I get an error while running my script and I know it is there, But I don't want Python to print the error. The error (it is a KeyError) is in a if statement. Is there a way to stop the if statement from printing error messages?
Try this.
while True:
try:
x = int(raw_input("Please enter a number: "))
break
except ValueError:
print "Oops! That was no valid number. Try again..."
Related
I have this piece of code here:
money_to_dep = int(input(f"\nHow much money do you want to deposit?(enter amount ONLY):\n>> "))
while True:
try:
print(money_to_dep)
break
except ValueError:
print(f"Wrong INPUT. Try again.")
break
When I type a string instead of an integer, an error still occurs even though I have done except ValueError? Why does this happen? This question really annoys me as I have never been able to have a clear understanding as to how try-except works with handling errors.
You have to put your code like this:
while True:
try:
money_to_dep = int(input(f"\nHow much money do you want to deposit?(enter amount ONLY):\n>> "))
print(money_to_dep)
break
except ValueError:
print(f"Wrong INPUT. Try again.")
The ValueError raise when you use int. So it has to be inside the try statement.
This means that your exception is occuring before your try catch. you need to put your input code inside try, so it can catch exception.
i.e
money_to_dep = int(input(f"\nHow much money do you want to deposit?(enter amount ONLY):\n>> "))
Exveption occurs, but not in try block so not catched. Change it to
while True:
try:
money_to_dep = int(input(f"\nHow much money do you want to deposit?(enter amount ONLY):\n>> "))
print(money_to_dep)
break
except Exception as E:
print(f"Wrong INPUT. Try again.")
break
I'm trying to write a basic tic-tac-toe in Python. One of my functions is meant to get the user's input as a number from 1 to 9. If the user enters a non-integer, or a number not between 1 and 9, it will return an error. It's nestled in a try-except to avoid worrying about type conversion.
def get_move():
print("Your move ...")
while True:
try:
move = input("Type a number from 1-9: ")
if move.isnumeric():
if (0 < int(move) < 10):
break
else:
print("The number is outside the range. Try again.")
else:
if (move.tolower == "quit") or (move.tolower == "exit"):
exit()
except:
print("Not a valid integer, try again.")
return move
This sort-of works when I run normally. But when I try to debug in VS Code, when the debugger reaches the line move = input("Type a number from 1-9: ") and I click "Step Into", it simply goes straight to the "except" clause. And the code proceeds in an infinite loop - it will never stop and wait for the user input, meaning that I have to manually stop the debugger.
Any idea why it might be doing this?
Edit:
Thanks for correcting my typo, but that hasn't solved the problem. I've changed the line from:
if (move.tolower == "quit") or (move.tolower == "exit"):
exit()
to:
if move.lower() == "quit" or move.lower() == "exit":
exit()
And also changed the except clause to except (ValueError, TypeError). Now I receive the following error:
Exception has occurred: EOFError
EOF when reading a line
File "[...]tictacpy.py", line 18, in get_move
move = input("Type a number from 1-9: ")
File "[...]tictacpy.py", line 46, in <module>
move = get_move()
Most likely this is happening because in the debugger you do not have a console for standard input, so calling input() will error (I don't know VS code specifically so I'm guessing here, but this is a reasonable cause).
In any case, I'd strongly suggest not using all-catching except clauses as this silences and wrongly handles errors that may happen that are not a part of your expected flow.
I'd start by changing your except to say except (ValueError, TypeError) so that it only catches errors resulting from bad input / type conversion issues. You'll then be able to see what the real error is.
Also, note that there is no such thing as move.tolower - you probably meant move.lower(). Maybe that's your bug?
I'm working on a school project and it specifically asks for a while not end-of-file loop which I don't know how to do in Python. I'm taking in user input (not from an external file) and computing some math in this infinite loop until they CTRL+C to break the loop. Any thoughts would really help me out. I've added part of the instruction if that helps clarify. Thanks.
Your loop is supposed to stop on two conditions:
An illegal value was entered, ie some value that couldn't be converted to a float. In this case, a ValueError will be raised.
