如何在Python中禁用标准错误流的日志记录?这行不通:
import logging
logger = logging.getLogger()
logger.removeHandler(sys.stderr)
logger.warning('foobar') # emits 'foobar' on sys.stderr
如何在Python中禁用标准错误流的日志记录?这行不通:
import logging
logger = logging.getLogger()
logger.removeHandler(sys.stderr)
logger.warning('foobar') # emits 'foobar' on sys.stderr
当前回答
使用上下文管理器-[最简单]
import logging
class DisableLogger():
def __enter__(self):
logging.disable(logging.CRITICAL)
def __exit__(self, exit_type, exit_value, exit_traceback):
logging.disable(logging.NOTSET)
使用示例:
with DisableLogger():
do_something()
如果你需要[更复杂的]细粒度的解决方案,你可以看看AdvancedLogger
AdvancedLogger can be used for fine grained logging temporary modifications
How it works:
Modifications will be enabled when context_manager/decorator starts working and be reverted after
Usage:
AdvancedLogger can be used
- as decorator `@AdvancedLogger()`
- as context manager `with AdvancedLogger():`
It has three main functions/features:
- disable loggers and it's handlers by using disable_logger= argument
- enable/change loggers and it's handlers by using enable_logger= argument
- disable specific handlers for all loggers, by using disable_handler= argument
All features they can be used together
AdvancedLogger的用例
# Disable specific logger handler, for example for stripe logger disable console
AdvancedLogger(disable_logger={"stripe": "console"})
AdvancedLogger(disable_logger={"stripe": ["console", "console2"]})
# Enable/Set loggers
# Set level for "stripe" logger to 50
AdvancedLogger(enable_logger={"stripe": 50})
AdvancedLogger(enable_logger={"stripe": {"level": 50, "propagate": True}})
# Adjust already registered handlers
AdvancedLogger(enable_logger={"stripe": {"handlers": "console"}
其他回答
我不是很了解日志模块,但是我使用它的方式是我通常只希望禁用调试(或信息)消息。您可以使用Handler.setLevel()将日志级别设置为CRITICAL或更高。
此外,您还可以替换sys。Stderr和sys。打开用于写入的文件的标准输出。见http://docs.python.org/library/sys.html # sys.stdout。但我不建议这样做。
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())
我使用:
logger = logging.getLogger()
logger.disabled = True
... whatever you want ...
logger.disabled = False
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()
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)