如何在Python中记录源文件名和行号

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?

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')

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)d %(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

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.

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.

# 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 ...
# An naive Sample of usage:
try:
    logger.info('Sample of info log')
    # your code here
except Exception as e:
    logger.error(e)

Different of the other answers, this will log 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.