You entered ctrl-Z, which means an EOF. (Sorry, having no Windows here, I just tested it on Linux, where I think that ctrl-D is the equivalent of the Windows ctrl-Z). An EOFError will be raised.
So, you should just create an infinite loop, and break out of it when one of these exceptions occur.
I separated the handling of the exceptions so that you can see what happens and print some meaningful error message:
while True:
try:
amount = float(input())
print(amount) # do your stuff here
except ValueError as err:
print('Terminating because', err)
break
except EOFError:
print('EOF!')
break
If you just want to quit without doing anything more, you can handle both exceptions the same way:
while True:
try:
amount = float(input())
print(amount) # do your stuff here
except (ValueError, EOFError):
break
Use the following:-
while True:
amount = raw_input("Dollar Amount: ")
try:
amount = float(amount)
# do your calculation with amount
# as per your needs here
except ValueError:
print 'Cannot parse'
break
I know this is basic but i actually don't even know what I've done wrong.
while True:
try:
end=int(input("If You Dont Want To Buy Anything Press 1 To Exit\nOr If You Would Like To Try Again Please Press 2"))
except ValueError:
print("\nPlease Enter Only 1 Or 2")
if end==1:
exit()
elif end==2:
continue
I have literally defined end at the start and yet the error is NameError: name 'end' is not defined I've even tried making end a global.
end is only assigned to if there was no ValueError. If int() raises an exception, then the assignment never takes place.
Either test for valid end values inside the try (so that you know no exception was raised), or assign a default value to end first.
For example, the following will not throw your exception and still prompt the user to re-enter the number if anything other than 1 or 2 was entered:
while True:
try:
end=int(input("If You Dont Want To Buy Anything Press 1 To Exit\nOr If You Would Like To Try Again Please Press 2"))
if end==1:
exit()
elif end==2:
break
except ValueError:
pass
print("\nPlease Enter Only 1 Or 2")
Note that I moved the print() to be outside the except block; it'll be printed if an exception was thrown or when no break or exit() was executed. Note that I used break here instead of continue to exit the while True loop; continue would just start the next iteration, prompting the user to enter a number again.
Others have explained the problem; here's a canonical solution. This matches the way many of us regard the process:
- grab a response
- until I get a legal response
- ... tell the user what's wrong
- ... get a new response
input_prompt = "If You Don't Want To Buy Anything Press 1 To Exit\nOr If You Would Like To Try Again Please Press 2"
response = input(input_prompt)
while response != '1' and response != '2':
print("\nPlease Enter Only 1 Or 2")
response = input(input_prompt)
end = int(reponse)
This has no unnatural loop exits (harder to follow and maintain), and no exception processing (slow).
It tries to assign a value to end, but catches a ValueError and after the except it tries to see what 'end' is however it was never assigned a value due to the Exception.
If int(input(...)) fails, a ValueError is raised. That happens before end is assigned. Add another control statement in the error handling, for instance
while True:
try:
end=int(input("If You Dont Want To Buy Anything Press 1 To Exit\nOr If You Would Like To Try Again Please Press 2"))
except ValueError:
print("\nPlease Enter Only 1 Or 2")
continue # ask again
if end==1:
exit()
elif end==2:
continue
I have a piece of code.
import sys
while(True):
print "Enter a number: "
try:
number = int(sys.stdin.readline())
except ValueError:
print "Error! Enter again an integer value"
continue
finally:
print number
break
Here I expect when I enter a non-integer number, the output should be
Error! Enter again an integer value
and then it should ask for input. But it is printing the message but asking for further inputs. Please explain it or if am thinking it wrong.
If I handle with NameError, then error message is not even being printed and the program is exiting with a traceback call.
The finally clause always runs, whether an exception was caught or not. You want else, which runs when there was no exception.
Also: you don't need parentheses for a while, and you probably want the raw_input function which is a little nicer to use than messing with sys.stdin directly.
So I would do:
while True:
try:
number = int(raw_input("Enter a number: "))
except ValueError:
print "Error! Enter again an integer value"
continue
else:
print number
break
Your finally should be else, otherwise it will execute regardless of whether or not there was an exception.