Simple Python Logging Exception from Future

exception, logging, python

Solution

To use `logger.exception` and get the traceback etc, you need to be inside an except block. Instead of checking the `future.exception()`, which returns the exception (if any), use the `future.result()` which raises the exception (if any).

So, `try` this instead (if you'll pardon the pun):

def future_callback_error_logger(future):
    try:
        future.result()
    except Exception:
        logger.exception("Executor Exception")

Problem

This should be a really simple question, but after googling, reading docs, and several other SO threads, I don't see the answer: How do I log an exception with Python standard logging? One small wrinkle is that I'm getting the exception from a Future. I'm not writing the `except` exception handler myself. Ideally, I would get the exception message, a stack trace, the extra message sent, and maybe the type of exception. Here's a simple program that shows my issue: ``` import logging from concurrent.futures import ThreadPoolExecutor logger = logging.getLogger(__name__) def test_f(a, b=-99, c=50): logger.info("test_f a={} b={} c={}".format(a, b, c)) def future_callback_error_logger(future): e = future.exception() if e is not None: # This log statement does not seem to do what I want. # It logs "Executor Exception" with no information about the exception. # I would like to see the exception type, message, and stack trace. logger.error("Executor Exception", exc_info=e) def submit_with_log_on_error(executor, func, *args, **kwargs): future = executor.submit(func, *args, **kwargs) future.add_done_callback(future_callback_error_logger) if __name__ == "__main__": logging.basicConfig(level="DEBUG") logger.info("start") executor = ThreadPoolExecutor(max_workers=5) # This will work. submit_with_log_on_error(executor, test_f, 10, c=20) # This will intentionally trigger an error due to too many arguments. # I would like that error to be properly logged. submit_with_log_on_error(executor, test_f, 10, 20, 30, 40) # This will work. submit_with_log_on_error(executor, test_f, 50, c=60) executor.shutdown(True) logger.info("shutdown") ```

Original source