Python logging module having a formatter causes AttributeError
Asked Answered
H

1

16

I am writing a terminal application, which, after passing in -v option, gets, unsurprisingly, verbose. I want to have the output available in the terminal, for easy testing (it gets redirected to a log file when running as cron anyways).

However, python logging module doesn't allow me to write out the messages with corresponding levels when using a formatter. (Formatter is copied directly from Python Logging Cookbok)

This behavior is not limited to Python3 only. Python2.7 raises the same exception under the given conditions.


one.py

from sys import stdout
import logging

if __name__ == '__main__':

    level = 20

    log = logging.getLogger()
    formatter = logging.Formatter('%(asctime)s - %(name)s - %(levelname)s - %(message)s')
    handler = logging.StreamHandler(stdout).setFormatter(formatter)
    log.addHandler(handler)
    log.setLevel(level)

    log.info("Blah")

one.py output

Traceback (most recent call last):
  File "/home/tlevi/PycharmProjects/untitled/main.py", line 14, in <module>
    log.info("Blah")
  File "/usr/lib/python3.4/logging/__init__.py", line 1279, in info
    self._log(INFO, msg, args, **kwargs)
  File "/usr/lib/python3.4/logging/__init__.py", line 1414, in _log
    self.handle(record)
  File "/usr/lib/python3.4/logging/__init__.py", line 1424, in handle
    self.callHandlers(record)
  File "/usr/lib/python3.4/logging/__init__.py", line 1485, in callHandlers
    if record.levelno >= hdlr.level:
AttributeError: 'NoneType' object has no attribute 'level'

two.py (works like a charm)

from sys import stdout
import logging

if __name__ == '__main__':

    level = 20

    log = logging.getLogger()
    handler = logging.StreamHandler(stdout)
    log.addHandler(handler)
    log.setLevel(level)

    log.info("Blah")

two.py output

Blah
Holily answered 16/12, 2015 at 18:30 Comment(0)
C
28

Instead of

handler = logging.StreamHandler(stdout).setFormatter(formatter)

Try:

handler = logging.StreamHandler(stdout)
handler.setFormatter(formatter)

What is happening is that in the first case you are assigning the return of setFormatter() to the handler variable, but setFormatter() does not return the handler (i.e. it returns None)

Carmelitacarmelite answered 16/12, 2015 at 18:42 Comment(2)
The lack of support for chaining in such a popular library is disappointing.Crowns
@Crowns - it begs the question "why is it popular?"Pacien

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