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Learn more about Teams except Exception as err: print(Exception, err) # I want to print the entire traceback here, # not just the exception name and details

I want to print the exact same output that is printed when the exception is raised without the try/except intercepting the exception, and I do not want it to exit my program.

Not a full answer, but someone might want to know that you can access lots of info looking into err.__traceback__ (at least in Python 3.x) Vito Gentile Feb 25, 2021 at 12:10 People viewed it 825k times while trying to find out how to print their stacktraces. That's another Zen of Python. Pavel Vlasov Apr 19, 2021 at 23:05 It seems I'm the only one in the world who wants to print the stack when there's no error (= only to see how I got here at this precise line (it's not my code, and it's so ugly I cant figure out how it did come here!)). Olivier Pons Jun 2, 2021 at 8:23 All the answers in this question are the ultimate beginners guide to debugging python code Iuri Guilherme May 13, 2022 at 15:41

traceback.format_exc() or sys.exc_info() will yield more info if that's what you want.

import traceback
import sys
    do_stuff()
except Exception:
    print(traceback.format_exc())
    print(sys.exc_info()[2])
                dont use exc... the traceback contains all the info stackoverflow.com/questions/4564559/…
– droid192
                Sep 8, 2019 at 18:11
                print(traceback.format_exc()) is better than traceback.print_tb(exc.__traceback__). print(sys.exc_info()) returns the whole tuple and looks like (<class 'UnicodeDecodeError'>, UnicodeDecodeError('utf-8', b'\x81', 0, 1, 'invalid start byte'), <traceback object at 0x7f179d64ae00>) So indeed traceback.format_exc() is really superior because that prints Traceback (most recent call last):   File "<ipython-input-15-9e3d6e01ef04>", line 2, in <module>     b"\x81".decode() UnicodeDecodeError: 'utf-8' codec can't decode byte 0x81 in position 0: invalid start byte
– Mark
                May 7, 2021 at 20:47

Some other answer have already pointed out the traceback module.

Please notice that with print_exc, in some corner cases, you will not obtain what you would expect. In Python 2.x:

import traceback
    raise TypeError("Oups!")
except Exception, err:
        raise TypeError("Again !?!")
    except:
    traceback.print_exc()

...will display the traceback of the last exception:

Traceback (most recent call last):
  File "e.py", line 7, in <module>
    raise TypeError("Again !?!")
TypeError: Again !?!

If you really need to access the original traceback one solution is to cache the exception infos as returned from exc_info in a local variable and display it using print_exception:

import traceback
import sys
    raise TypeError("Oups!")
except Exception, err:
        exc_info = sys.exc_info()
        # do you usefull stuff here
        # (potentially raising an exception)
            raise TypeError("Again !?!")
        except:
        # end of useful stuff
    finally:
        # Display the *original* exception
        traceback.print_exception(*exc_info)
        del exc_info

Producing:

Traceback (most recent call last):
  File "t.py", line 6, in <module>
    raise TypeError("Oups!")
TypeError: Oups!

Few pitfalls with this though:

  • From the doc of sys_info:

    Assigning the traceback return value to a local variable in a function that is handling an exception will cause a circular reference. This will prevent anything referenced by a local variable in the same function or by the traceback from being garbage collected. [...] If you do need the traceback, make sure to delete it after use (best done with a try ... finally statement)

  • but, from the same doc:

    Beginning with Python 2.2, such cycles are automatically reclaimed when garbage collection is enabled and they become unreachable, but it remains more efficient to avoid creating cycles.

    On the other hand, by allowing you to access the traceback associated with an exception, Python 3 produce a less surprising result:

    import traceback
        raise TypeError("Oups!")
    except Exception as err:
            raise TypeError("Again !?!")
        except:
        traceback.print_tb(err.__traceback__)
    

    ... will display:

      File "e3.py", line 4, in <module>
        raise TypeError("Oups!")
    

    If you're debugging and just want to see the current stack trace, you can simply call:

    traceback.print_stack()

    There's no need to manually raise an exception just to catch it again.

