python - 在自定义类中包装记录器功能时显示正确的 funcName

Question

这是我用于记录的格式化字符串:

'%(asctime)s - %(levelname)-10s - %(funcName)s - %(message)s'

但是为了显示日志消息，我有一个包装器做更多的事情(我设置了不同的日志级别，配置不同的日志后端，提供方便的函数来访问自定义级别等):

class MyLogger(logging.Logger):

    def split_line(self, level, message):
        ....
        self.log.(level, line)

    def progress(self, message):
        self.split_line(PROGRESS, message)

有了这个设置，每当我记录一些东西时:

def myfunc():
    log.progress('Hello')

我明白了:

013-10-27 08:47:30,130 - PROGRESS   - split_line - Hello

这不是我想要的，即:

013-10-27 08:47:30,130 - PROGRESS   - myfunc     - Hello

如何告诉记录器为函数名称使用正确的上下文？我认为这实际上会比 stackframe 高两个级别。

编辑

这是一个显示问题的测试程序:

import sys
import logging

PROGRESS = 1000

class MyLogger(logging.Logger):

    PROGRESS = PROGRESS
    LOG_FORMATTER = '%(asctime)s - %(levelname)-10s - %(funcName)s - %(message)s'
    DEF_LOGGING_LEVEL = logging.WARNING

    def __init__(self, log_name, level=None):
        logging.Logger.__init__(self, log_name)
        self.formatter = logging.Formatter(self.LOG_FORMATTER)
        self.initLogger(level)

    def initLogger(self, level=None):
        self.setLevel(level or self.DEF_LOGGING_LEVEL)
        self.propagate = False

    def add_handler(self, log_file, use_syslog):
        if use_syslog : hdlr = logging.handlers.SysLogHandler(address='/dev/log')
        elif log_file : hdlr = logging.FileHandler(log_file)
        else          : hdlr = logging.StreamHandler(sys.stderr)
        hdlr.setFormatter(self.formatter)
        self.addHandler(hdlr)
        return hdlr

    def addHandlers(self, log_file=None, progress_file=None, use_syslog=False):
        self.logger_hdlr = self.add_handler(log_file, use_syslog)
        if progress_file:
            self.progress_hdlr = self.add_handler(progress_file, use_syslog)
            self.progress_hdlr.setLevel(self.PROGRESS)
        else:
            self.progress_hdlr = None

    def split_line(self, level, txt, *args):
        txt = txt % (args)
        for line in txt.split('\n'):
            self.log(level, line)

    def progress(self, txt, *args):
        self.split_line(self.PROGRESS, txt, *args)

logging.setLoggerClass(MyLogger)
logging.addLevelName(PROGRESS, 'PROGRESS')
logger = logging.getLogger(__name__)
logger.addHandlers()

name = 'John'
logger.progress('Hello %s\nHow are you doing?', name)

生产:

2013-10-27 09:47:39,577 - PROGRESS   - split_line - Hello John
2013-10-27 09:47:39,577 - PROGRESS   - split_line - How are you doing?

Answer 1

本质上，应该归咎于 Logger 类的代码:

这个方法

def findCaller(self):
    """
    Find the stack frame of the caller so that we can note the source
    file name, line number and function name.
    """
    f = currentframe()
    #On some versions of IronPython, currentframe() returns None if
    #IronPython isn't run with -X:Frames.
    if f is not None:
        f = f.f_back
    rv = "(unknown file)", 0, "(unknown function)"
    while hasattr(f, "f_code"):
        co = f.f_code
        filename = os.path.normcase(co.co_filename)
        if filename == _srcfile:
            f = f.f_back
            continue
        rv = (co.co_filename, f.f_lineno, co.co_name)
        break
    return rv

返回调用者链中不属于当前模块的第一个函数。

您可以继承 Logger 并通过添加稍微复杂的逻辑来覆盖此方法。跳过另一个级别的调用深度或添加另一个条件。

---

在您非常特殊的情况下，避免自动行拆分可能会更简单

logger.progress('Hello %s', name)
logger.progress('How are you doing?')

或去做

def splitter(txt, *args)
    txt = txt % (args)
    for line in txt.split('\n'):
        yield line

for line in splitter('Hello %s\nHow are you doing?', name):
    logger.progress(line)

并且有一个

def progress(self, txt, *args):
    self.log(self.PROGRESS, txt, *args)

可能会省去很多麻烦。

编辑 2:不，这无济于事。它现在会将 progress 显示为您的调用者函数名称...