In my program, I want to take in an input which should be a number.If the user inputs a string, however, the program returns an exception.I want the program set in such a way that the input is converted to int and then if the string is anything other than an int the program prints that "Stop writing please".Like for example:
x=input("Enter a number")
if int(x)=?: #This line should check whether the integer conversion is possible
print("YES")
else #This line should execute if the above conversion couldn't take place
print("Stop writing stuff")
You'll need to use try-except blocks:
x=input("Enter a number")
try:
x = int(x) # If the int conversion fails, the program jumps to the exception
print("YES") # In that case, this line will not be reached
except ValueError:
print("Stop writing stuff")
You can simply use a try-except block to catch the exceptional case, and inside there, print your statement. Something like this:
x=input("Enter a number")
try:
x=int(x)
print("YES")
except:
print("Stop writing stuff")
Related
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'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 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