most - Python Cuando encuentro una excepción, ¿cómo obtengo el tipo, archivo y número de línea?

traceback python 3

Question

Capturando una excepción que se imprimiría así:

Traceback (most recent call last):
File "c:/tmp.py", line 1, in <module>
4 / 0
ZeroDivisionError: integer division or modulo by zero

Quiero formatearlo en:

ZeroDivisonError, tmp.py, 1

Answer 1

Aquí hay un ejemplo de mostrar el número de línea de donde ocurre la excepción.

import sys
try:
print(5/0)
except Exception as e:
print(''Error on line {}''.format(sys.exc_info()[-1].tb_lineno), type(e).__name__, e)
print(''And the rest of program continues'')

Answer 2

La forma más simple que funcionó para mí.

import traceback
try:
print(4/0)
except ZeroDivisionError:
print(traceback.format_exc())

Salida

Traceback (most recent call last):
File "/path/to/file.py", line 51, in <module>
print(4/0)
ZeroDivisionError: division by zero
Process finished with exit code 0

Answer 3

Source (Py v2.7.3) para traceback.format_exception() y llamadas / funciones relacionadas ayuda en gran medida. De manera vergonzosa, siempre me olvido de leer la fuente . Solo lo hice después de buscar detalles similares en vano. Una pregunta simple, "¿Cómo recrear el mismo resultado que Python para una excepción, con todos los mismos detalles?" Esto le daría a cualquiera 90 +% de lo que sea que esté buscando. Frustrado, se me ocurrió este ejemplo. Espero que ayude a otros. (¡Seguro que me ayudó! ;-)

import sys, traceback
traceback_template = ''''''Traceback (most recent call last):
File "%(filename)s", line %(lineno)s, in %(name)s
%(type)s: %(message)s/n'''''' # Skipping the "actual line" item
# Also note: we don''t walk all the way through the frame stack in this example
# see hg.python.org/cpython/file/8dffb76faacc/Lib/traceback.py#l280
# (Imagine if the 1/0, below, were replaced by a call to test() which did 1/0.)
try:
1/0
except:
# http://docs.python.org/2/library/sys.html#sys.exc_info
exc_type, exc_value, exc_traceback = sys.exc_info() # most recent (if any) by default
''''''
Reason this _can_ be bad: If an (unhandled) exception happens AFTER this,
or if we do not delete the labels on (not much) older versions of Py, the
reference we created can linger.
traceback.format_exc/print_exc do this very thing, BUT note this creates a
temp scope within the function.
''''''
traceback_details = {
''filename'': exc_traceback.tb_frame.f_code.co_filename,
''lineno'' : exc_traceback.tb_lineno,
''name'' : exc_traceback.tb_frame.f_code.co_name,
''type'' : exc_type.__name__,
''message'' : exc_value.message, # or see traceback._some_str()
}
del(exc_type, exc_value, exc_traceback) # So we don''t leave our local labels/objects dangling
# This still isn''t "completely safe", though!
# "Best (recommended) practice: replace all exc_type, exc_value, exc_traceback
# with sys.exc_info()[0], sys.exc_info()[1], sys.exc_info()[2]
print
print traceback.format_exc()
print
print traceback_template % traceback_details
print

En respuesta específica a esta consulta:

sys.exc_info()[0].__name__, os.path.basename(sys.exc_info()[2].tb_frame.f_code.co_filename), sys.exc_info()[2].tb_lineno

Answer 4

import sys, os
try:
raise NotImplementedError("No error")
except Exception as e:
exc_type, exc_obj, exc_tb = sys.exc_info()
fname = os.path.split(exc_tb.tb_frame.f_code.co_filename)[1]
print(exc_type, fname, exc_tb.tb_lineno)