python - 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:

their_md5 = 'c38f03d2b7160f891fc36ec776ca4685'
my_md5 = 'c64e53bbb108a1c65e31eb4d1bb8e3b7' 
if their_md5 != my_md5:
    raise ValueError('md5 sum does not match!')

Existing output now, but only desired when called with foo.py --debug:

Traceback (most recent call last):
  File "b:codeaptapt.py", line 1647, in <module>
    __main__.__dict__[command] (packages)
  File "b:codeaptapt.py", line 399, in md5
    raise ValueError('md5 sum does not match!')
ValueError: md5 sum does not match!

Desired normal output:

ValueError: md5 sum does not match!

Here's a test script: https://gist.github.com/maphew/e3a75c147cca98019cd8

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.