$ ./runtests.py -v tests/managers/test_customer.py:CustomerManagerTest.test_register_without_subscription --ipdb
...
test_register_without_subscription (tests.managers.test_customer.CustomerManagerTest) ...
- TRACEBACK --------------------------------------------------------------------
Traceback (most recent call last):
File "/usr/lib/python2.7/unittest/case.py", line 331, in run
testMethod()
File "*****/tests/managers/test_customer.py", line 198, in test_register_without_subscription
1/0
ZeroDivisionError: integer division or modulo by zero
--------------------------------------------------------------------------------
> *****/tests/managers/test_customer.py(198)test_register_without_subscription()
197 def test_register_without_subscription(self):
--> 198 1/0
199 ...
ipdb> import sys
ipdb> sys.exc_info()
(<type 'exceptions.AttributeError'>, AttributeError("Pdb instance has no attribute 'do_sys'",), <traceback object at 0x47eb908>)
ipdb>
I could not find any command in ipdb help
that shows me current exception.
Doing import sys; print sys.exc_info()
doesn't work.
Currently I do:
try:
do_something_that_raises_an_exception()
except Exception as exc:
import ipdb; ipdb.set_trace()
then I can work with exc
to analyze it.
How to easily get a reference to the currently effective exception?
This has frustrated me too for a while. I eventually found the answer here, along with a good detailed explanation.
The short answer is, use the
!
magic prefix (!sys.exc_info()
):This basically tells the debugger: "guessing wouldn't be necessary. it is python code I'm typing", thus preventing it from trying to guess what you mean by "sys", a process during which some internal exception is raised, overwriting
sys.exc_info()
, which used to hold your original exception.