How to use loguru.catch in class

Question

I have a template return logger like this

class MyLog(object):
    def __init__(self, call_filename=None):
        self._caller_filename = call_filename

    def producer_log(self):
        logger.add(f"Log/{self.caller_filename}/{{time:YYYY-MM-DD}}.log", level="DEBUG",
                   backtrace=True, diagnose=True, rotation='00:00', enqueue=True, compression="zip")
        logger.add(f"Log/task_wrong/{{time:YYYY-MM-DD}}.log", level="DEBUG",
                   backtrace=True, diagnose=True, rotation='00:00', enqueue=True, compression="zip",
                   filter=lambda record: "special" in record["extra"],
                   format="{time:YYYY-MM-DD HH:mm:ss.SSS} | {level: <8} | {file}:{function}:{line} - {message}")
        task_error_log = logger.bind(special=True)
        return logger, task_error_log

And another py file 'test.py' use this function producer_log.As you see,i need send a call_filename to MyLog,so i take it in class of test.py.However,i cant use the decorator logger.catch in test.py which return by producer_log function

class Task(object):
    def __init__(self,filepath):
        self.logger, self.task_error_log = MyLog(filepath).producer_log()

I have tried take function producer_log out of class ,but i need send call_filename,so it's not work

score 0 · Answer 1 · answered Jul 04 '23 at 03:55

I derive from Task and it looks like work.

logger, task_error_log = MyLog(os.path.splitext(os.path.basename(__file__))[0]).producer_log()


class TaskInstance(Task):
    def __init__(self):
        super().__init__()
        self.logger, self.task_error_log = logger, task_error_log

    @logger.catch
    def main(self):
        super().main()

If you have better idea,please tell me. Thanks

How to use loguru.catch in class

1 Answers1