How to print the full traceback without halting th

Question

I'm writing a program that parses 10 websites, locates data files, saves the files, and then parses them to make data that can be readily used in the NumPy library. There are tons of errors this file encounters through bad links, poorly formed XML, missing entries, and other things I've yet to categorize. I initially made this program to handle errors like this:

try:
    do_stuff()
except:
    pass

But now I want to log errors:

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

Note this is printing to a log file for later review. This usually prints very useless data. What I want is to print the exact same lines printed when the error triggers without the try-except intercepting the exception, but I don't want it to halt my program since it is nested in a series of for loops that I would like to see to completion.

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()[0])

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

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:
            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 4

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 5

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 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.

Answer 6

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')