how to print error in try except python
try:
# some code
except Exception as e:
print("ERROR : "+str(e))
how to print error in try except python
try:
# some code
except Exception as e:
print("ERROR : "+str(e))
python try catch
try:
# Dangerous stuff
except ValueError:
# If you use try, at least 1 except block is mandatory!
# Handle it somehow / ignore
except (BadThingError, HorrbileThingError) as e:
# Hande it differently
except:
# This will catch every exception.
else:
# Else block is not mandatory.
# Dangerous stuff ended with no exception
finally:
# Finally block is not mandatory.
# This will ALWAYS happen after the above blocks.
catch error python
import traceback
dict = {'a':3,'b':5,'c':8}
try:
print(dict[q])
except:
traceback.print_exc()
# This will trace you back to the line where everything went wrong.
# So in this case you will get back line 5
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
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"
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