    Output goes to STDERR by default BTW. Wasn't appearing in my logs because it was being redirected somewhere else. – mpen Dec 11, 2018 at 0:18 @pppery I can't see it with python 3.8. And the thing with try and catch is that it doesn't display the full traceback, Only from raise to except. – x-yuri Jun 25, 2020 at 19:12

    How to print the full traceback without halting the program?

    When you don't want to halt your program on an error, you need to handle that error with a try/except:

    do_something_that_might_error() except Exception as error: handle_the_error(error)

    To extract the full traceback, we'll use the traceback module from the standard library:

    import traceback
    

    And to create a decently complicated stacktrace to demonstrate that we get the full stacktrace:

    def raise_error():
        raise RuntimeError('something bad happened!')
    def do_something_that_might_error():
        raise_error()
    

    Printing

    To print the full traceback, use the traceback.print_exc method:

    do_something_that_might_error() except Exception as error: traceback.print_exc()

    Which prints:

    Traceback (most recent call last):
      File "<stdin>", line 2, in <module>
      File "<stdin>", line 2, in do_something_that_might_error
      File "<stdin>", line 2, in raise_error
    RuntimeError: something bad happened!
    

    Better than printing, logging:

    However, a best practice is to have a logger set up for your module. It will know the name of the module and be able to change levels (among other attributes, such as handlers)

    import logging
    logging.basicConfig(level=logging.DEBUG)
    logger = logging.getLogger(__name__)
    

    In which case, you'll want the logger.exception function instead:

    do_something_that_might_error() except Exception as error: logger.exception(error)

    Which logs:

    ERROR:__main__:something bad happened!
    Traceback (most recent call last):
      File "<stdin>", line 2, in <module>
      File "<stdin>", line 2, in do_something_that_might_error
      File "<stdin>", line 2, in raise_error
    RuntimeError: something bad happened!
    

    Or perhaps you just want the string, in which case, you'll want the traceback.format_exc function instead:

    do_something_that_might_error() except Exception as error: logger.debug(traceback.format_exc())

    Which logs:

    DEBUG:__main__:Traceback (most recent call last):
      File "<stdin>", line 2, in <module>
      File "<stdin>", line 2, in do_something_that_might_error
      File "<stdin>", line 2, in raise_error
    RuntimeError: something bad happened!
    

    Conclusion

    And for all three options, we see we get the same output as when we have an error:

    >>> do_something_that_might_error()
    Traceback (most recent call last):
      File "<stdin>", line 1, in <module>
      File "<stdin>", line 2, in do_something_that_might_error
      File "<stdin>", line 2, in raise_error
    RuntimeError: something bad happened!
    

    Which to use

    Performance concerns aren't important here as IO usually dominates. I'd prefer, since it does precisely what's being requested in a forward compatible way:

    logger.exception(error)
    

    Logging levels and outputs can be adjusted, making it easy to turn off without touching the code. And usually doing what's directly needed is the most efficient way to do it.

    as said above and for me too, traceback.print_exc() returns only the last call : how do you succeed to return several level of the stack (and possibly all levele s?) – herve-guerin Nov 22, 2017 at 16:08 @geekobi I'm not sure what you're asking here. I demonstrate that we get the traceback up to the entry point of the program/interpreter. What are you not clear on? – Russia Must Remove Putin Nov 22, 2017 at 16:59 What @geekobi is saying is if you catch and re-raise, traceback.print_exc() will just return the re-raise stack, not the original stack. – fizloki Aug 22, 2018 at 1:38 @fizloki how are you "reraising"? Are you doing a bare raise or exception chaining, or are you hiding the original traceback? see stackoverflow.com/questions/2052390/… – Russia Must Remove Putin Aug 22, 2018 at 2:35 Thanks for that followup question - I think the exc_info=True argument is actually better for logging, a keyword argument is more maintainable than custom code that puts the traceback into a string. I'll get around to updating my answer. – Russia Must Remove Putin Jul 16, 2020 at 13:43

    First, do not use prints for logging, there is a stable, proven and well-thought out stdlib module to do that: logging. You definitely should use it instead.

