如果代码错误，如何打印异常？

Question

If have a function were I request some data from a website but when I get an error I want to print the error and restart my code. But I don't know the exact code, can please someone help me? This is a code example:

import time
input1 = input("Blabla: ")

def repeat():
    try:
        if input1 == "123":
            raise "Error: 123"
    except Exception as e:
        print(e)
        time.sleep(5) # Wait 5 seconds
        repeat() # Rerun code

repeat()

When I run this code I get the error 'exceptions must derive from BaseException'. Can someone help me?

Answer 1

You can't just raise random strings as an exception. If you want to raise a general exception without defining a relevant type, just raise Exception , replacing:

raise "Error: 123"

with:

raise Exception("Error: 123")  # The "Error: " should probably be removed

Alternatively, if you can use a more specific error, do so. If 123 is invalid because the value is wrong, use ValueError instead of Exception . If there is some more specific reason, make a subclass to make it easier for others to catch, eg (at top level in the module):

class SpecialValueError(ValueError):
    pass

so you can do:

raise SpecialValueError("Error: 123")

and people can catch it specifically, or via plain except ValueError: , except Exception: , etc.

Answer 2

Right now you are printing the error object's str while you need to print it's representation

Try this:

def repeat():
    try:
        if input1 == "123":
            raise Exception("Error: 123") # You need to use an Exception class
    except Exception as e:
        print(repr(e)) # Notice I added repr()
        time.sleep(5)
        repeat()

Exception's Representation vs Exception's String

String:

try: 
    raise Exception("Exception I am!") 
except Exception as e: 
    print(e) 
# Output: Exception I am!

Representation:

try: 
    raise Exception("Exception I am!") 
except Exception as e: 
    print(repr(e)) 

# Output: Exception('Exception I am!')