Reputation: 71
I feel stupid asking this since the answer is probably simple but I couldn't really find the answer by googling it... I have a lot of code which I don't want help with, I want to debug on my own, and I use try/except for it, but except doesn't print the error messsage and I couldn't find a way to print that. Basically, how do I check what my error is?
Upvotes: 1
Views: 7245
Reputation: 29
You CAN use try except to catch ANY error without worrying for its types. Use this code
try:
blah blah
except Exception as A: #(Where A is a temporary variable)
print(a)
Tadaaa, you just captured the error without blowing your whole code.
Upvotes: 0
Reputation: 660
You can use traceback
, more specifically traceback.print_exc()
. This prints the full error traceback, rather than just the one line produced from using as
in your except
block.
I've put an example below using a ValueError
.
import traceback
try:
int("s")
except ValueError:
traceback.print_exc()
This produces the same print result as if there was no try-except, the difference being that this solution allows the code to continue running.
Upvotes: 1
Reputation: 71
Firstly to check what is your error, you can watch into your terminal there will be error's name:
print(5+"5")
>>>
Traceback (most recent call last):
File "c:\Users\USER\Desktop\how_to_use_try_except.py",
line 1, in <module>
print(5+"5")
TypeError: unsupported operand type(s) for +: 'int' and 'str'
and to catch it, you need to copy error's name, and except it
try:
print(5+"5")
except TypeError:
print("You can't add string value to int!")
also you can write
try:
print(5+"5")
except TypeError as e:
print(e)
>>>
unsupported operand type(s) for +: 'int' and 'str'
Upvotes: 2