    Second, do not be tempted to do a mess with unrelated tools when there is a native and simple approach. Here it is:

    log = logging.getLogger(__name__)
        call_code_that_fails()
    except MyError:
        log.exception('Any extra info you want to see in your logs')
    

    That's it. You are done now.

    Explanation for anyone who is interested in how things work under the hood

    What log.exception is actually doing is just a call to log.error (that is, log event with level ERROR) and print traceback then.

    Why is it better?

    Well, here are some considerations:

  • it is just right;
  • it is straightforward;
  • it is simple.
  • Why should nobody use traceback or call logger with exc_info=True or get their hands dirty with sys.exc_info?

    Well, just because! They all exist for different purposes. For example, traceback.print_exc's output is a little bit different from tracebacks produced by the interpreter itself. If you use it, you will confuse anyone who reads your logs, they will be banging their heads against them.

    Passing exc_info=True to log calls is just inappropriate. But, it is useful when catching recoverable errors and you want to log them (using, e.g INFO level) with tracebacks as well, because log.exception produces logs of only one level - ERROR.

    And you definitely should avoid messing with sys.exc_info as much as you can. It's just not a public interface, it's an internal one - you can use it if you definitely know what you are doing. It is not intended for just printing exceptions.

    It also doesn't work as-is. That's not it. I'm not done now: this answer just wastes time. – A. Rager Apr 27, 2019 at 16:07 I would also add that you can just do logging.exception(). No need to create instance of log unless you have special requirements. – Shital Shah May 18, 2019 at 6:35 I find this answer kind of ridiculous. It's full of "do/don't do this just because" without explaining why. Your points in "why is it better?" is practically just all saying the same thing: "because it is." Which I do not find helpful. You did explain a little bit at least. – Break Dec 18, 2020 at 1:28 Good information (I didn't know about logging.exception) but a bit condescending. I think this is due to language barrier rather than malicious intent. – rjh Sep 5, 2021 at 14:49

    traceback.format_exception(exception_object)

    If you only have the exception object, you can get the traceback as a string from any point of the code in Python 3 with:

    import traceback
    ''.join(traceback.format_exception(None, exc_obj, exc_obj.__traceback__))
    

    Full example:

    #!/usr/bin/env python3
    import traceback
    def f():
    def g():
        raise Exception('asdf')
    except Exception as e:
        exc_obj = e
    tb_str = ''.join(traceback.format_exception(None, exc_obj, exc_obj.__traceback__))
    print(tb_str)
    

    Output:

    Traceback (most recent call last):
      File "./main.py", line 12, in <module>
      File "./main.py", line 9, in g
        raise Exception('asdf')
    Exception: asdf
    

    Documentation: https://docs.python.org/3.9/library/traceback.html#traceback.format_exception

    See also: Extract traceback info from an exception object

    Tested in Python 3.9

    Why does the only way to do this in the language involving passing the same information twice (both exc_obj and exc_obj.__traceback__), and an irrelevant third argument None? – Jack M May 19, 2022 at 15:02

    In addition to Aaron Hall's answer, if you are logging, but don't want to use logging.exception() (since it logs at the ERROR level), you can use a lower level and pass exc_info=True. e.g.

    do_something_that_might_error() except Exception: logging.info('General exception noted.', exc_info=True) This is good too when dealing with a detected failure of logging... i.e. when for some reason you have failed to create an actual Logger object. – mike rodent Dec 21, 2021 at 17:09

    I don't see this mentioned in any of the other answers. If you're passing around an Exception object for whatever reason...

    In Python 3.5+ you can get a trace from an Exception object using traceback.TracebackException.from_exception(). For example:

    import traceback
    def stack_lvl_3():
        raise Exception('a1', 'b2', 'c3')
    def stack_lvl_2():
            stack_lvl_3()
        except Exception as e:
            # raise
            return e
    def stack_lvl_1():
        e = stack_lvl_2()
        return e
    e = stack_lvl_1()
    tb1 = traceback.TracebackException.from_exception(e)
    print(''.join(tb1.format()))
    

    However, the above code results in:

    Traceback (most recent call last):
      File "exc.py", line 10, in stack_lvl_2
        stack_lvl_3()
      File "exc.py", line 5, in stack_lvl_3
        raise Exception('a1', 'b2', 'c3')
    Exception: ('a1', 'b2', 'c3')
    

    This is just two levels of the stack, as opposed to what would have been printed on screen had the exception been raised in stack_lvl_2() and not intercepted (uncomment the # raise line).

