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:\code\apt\apt.py", line 1647, in <module>     __main__.__dict__[command] (packages)   File "b:\code\apt\apt.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.