我找到了一个解决方案:

logger = logging.getLogger('my-logger')
logger.propagate = False
# now if you use logger it will not log to console.

这将防止日志被发送到包含控制台日志的上层记录器。

不需要转移标准输出。这里有一个更好的方法:

import logging
class MyLogHandler(logging.Handler):
    def emit(self, record):
        pass

logging.getLogger().addHandler(MyLogHandler())

一个更简单的方法是:

logging.getLogger().setLevel(100)

使用装饰器找到了一个优雅的解决方案，它解决了以下问题:如果您正在编写一个具有多个函数的模块，每个函数都有几个调试消息，并且您希望禁用除当前关注的函数之外的所有函数的登录，该怎么办?

你可以使用装饰器:

import logging, sys
logger = logging.getLogger()
logging.basicConfig(stream=sys.stderr, level=logging.DEBUG)


def disable_debug_messages(func):
    def wrapper(*args, **kwargs):
        prev_state = logger.disabled
        logger.disabled = True
        result = func(*args, **kwargs)
        logger.disabled = prev_state
        return result
    return wrapper

然后，你可以这样做:

@disable_debug_messages
def function_already_debugged():
    ...
    logger.debug("This message won't be showed because of the decorator")
    ...

def function_being_focused():
    ...
    logger.debug("This message will be showed")
    ...

即使从function_being_focused内部调用function_already_debug，也不会显示来自function_already_debug的调试消息。 这确保您将只看到您所关注的函数的调试消息。

希望能有所帮助!

这将防止所有来自第三个库的日志记录，就像这里描述的那样 https://docs.python.org/3/howto/logging.html#configuring-logging-for-a-library

logging.getLogger('somelogger').addHandler(logging.NullHandler())

子类化你想要暂时禁用的处理程序:

class ToggledHandler(logging.StreamHandler):
"""A handler one can turn on and off"""

def __init__(self, args, kwargs):
    super(ToggledHandler, self).__init__(*args, **kwargs)
    self.enabled = True  # enabled by default

def enable(self):
    """enables"""
    self.enabled = True

def disable(self):
    """disables"""
    self.enabled = False

def emit(self, record):
    """emits, if enabled"""
    if self.enabled:
        # this is taken from the super's emit, implement your own
        try:
            msg = self.format(record)
            stream = self.stream
            stream.write(msg)
            stream.write(self.terminator)
            self.flush()
        except Exception:
            self.handleError(record)

通过名称查找处理程序非常简单:

_handler = [x for x in logging.getLogger('').handlers if x.name == your_handler_name]
if len(_handler) == 1:
    _handler = _handler[0]
else:
    raise Exception('Expected one handler but found {}'.format(len(_handler))

一旦发现:

_handler.disable()
doStuff()
_handler.enable()

您可以更改特定处理程序的调试模式级别，而不是完全禁用它。

因此，如果你有一个情况，你想停止调试模式的控制台，但你仍然需要保持其他级别，如错误。你可以像下面这样做

# create logger
logger = logging.getLogger(__name__)

def enableConsoleDebug (debug = False):
    #Set level to logging.DEBUG to see CRITICAL, ERROR, WARNING, INFO and DEBUG statements
    #Set level to logging.ERROR to see the CRITICAL & ERROR statements only
    logger.setLevel(logging.DEBUG)

    debugLevel = logging.ERROR
    if debug:
        debugLevel = logging.DEBUG

    for handler in logger.handlers:
        if type(handler) is logging.StreamHandler:
            handler.setLevel (debugLevel)