    As I understand it, that's because an exception records only the current level of the stack when it is raised, stack_lvl_3() in this case. As it's passed back up through the stack, more levels are being added to its __traceback__. But we intercepted it in stack_lvl_2(), meaning all it got to record was levels 3 and 2. To get the full trace as printed on stdout we'd have to catch it at the highest (lowest?) level:

    import traceback
    def stack_lvl_3():
        raise Exception('a1', 'b2', 'c3')
    def stack_lvl_2():
        stack_lvl_3()
    def stack_lvl_1():
        stack_lvl_2()
        stack_lvl_1()
    except Exception as exc:
        tb = traceback.TracebackException.from_exception(exc)
    print('Handled at stack lvl 0')
    print(''.join(tb.stack.format()))
    

    Which results in:

    Handled at stack lvl 0
      File "exc.py", line 17, in <module>
        stack_lvl_1()
      File "exc.py", line 13, in stack_lvl_1
        stack_lvl_2()
      File "exc.py", line 9, in stack_lvl_2
        stack_lvl_3()
      File "exc.py", line 5, in stack_lvl_3
        raise Exception('a1', 'b2', 'c3')
    

    Notice that the stack print is different, the first and last lines are missing. Because it's a different format().

    Intercepting the exception as far away from the point where it was raised as possible makes for simpler code while also giving more information.

    This is a lot better than the previous method(s), but is still ridiculously convoluted just to print out a stacktrace. Java takes less code FGS. – elhefe Feb 20, 2020 at 22:37

    In python3 (works in 3.9) we can define a function and can use that where ever we want to print the details.

    import traceback
    def get_traceback(e):
        lines = traceback.format_exception(type(e), e, e.__traceback__)
        return ''.join(lines)
    except Exception as e:
        print('------Start--------')
        print(get_traceback(e))
        print('------End--------')
        spam(1,2)
    except Exception as e:
        print('------Start--------')
        print(get_traceback(e))
        print('------End--------')
    

    The output would be like:

    bash-3.2$ python3 /Users/soumyabratakole/PycharmProjects/pythonProject/main.py
    ------Start--------
    Traceback (most recent call last):
      File "/Users/soumyabratakole/PycharmProjects/pythonProject/main.py", line 26, in <module>
    ZeroDivisionError: division by zero
    ------End--------
    ------Start--------
    Traceback (most recent call last):
      File "/Users/soumyabratakole/PycharmProjects/pythonProject/main.py", line 33, in <module>
        spam(1,2)
    NameError: name 'spam' is not defined
    ------End--------
    

    If you have an Error object already, and you want to print the whole thing, you need to make this slightly awkward call:

    import traceback
    traceback.print_exception(type(err), err, err.__traceback__)
    

    That's right, print_exception takes three positional arguments: The type of the exception, the actual exception object, and the exception's own internal traceback property.

    In python 3.5 or later, the type(err) is optional... but it's a positional argument, so you still have to explicitly pass None in its place.

    traceback.print_exception(None, err, err.__traceback__)
    

    I have no idea why all of this isn't just traceback.print_exception(err). Why you would ever want to print out an error, along with a traceback other than the one that belongs to that error, is beyond me.

    To get the precise stack trace, as a string, that would have been raised if no try/except were there to step over it, simply place this in the except block that catches the offending exception.

    desired_trace = traceback.format_exc(sys.exc_info())
    

    Here's how to use it (assuming flaky_func is defined, and log calls your favorite logging system):

    import traceback
    import sys
        flaky_func()
    except KeyboardInterrupt:
        raise
    except Exception:
        desired_trace = traceback.format_exc(sys.exc_info())
        log(desired_trace)
    

    It's a good idea to catch and re-raise KeyboardInterrupts, so that you can still kill the program using Ctrl-C. Logging is outside the scope of the question, but a good option is logging. Documentation for the sys and traceback modules.

