Is it possible to decorate/extend the python standard logging system, so that when a logging method is invoked it also logs the file and the line number where it was invoked or maybe the method that invoked it?
6 Answers
Sure, check formatters in logging docs. Specifically the lineno and pathname variables.
%(pathname)s Full pathname of the source file where the logging call was issued(if available).
%(filename)s Filename portion of pathname.
%(module)s Module (name portion of filename).
%(funcName)s Name of function containing the logging call.
%(lineno)d Source line number where the logging call was issued (if available).
Looks something like this:
formatter = logging.Formatter('[%(asctime)s] p%(process)s {%(pathname)s:%(lineno)d} %(levelname)s - %(message)s','%m-%d %H:%M:%S')
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5And, yes, the **upper/lower case mess** in the variables needs to be considered. – Tom Pohl May 15 '19 at 09:05
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3Otherwise referred to as "very poorly implemented camel case". – Jon Spencer Aug 06 '19 at 21:38
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2I'm using Logger.py file to share my logger between my files and when I use this syntax I get the filename and lineno of the Logger and not the file and line used to call the logger method. Any way of solving that ? – user2396640 May 12 '22 at 15:54
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why do you have the `{}` and the `...)d}`? – Charlie Parker Jun 17 '22 at 17:02
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1what does `p%(process)s` do? especially the p and s. – Charlie Parker Jun 20 '22 at 15:30
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@CharlieParker `p` is just the letter p `(process)` is the PID of the process that created that log and `s` is because it is a string – LombardiD Mar 06 '23 at 22:52
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It would help if you explained what you do with a formatter in terms of Python's `logging` module in a file. – rjurney Apr 19 '23 at 21:21
On top of Seb's very useful answer, here is a handy code snippet that demonstrates the logger usage with a reasonable format:
#!/usr/bin/env python
import logging
logging.basicConfig(format='%(asctime)s,%(msecs)03d %(levelname)-8s [%(filename)s:%(lineno)d] %(message)s',
datefmt='%Y-%m-%d:%H:%M:%S',
level=logging.DEBUG)
logger = logging.getLogger(__name__)
logger.debug("This is a debug log")
logger.info("This is an info log")
logger.critical("This is critical")
logger.error("An error occurred")
Generates this output:
2017-06-06:17:07:02,158 DEBUG [log.py:11] This is a debug log
2017-06-06:17:07:02,158 INFO [log.py:12] This is an info log
2017-06-06:17:07:02,158 CRITICAL [log.py:13] This is critical
2017-06-06:17:07:02,158 ERROR [log.py:14] An error occurred

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7Use this for more details: formatter = logging.Formatter( '%(asctime)s, %(levelname)-8s [%(filename)s:%(module)s:%(funcName)s:%(lineno)d] %(message)s') – Girish Gupta Mar 08 '18 at 09:43
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is there a way to change just in one place at the top of the code whether or not the logging messages get printed? I would like two modes, one with lots of prints to see what exactly the program does; and one, for when it's stable enough, where no output is shown. – Marie. P. Mar 10 '20 at 16:03
import logging
# your imports above ...
logging.basicConfig(
format='%(asctime)s,%(msecs)d %(levelname)-8s [%(pathname)s:%(lineno)d in ' \
'function %(funcName)s] %(message)s',
datefmt='%Y-%m-%d:%H:%M:%S',
level=logging.DEBUG
)
logger = logging.getLogger(__name__)
# your classes and methods below ...
# A very naive sample of usage:
try:
logger.info('Sample of info log')
# your code here
except Exception as e:
logger.error(e)
Different from the other answers, this will log the full path of file and the function name that might have occurred an error. This is useful if you have a project with more than one module and several files with the same name distributed in these modules.
Example output:
2022-12-02:10:00:00,000 INFO [<stdin>:2 in function <module>] Sample of info log
2022-12-02:10:00:00,000 INFO [<full path>/logging_test_file.py:15 in function <module>] Sample of info log

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To build on the above in a way that sends debug logging to standard out:
import logging
import sys
root = logging.getLogger()
root.setLevel(logging.DEBUG)
ch = logging.StreamHandler(sys.stdout)
ch.setLevel(logging.DEBUG)
FORMAT = "[%(filename)s:%(lineno)s - %(funcName)20s() ] %(message)s"
formatter = logging.Formatter(FORMAT)
ch.setFormatter(formatter)
root.addHandler(ch)
logging.debug("I am sent to standard out.")
Putting the above into a file called debug_logging_example.py
produces the output:
[debug_logging_example.py:14 - <module>() ] I am sent to standard out.
Then if you want to turn off logging comment out root.setLevel(logging.DEBUG)
.
For single files (e.g. class assignments) I've found this a far better way of doing this as opposed to using print()
statements. Where it allows you to turn the debug output off in a single place before you submit it.

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For devs using PyCharm or Eclipse pydev, the following will produce a link to the source of the log statement in the console log output:
import logging, sys, os
logging.basicConfig(stream=sys.stdout, level=logging.DEBUG, format='%(message)s | \'%(name)s:%(lineno)s\'')
log = logging.getLogger(os.path.basename(__file__))
log.debug("hello logging linked to source")
See Pydev source file hyperlinks in Eclipse console for longer discussion and history.

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If the logger is set with the GetLogger(name) option, where the name is a name that you specified you can also format the logger using %(name)s
. You can specify a different name in every file with the GetLogger function, and when a log is produced you will know from which file comes through the name you set.
Example:
import logging
logging.getLogger("main")
logging.basicConfig(#filename=log_fpath,
level=log_level,
format='[%(asctime)s] src:%(name)s %(levelname)s:%(message)s',
handlers=[logging.FileHandler(log_fpath)])

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