how to handle exceptions in python

Question

I get errors like this

return SegmentWriter(self, **kwargs)
  File "/usr/local/lib/python2.7/site-packages/whoosh/writing.py", line 502, in __init__
    raise LockError
whoosh.index.LockError

I would like to catch these errors with a try/except statement. So I wrote

try:
    do whatever causes the error
except LockError:
    print "LockError..."
    handle error

but this leads to a NameError, since LockError is unknown?

    except LockError:
NameError: global name 'LockError' is not defined

how do I handle these Lock errors?

u can try except Exception as e: in the next line print e — manoj prashant k
– manoj prashant k, Commented May 23, 2016 at 8:53
@manojprashantk please don't suggest things like that. It's a terrible idea to catch all exceptions; only ever catch the ones you are expecting and know how to deal with. — Daniel Roseman
– Daniel Roseman, Commented May 23, 2016 at 9:11
@DanielRoseman: I think they meant to help with debugging; catch all exceptions just to be able to see the type. — Martijn Pieters
– Martijn Pieters, Commented May 23, 2016 at 9:11
@Daniel Roseman thanks for your suggestion. But that is why i have given it as a comment rather than posting it as an answer :) — manoj prashant k
– manoj prashant k, Commented May 23, 2016 at 9:56

SparkAndShine · Accepted Answer · 2016-05-23 09:00:06Z

4

First import exception in your namespace. Add this to your module:

from whoosh.index import LockError

edited May 23, 2016 at 9:00

SparkAndShine

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answered May 23, 2016 at 8:52

Daniil Ryzhkov

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