如何在不停止/退出程序的情况下捕获并打印完整的异常回溯？

Question

I want to catch and log exceptions without exiting, eg,

try:
    do_stuff()
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.

Answer 1

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

import traceback
import sys

try:
    do_stuff()
except Exception:
    print(traceback.format_exc())
    # or
    print(sys.exc_info()[2])

Answer 2

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

try:
    raise TypeError("Oups!")
except Exception, err:
    try:
        raise TypeError("Again !?!")
    except:
        pass

    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

try:
    raise TypeError("Oups!")
except Exception, err:
    try:
        exc_info = sys.exc_info()

        # do you usefull stuff here
        # (potentially raising an exception)
        try:
            raise TypeError("Again !?!")
        except:
            pass
        # 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

try:
    raise TypeError("Oups!")
except Exception as err:
    try:
        raise TypeError("Again !?!")
    except:
        pass

    traceback.print_tb(err.__traceback__)

... will display:

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

Answer 3

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.

Answer 4

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:

try:
    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:

try:
    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:

try:
    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:

try:
    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.

Answer 5

First, do not use print s 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__)

try:
    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, eg 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.

Answer 6

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():
    g()

def g():
    raise Exception('asdf')

try:
    g()
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>
    g()
  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

Answer 7

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 . eg

try:
    do_something_that_might_error()
except Exception:
    logging.info('General exception noted.', exc_info=True)

Answer 8

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():
    try:
        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()


try:
    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.

Answer 9

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)

try:
    1/0
except Exception as e:
    print('------Start--------')
    print(get_traceback(e))
    print('------End--------')

try:
    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>
    1/0
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--------

Answer 10

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.

Answer 11

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

try:
    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 KeyboardInterrupt s, 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.

Answer 12

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

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

Answer 13

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
1/0
ZeroDivisionError: integer division or modulo by zero
o
[Finished in 0.1s]

So I use:

import traceback, sys

def do_stuff():
    1/0

try:
    do_stuff()
except Exception:
    print(traceback.format_exc())
    print('hello')

Answer 14

import io
import traceback

try:
    call_code_that_fails()
except:

    errors = io.StringIO()
    traceback.print_exc(file=errors)
    contents = str(errors.getvalue())
    print(contents)
    errors.close()

Answer 15

You want the traceback module. It will let you print stack dumps like Python normally does. In particular, the print_last function will print the last exception and a stack trace.

Answer 16

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

Answer 17

python 3 solution

stacktrace_helper.py :

from linecache import getline
import sys
import traceback


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 = ""
    try:
        line_code = getline(ex_file, ex_line).strip()
    except:
        pass

    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

Answer 18

One liner:

import traceback
traceback.print_exc()

Use the traceback module which automatically grabs the current exception when inside an except:

https://stackoverflow.com/a/9555145/2217801

Answer 19

You could do:

try:
    do_stuff()
except Exception, err:
    print(Exception, err)
    raise err