
if not name:
logging.basicConfig()
return logging.getLogger()
elif name not in LoggerManager._loggers.keys():
logging.basicConfig()
LoggerManager._loggers[name] = logging.getLogger(str(name))
return LoggerManager._loggers[name]
log=LoggerManager().getLogger("Hello")
log.setLevel(level=logging.DEBUG)
So LoggerManager can be a pluggable to the entire application.
Hope it makes sense and value.
There are several answers. i ended up with a similar yet different solution that makes sense to me, maybe it will make sense to you as well.
My main objective was to be able to pass logs to handlers by their level (debug level logs to the console, warnings and above to files):
from flask import Flask
import logging
from logging.handlers import RotatingFileHandler
app = Flask(__name__)
# make default logger output everything to the console
logging.basicConfig(level=logging.DEBUG)
rotating_file_handler = RotatingFileHandler(filename="logs.log")
rotating_file_handler.setLevel(logging.INFO)
app.logger.addHandler(rotating_file_handler)
created a nice util file named logger.py:
import logging
def get_logger(name):
return logging.getLogger("flask.app." + name)
the flask.app is a hardcoded value in flask. the application logger is always starting with flask.app as its the module's name.
now, in each module, i'm able to use it in the following mode:
from logger import get_logger
logger = get_logger(__name__)
logger.info("new log")
This will create a new log for "app.flask.MODULE_NAME" with minimum effort.
The best practice would be to create a module separately which has only one method whose task we be to give a logger handler to the the calling method. Save this file as m_logger.py
import logger, logging
def getlogger():
# logger
logger = logging.getLogger(__name__)
logger.setLevel(logging.DEBUG)
# create console handler and set level to debug
#ch = logging.StreamHandler()
ch = logging.FileHandler(r'log.txt')
ch.setLevel(logging.DEBUG)
# create formatter
formatter = logging.Formatter('%(asctime)s - %(levelname)s - %(message)s')
# add formatter to ch
ch.setFormatter(formatter)
# add ch to logger
logger.addHandler(ch)
return logger
Now call the getlogger() method whenever logger handler is needed.
from m_logger import getlogger
logger = getlogger()
logger.info('My mssg')
New to python so I don't know if this is advisable, but it works great for not re-writing boilerplate.
Your project must have an init.py so it can be loaded as a module
# Put this in your module's __init__.py
import logging.config
import sys
# I used this dictionary test, you would put:
# logging.config.fileConfig('logging.conf')
# The "" entry in loggers is the root logger, tutorials always
# use "root" but I can't get that to work
logging.config.dictConfig({
"version": 1,
"formatters": {
"default": {
"format": "%(asctime)s %(levelname)s %(name)s %(message)s"
},
},
"handlers": {
"console": {
"level": 'DEBUG',
"class": "logging.StreamHandler",
"stream": "ext://sys.stdout"
}
},
"loggers": {
"": {
"level": "DEBUG",
"handlers": ["console"]
}
}
})
def logger():
# Get the name from the caller of this function
return logging.getLogger(sys._getframe(1).f_globals['__name__'])
sys._getframe(1) suggestion comes from here
Then to use your logger in any other file:
from [your module name here] import logger
logger().debug("FOOOOOOOOO!!!")
Caveats:
You must run your files as modules, otherwise import [your module] won't work:
python -m [your module name].[your filename without .py]
The name of the logger for the entry point of your program will be __main__, but any solution using __name__ will have that issue.