I saw a post from stackoverflow asking why his logger does not work as expect? The code is like:

import logging


logger = logging.getLogger()

logger.info("some message")

There is no message coming out. Okay, let’s check the level of this logger:

print(logger.level)

# output is 30 (logging.WARNING)

Then we can just change its level to DEBUG, it will certain work, right?

logger.setLevel(logging.DEBUG)

However, still no output! W*F, what happened?

According to official doc, if you do not set an explicit handler for the logger, a special handler called lastResort will be used. See the code here. By default the logging level of lastResort (it is stream handler) is 30. We can change its level to output info message.

# setting both the logger and handler's level will work as expected.
logger.setLevel(logging.DEBUG)
logging.lastResort.setLevel(logging.DEBUG)

However, this is like a hack and never an encouraged solution.

Using logging.basicConfig()

If we want to do logging real quick, we can use the method logging.basicConfig.

logging.basicConfig(level=logging.DEBUG)

This will create the logger. The logger.level will be the level we set here. Without a level param, the default level for root logger is WARNING.

A stream handler will be also created, with level NOTSET. This can be verified by printing the logger’s handlers:

print(logger.handlers)
# you will see: [<StreamHandler <stderr> (NOTSET)>]

reference

• logger logging without handler: https://stackoverflow.com/q/64570918/6064933

Author jdhao

LastMod

2022-05-27

License CC BY-NC-ND 4.0

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