PYTHON

how to print error in try except python

try:
  # some code
except Exception as e:
	print("ERROR : "+str(e))

print type of exception python

try:
    someFunction()
except Exception as ex:
    template = "An exception of type {0} occurred. Arguments:
{1!r}"
    message = template.format(type(ex).__name__, ex.args)
    print (message)

exception pyton print

except Exception as e: print(e)

python except print error type

>>> 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 be printed directly,
...                          # but may be overridden in exception subclasses
...     x, y = inst.args     # unpack args
...     print('x =', x)
...     print('y =', y)
...
<class 'Exception'>
('spam', 'eggs')
('spam', 'eggs')
x = spam
y = eggs

print type error python

try:
    #Code to execute
except Exception as err:
    print(f"{type(err).__name__} was raised: {err}")

python try except: print error

except Exception as e:

python try except print error

   1 (x,y) = (5,0)
   2 try:
   3   z = x/y
   4 except ZeroDivisionError as e:
   5   z = e # representation: "<exceptions.ZeroDivisionError instance at 0x817426c>"
   6 print z # output: "integer division or modulo by zero"