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
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 am trying to make an input that would prevent a user entering an input like this "QWERTY" this "qwerty" or this "QW3RTY"
The input is for names so I want to make sure that it would need the user to have a capital letter at the start of their name "John" not "john"
I have tried to loop the question using while True: try and attempted to use .isalpha and .title but I couldnt seem to make it work
while True:
try:
name = input(str("What is your name? "))
if name is not name.isalpha:
print("Please Enter a Valid name")
continue
if name is not name.title:
print("Please have a capital letter at the start of your name!")
continue
else:
break
I expected for the if statements to work but it comes up with invalid syntax.
Your logic is faulty ... but you've replicated it, even though the first if failed. Correct your problems one at a time, rather than trying to write the whole program at once.
name is a string; name.isalpha is a function.
A string cannot ever be identical to a function.
I think what you want is
if name.isalpha():
Also, try requires an except clause to catch exceptions. Again, add program features individually. That way, when you hit an error, you'll be fixing only that one error.
See this lovely debug blog for help.
Implementing features one at a time is a much better place to start than to try to catch everything all at once. Also, make sure to call your functions, otherwise, they will be truthy:
if str:
print("True!")
else:
print("False!")
True!
# compared to
if str():
print("True!")
else:
print("False!")
False!
Functions are objects, and will not act Falsey.
while True:
name = input("What is your name? ") # no need for str function here
# you can wrap this whole thing in a
# try/except for ValueError
try:
if name.isupper() or name.islower() or not name == name.title():
raise ValueError("Please include proper capitalization")
elif not name.isalpha():
raise ValueError("Use only alphabetical characters, please")
else:
break
except ValueError as e:
print(e)
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.
Just a quick question because I really can't find a simple solution to my problem.
Is there a way to get a user input that is meant to be a integer, but when a string is entered
the program will not break and instead displays "Error"
I've been trying to work around it by converting strings to integers and vice-versa, but I constantly get "invalid literal for int() with base 10" error, or when it displays "Error" it does so in an infinite loop.
Here's my code just to help clear the question
choice = input("Enter your choice: ")
while choice != 3:
if choice == 1:
get_songs()
print
main()
elif choice == 2:
read_songs()
print
main()
else:
print "Invalid choice"
So essentially I want the else operation to work for strings as well as for an integer that is greater than 3 or less than 1.
but I constantly get "invalid literal for int() with base 10" error
You get an exception, specifically a ValueError. You can catch the exception using an except block. For more info, refer to whatever language tutorial you've been using thus far, or try Google for except block Python.
when it displays "Error" it does so in an infinite loop.
When you detect that the input isn't correct, you need to get new input before you try the loop again. Put the input-getting stuff inside your loop. Do not use recursion (calling main() from within main) in addition to the loop; you're only going to confuse yourself this way. Because you don't get a value for choice until you're inside the loop, it's easier to explicitly break out of the loop when you find the appropriate choice value, instead of trying to control the loop with it (i.e. testing for it in the while condition).
We can also use continue to simplify the loop structure: instead of doing all the work in the try block, we limit that to the part where we extract a number. We use continue in the except block to skip the rest of the loop when we don't have an actual number, and only do the rest of the loop when we do. (After all, maybe the code we call for choice == 1 or choice == 2 could raise ValueError for some totally different reason, and we'd want to do something different about that.)
while True:
try:
choice = int(raw_input("Give me a number"))
except ValueError:
print "Could you at least give me an actual number?"
continue
if choice == 1:
do_something()
elif choice == 2:
do_something_else()
elif choice == 3:
break
else:
print "Try a different number"
Since that choice always != 3, you will get an infinite loop.
You should get input again if it jumps into the "else" condition.
In your code, the while loop should encapsulate the input() line. The following is a clearer alternative, however:
Create a function which gets user input:
def getInteger(prompt):
while True:
userIn = input(prompt)
try:
return int(userIn)
except ValueError:
print "Error"
Here's something: I dislike making the user enter "" around the strings.
ch = raw_input("Enter choice: ")
#ch *is* a string at this point
if ch.isdigit():
choice = int(ch)
else:
print "Invalid choice"
Edit (from comments):
isdigit may not handle locale encoding correctly. In Python 3, you can use isdecimal instead.
— J.F. Sebastian