Hide traceback unless a debug flag is set

Question

What is the idiomatic python way to hide traceback errors unless a verbose or debug flag is set?

Example code:

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Answer 1

The short way is using the sys module and use this command:

sys.tracebacklimit = 0

Use your flag to determine the behaviour.

Example:

>>> import sys
>>> sys.tracebacklimit=0
>>> int('a')
ValueError: invalid literal for int() with base 10: 'a'

The nicer way is to use and exception hook:

def exception_handler(exception_type, exception, traceback):
    # All your trace are belong to us!
    # your format
    print "%s: %s" % (exception_type.__name__, exception)

sys.excepthook = exception_handler

Edit:

If you still need the option of falling back to the original hook:

def exception_handler(exception_type, exception, traceback, debug_hook=sys.excepthook):
    if _your_debug_flag_here:
        debug_hook(exception_type, exception, traceback)
    else:
        print "%s: %s" % (exception_type.__name__, exception)

Now you can pass a debug hook to the handler, but you'll most likely want to always use the one originated in sys.excepthook (so pass nothing in debug_hook). Python binds default arguments once in definition time (common pitfall...) which makes this always work with the same original handler, before replaced.

Answer 2

try:
    pass # Your code here
except Exception as e:
    if debug:
        raise # re-raise the exception
              # traceback gets printed
    else:
        print("{}: {}".format(type(e).__name__, e))