python - 子记录器的默认处理程序是什么？

Question

我知道根记录器具有 lastResort 处理程序，如果未使用 logging.basicConfig 配置根记录器处理程序并且用户使用logging.info/debug/warning/error/ritic("some msg")或者 logger =logging.gerLogger() 后跟 logger.info/debug/warning/error/ritic("some msg")

但是对于使用logger =logging.getLogger("someName")创建的子记录器，当我们编写logger.info/debug/warning/error/ritic("some msg")，并且没有配置处理程序，当 propagate = true 时，子记录器是否有默认处理程序？或者它是否回退到祖先的处理程序，然后返回到根记录器的处理程序，如果祖先/根记录器没有处理程序，则返回到根记录器 lastResort 处理程序？

如果传播设置为 false，那么是否会有一些默认处理程序分配给子记录器？

Answer 1

简短回答:logging.lastResort

长答案:

根据doc ，如果没有提供日志配置，则:

The event is output using a ‘handler of last resort’, stored in logging.lastResort. This internal handler is not associated with any logger, and acts like a StreamHandler which writes the event description message to the current value of sys.stderr (therefore respecting any redirections which may be in effect). No formatting is done on the message - just the bare event description message is printed. The handler’s level is set to WARNING, so all events at this and greater severities will be output.

同时根据logging source code :

class Logger(Filterer):
    # ...
    def callHandlers(self, record):
        # ...
        # found is the number of handlers
        if (found == 0):
            if lastResort:
                if record.levelno >= lastResort.level:
                    lastResort.handle(record)

因此，如果没有处理程序，它将调用 lastResort，无论它是根记录器还是子记录器。 lastResort 实际上是 _StderrHandler(WARNING) ，这是:

class _StderrHandler(StreamHandler):
    """
    This class is like a StreamHandler using sys.stderr, but always uses
    whatever sys.stderr is currently set to rather than the value of
    sys.stderr at handler construction time.
    """
    def __init__(self, level=NOTSET):
        """
        Initialize the handler.
        """
        Handler.__init__(self, level)

    @property
    def stream(self):
        return sys.stderr

所以你可以看到它实际上是一个具有警告级别的StreamHandler。