Logging uncaught exceptions in Python

2019-01-03 21:24发布

How do you cause uncaught exceptions to output via the logging module rather than to stderr?

I realize the best way to do this would be:

try:
    raise Exception, 'Throwing a boring exception'
except Exception, e:
    logging.exception(e)

But my situation is such that it would be really nice if logging.exception(...) were invoked automatically whenever an exception isn't caught.

8条回答
成全新的幸福
2楼-- · 2019-01-03 21:55

Here's a complete small example that also includes a few other tricks:

import sys
import logging
logger = logging.getLogger(__name__)
handler = logging.StreamHandler(stream=sys.stdout)
logger.addHandler(handler)

def handle_exception(exc_type, exc_value, exc_traceback):
    if issubclass(exc_type, KeyboardInterrupt):
        sys.__excepthook__(exc_type, exc_value, exc_traceback)
        return

    logger.error("Uncaught exception", exc_info=(exc_type, exc_value, exc_traceback))

sys.excepthook = handle_exception

if __name__ == "__main__":
    raise RuntimeError("Test unhandled")
  • Ignore KeyboardInterrupt so a console python program can exit with Ctrl + C.

  • Rely entirely on python's logging module for formatting the exception.

  • Use a custom logger with an example handler. This one changes the unhandled exception to go to stdout rather than stderr, but you could add all sorts of handlers in this same style to the logger object.

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兄弟一词,经得起流年.
3楼-- · 2019-01-03 21:55

Maybe you could do something at the top of a module that redirects stderr to a file, and then logg that file at the bottom

sock = open('error.log', 'w')               
sys.stderr = sock

doSomething() #makes errors and they will log to error.log

logging.exception(open('error.log', 'r').read() )
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