Python get an error code from exception

Question

In python, I have code that handles exceptions and prints error codes and messages.

try:
    somecode() #raises NameError
except Exception as e:
    print('Error! Code: {c}, Message, {m}'.format(c = e.code, m = str(e))

However, e.code is not the proper way to get the error name (NameError), and I cannot find the answer to this. How am I supossed to get the error code?

Error code (whatever this means) or error name? For the latter, try `type(e).__name__` — Michael Butscher, Dec 09 '17 at 00:21
Did you look at the Python documentation for the Exception class? — OneCricketeer, Dec 09 '17 at 00:23
use `print( dir(e) )` to see what fields you have in `e`. Every exception may have different fields. — furas, Dec 09 '17 at 00:23

score 8 · Answer 1 · answered Oct 01 '21 at 20:14

8

This has worked for me.

  except Exception as e:
    errnum = e.args[0]

answered Oct 01 '21 at 20:14

PerkHaus

89
1
2

this dont return a code due to #Dustin Wyatt explanation. Just returns the error message – JRichardsz Feb 05 '23 at 04:51

score 4 · Answer 2 · answered Jan 22 '20 at 22:39

4

Python exceptions do not have "codes".

You can create a custom exception that does have a property called code and then you can access it and print it as desired.

This answer has an example of adding a code property to a custom exception.

answered Jan 22 '20 at 22:39

Dustin Wyatt

4,046
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François M. · Answer 3 · 2020-11-13T03:06:14.210

Your question is unclear, but from what I understand, you do not want to find the name of the error (NameError), but the error code. Here is how to do it. First, run this:

try:
    # here, run some version of your code that you know will fail, for instance:
    this_variable_does_not_exist_so_this_code_will_fail
except Exception as e:
    print(dir(e))

You can now see what is in e. You will get something like this:

['__cause__', '__class__', '__context__', '__delattr__', '__dict__', '__dir__', '__doc__', '__eq__', '__format__', '__ge__', '__getattribute__', '__gt__', '__hash__', '__init__', '__init_subclass__', '__le__', '__lt__', '__ne__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__setstate__', '__sizeof__', '__str__', '__subclasshook__', '__suppress_context__', '__traceback__', 'args', 'with_traceback']

This list will include special methods (__x__ stuff), but will end with things without underscores. You can try them one by one to find what you want, like this:

try:
    # here, run some version of your code that you know will fail, for instance:
    this_variable_does_not_exist_so_this_code_will_fail
except Exception as e:
    print(e.args)
    print(e.with_traceback)

In the case of this specific error, print(e.args) is the closest you'll get to an error code, it will output ("name 'this_variable_does_not_exist_so_this_code_will_fail' is not defined",).

In this case, there is only two things to try, but in your case, your error might have more. For example, in my case, a Tweepy error, the list was:

['__cause__', '__class__', '__context__', '__delattr__', '__dict__', '__dir__', '__doc__', '__eq__', '__format__', '__ge__', '__getattribute__', '__gt__', '__hash__', '__init__', '__init_subclass__', '__le__', '__lt__', '__module__', '__ne__', '__new__', '__reduce__', '__reduce_ex__', '__repr__', '__setattr__', '__setstate__', '__sizeof__', '__str__', '__subclasshook__', '__suppress_context__', '__traceback__', '__weakref__', 'api_code', 'args', 'reason', 'response', 'with_traceback']

I tried the last five one by one. From print(e.args), I got ([{'code': 187, 'message': 'Status is a duplicate.'}],), and from print(e.api_code), I got 187. So I figured out that either e.args[0][0]["code"] or e.api_code would give me the error code I'm searching for.

score -2 · Accepted Answer · answered Dec 09 '17 at 00:46

-2

Try this:

try:
    somecode() #raises NameError
except Exception as e:
    print('Error! Code: {c}, Message, {m}'.format(c = type(e).__name__, m = str(e)))

Read this for a more detailed explanation.

answered Dec 09 '17 at 00:46

63677

282
3
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5

This does *not* print the error code, only the name of the exception `NameError` and the exception's message – Hamman Samuel Oct 07 '18 at 11:19
3

Mods: this should not be marked as the correct answer as it does not answer the question. – rjurney Jan 21 '20 at 19:52
See my answer for (what I believe to be) the correct one. – François M. Nov 13 '20 at 03:03

score -2 · Answer 5 · edited Sep 11 '20 at 12:19

-2

Since it return the object of tuple of tuple of dictionary, we can extract the code as

try:
  
  pass

except Exception as e:

  print(e[0][0]['code'] + e[0][0]['message'])

edited Sep 11 '20 at 12:19

Brian Tompsett - 汤莱恩

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answered Sep 11 '20 at 10:25

M.G

1
1

Python get an error code from exception

5 Answers5