[python] How to print an exception in Python?

try:
    something here
except:
    print('the whatever error occurred.')

How can I print the error/exception in my except: block?

The traceback module provides methods for formatting and printing exceptions and their tracebacks, e.g. this would print exception like the default handler does:

import traceback

try:
    1/0
except Exception:
    traceback.print_exc()

Output:

Traceback (most recent call last):
  File "C:\scripts\divide_by_zero.py", line 4, in <module>
    1/0
ZeroDivisionError: division by zero

One has pretty much control on which information from the traceback to be displayed/logged when catching exceptions.

The code

with open("not_existing_file.txt", 'r') as text:
    pass

would produce the following traceback:

Traceback (most recent call last):
  File "exception_checks.py", line 19, in <module>
    with open("not_existing_file.txt", 'r') as text:
FileNotFoundError: [Errno 2] No such file or directory: 'not_existing_file.txt'

Print/Log the full traceback

As others already mentioned, you can catch the whole traceback by using the traceback module:

import traceback
try:
    with open("not_existing_file.txt", 'r') as text:
        pass
except Exception as exception:
    traceback.print_exc()

This will produce the following output:

Traceback (most recent call last):
  File "exception_checks.py", line 19, in <module>
    with open("not_existing_file.txt", 'r') as text:
FileNotFoundError: [Errno 2] No such file or directory: 'not_existing_file.txt'

You can achieve the same by using logging:

try:
    with open("not_existing_file.txt", 'r') as text:
        pass
except Exception as exception:
    logger.error(exception, exc_info=True)

Output:

__main__: 2020-05-27 12:10:47-ERROR- [Errno 2] No such file or directory: 'not_existing_file.txt'
Traceback (most recent call last):
  File "exception_checks.py", line 27, in <module>
    with open("not_existing_file.txt", 'r') as text:
FileNotFoundError: [Errno 2] No such file or directory: 'not_existing_file.txt'

Print/log error name/message only

You might not be interested in the whole traceback, but only in the most important information, such as Exception name and Exception message, use:

try:
    with open("not_existing_file.txt", 'r') as text:
        pass
except Exception as exception:
    print("Exception: {}".format(type(exception).__name__))
    print("Exception message: {}".format(exception))

Output:

Exception: FileNotFoundError
Exception message: [Errno 2] No such file or directory: 'not_existing_file.txt'

In Python 2.6 or greater it's a bit cleaner:

except Exception as e: print(e)

In older versions it's still quite readable:

except Exception, e: print e

Python 3: logging

Instead of using the basic print() function, the more flexible logging module can be used to log the exception. The logging module offers a lot extra functionality, e.g. logging messages into a given log file, logging messages with timestamps and additional information about where the logging happened. (For more information check out the official documentation.)

Logging an exception can be done with the module-level function logging.exception() like so:

import logging

try:
    1/0
except BaseException:
    logging.exception("An exception was thrown!")

Output:

ERROR:root:An exception was thrown!
Traceback (most recent call last):
  File ".../Desktop/test.py", line 4, in <module>
    1/0
ZeroDivisionError: division by zero 

Notes:

• the function logging.exception() should only be called from an exception handler

• the logging module should not be used inside a logging handler to avoid a RecursionError (thanks @PrakharPandey)

Alternative log-levels

It's also possible to log the exception with another log-level by using the keyword argument exc_info=True like so:

logging.debug("An exception was thrown!", exc_info=True)
logging.info("An exception was thrown!", exc_info=True)
logging.warning("An exception was thrown!", exc_info=True)

In case you want to pass error strings, here is an example from Errors and Exceptions (Python 2.6)

>>> try:
...    raise Exception('spam', 'eggs')
... except Exception as inst:
...    print type(inst)     # the exception instance
...    print inst.args      # arguments stored in .args
...    print inst           # __str__ allows args to printed directly
...    x, y = inst          # __getitem__ allows args to be unpacked directly
...    print 'x =', x
...    print 'y =', y
...
<type 'exceptions.Exception'>
('spam', 'eggs')
('spam', 'eggs')
x = spam
y = eggs

(I was going to leave this as a comment on @jldupont's answer, but I don't have enough reputation.)

I've seen answers like @jldupont's answer in other places as well. FWIW, I think it's important to note that this:

except Exception as e:
    print(e)

will print the error output to sys.stdout by default. A more appropriate approach to error handling in general would be:

except Exception as e:
    print(e, file=sys.stderr)

(Note that you have to import sys for this to work.) This way, the error is printed to STDERR instead of STDOUT, which allows for the proper output parsing/redirection/etc. I understand that the question was strictly about 'printing an error', but it seems important to point out the best practice here rather than leave out this detail that could lead to non-standard code for anyone who doesn't eventually learn better.

I haven't used the traceback module as in Cat Plus Plus's answer, and maybe that's the best way, but I thought I'd throw this out there.

One liner error raising can be done with assert statements if that's what you want to do. This will help you write statically fixable code and check errors early.

assert type(A) is type(""), "requires a string"