    This doesn't work in Python 3 and needs to be changed to desired_trace = traceback.format_exc(). Passing sys.exc_info() as the argument was never the correct thing to do, but gets silently ignored in Python 2—but not in Python 3 (3.6.4 anyway). – martineau Feb 6, 2018 at 18:54 KeyboardInterrupt is not derived (directly or indirectly) from Exception. (Both are derived from BaseException.) This means except Exception: will never catch a KeyboardInterrupt, and thus the except KeyboardInterrupt: raise is completely unnecessary. – AJNeufeld Aug 17, 2018 at 20:56

    You will need to put the try/except inside the most innerloop where the error may occur, i.e.

    for i in something:
        for j in somethingelse:
            for k in whatever:
                    something_complex(i, j, k)
                except Exception, e:
                    print e
                something_less_complex(i, j)
            except Exception, e:
                print e
    

    ... and so on

    In other words, you will need to wrap statements that may fail in try/except as specific as possible, in the most inner-loop as possible.

    A remark about this answer's comments: print(traceback.format_exc()) does a better job for me than traceback.print_exc(). With the latter, the hello is sometimes strangely "mixed" with the traceback text, like if both want to write to stdout or stderr at the same time, producing weird output (at least when building from inside a text editor and viewing the output in the "Build results" panel).

    Traceback (most recent call last):
    File "C:\Users\User\Desktop\test.py", line 7, in
    hell do_stuff()
    File "C:\Users\User\Desktop\test.py", line 4, in do_stuff
    ZeroDivisionError: integer division or modulo by zero
    [Finished in 0.1s]

    So I use:

    import traceback, sys
    def do_stuff():
        do_stuff()
    except Exception:
        print(traceback.format_exc())
        print('hello')
        errors = io.StringIO()
        traceback.print_exc(file=errors)  # Instead of printing directly to stdout, the result can be further processed
        contents = str(errors.getvalue())
        print(contents)
        errors.close()
                    2 comments: The use of traceback.print_exc() was already discussed in previous answers. More importantly, why all that mucking about with io.StringIO when those last five lines are exactly equivalent to traceback.print_exc()?
    – joanis
                    Nov 19, 2021 at 19:01
                    @joanis I believe those lines are useful if you want to get access to the error body and not just print it. Personally I found it useful because I am recording the stack trace to a database.
    – pcko1
                    Apr 5, 2022 at 19:47
    def get_stack_trace():
        exc_type, exc_value, exc_tb = sys.exc_info()
        trace = traceback.format_stack()
        trace = list(filter(lambda x: ("\\lib\\" not in x and "/lib/" not in x and "stacktrace_helper.py" not in x), trace))
        ex_type = exc_type.__name__
        ex_line = exc_tb.tb_lineno
        ex_file = exc_tb.tb_frame.f_code.co_filename
        ex_message = str(exc_value)
        line_code = ""
            line_code = getline(ex_file, ex_line).strip()
        except:
        trace.insert(
            0, f'File "{ex_file}", line {ex_line}, line_code: {line_code} , ex: {ex_type} {ex_message}',
        return trace
    def get_stack_trace_str(msg: str = ""):
        trace = list(get_stack_trace())
        trace_str = "\n".join(list(map(str, trace)))
        trace_str = msg + "\n" + trace_str
        return trace_str
    

    This is my solution to write the error in a log file and also on console:

    import logging, sys
    import traceback
    logging.basicConfig(filename='error.log', level=logging.DEBUG)
    def handle_exception(exc_type, exc_value, exc_traceback):
        if issubclass(exc_type, KeyboardInterrupt):
            sys.__excepthook__(exc_type, exc_value, exc_traceback)
            return
        exc_info=(exc_type, exc_value, exc_traceback)
        logging.critical("\nDate:" + str(datetime.datetime.now()), exc_info=(exc_type, exc_value, exc_traceback))
        print("An error occured, check error.log to see the error details")
        traceback.print_exception(*exc_info)
    sys.excepthook = handle_exception
    
  •