i am trying to learn about the while loop and am trying to build something with it.
i have a function that will hold a choice that the user will make but every time i run it the script is asking for the input twice..
can someone please explain to me what is wrong with my code?
thanks in advance!! :)
code:
def choice():
choice = int(input("what is your choice? "))
valid_choice = False
while not valid_choice:
if choice >= 1 and choice <= 4:
return choice
valid_choice = True
else:
print("Please enter a valid response..")
choice = int(input("what is your choice? "))
def beginning():
while True:
display_menu()
choose = choice()
if choose == 1: print("Your status")
print(Michael.status())
print()
print("Banks's status:")
print(Prod.status()) elif choose == 2: withraw = float(input("How much money do you want to withraw from the bank?: "))
Prod.withrawl(withraw)
Michael.withraw(withraw)
You were calling choice twice
Related
I am making a program that generates a random number and asks you to guess the number out of the range 1-100. Once you put in a number, it will generate a response based on the number. In this case, it is Too high, Too low, Correct, or Quit too soon if the input is 0, which ends the program(simplified, but basically the same thing).
It counts the number of attempts based on how many times you had to do the input function, and it uses a while loop to keep asking for the number until you get it correct. The problem that I am facing is that I have to make it break out of the while loop once the guess is either equal to the random number or 0. This normally isn't an issue, because you could use sys.exit() or some other function, but according to the instructions I can't use break, quit, exit, sys.exit, or continue. The problem is most of the solutions I've found for breaking the while loop implement break, sys.exit, or something similar and I can't use those. I used sys.exit() as a placeholder, though, so that it would run the rest of the code, but now I need to figure out a way to break the loop without using it. This is my code:
import random
import sys
def main():
global attempts
attempts = 0
guess(attempts)
keep_playing(attempts)
def guess(attempts):
number = random.randint(1,100)
guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
while guess != 0:
if guess != number:
if guess < number:
print("Too low, try again")
attempts += 1
guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
elif guess > number:
print("Too high, try again")
attempts += 1
guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
else:
print()
print("Congratulations! You guessed the right number!")
print("There were", attempts,"attempts")
print()
#Ask if they want to play again
sys.exit()#<---- using sys.exit as a placeholder currently
else:
print()
print("You quit too early")
print("The number was ",number,sep='')
#Ask if they want to play again
sys.exit()#<----- using sys.exit as a placeholder currently
def keep_playing(attempts):
keep_playing = 'y'
if keep_playing == 'y' or keep_playing == 'n':
if keep_playing == 'y':
guess(attempts)
keep_playing = input("Another game (y to continue)? ")
elif keep_playing == 'n':
print()
print("You quit too early")
print("Number of attempts", attempts)
main()
If anyone has any suggestions or solutions for how to fix this, please let me know.
Try to implement this solution to your code:
is_playing = True
while is_playing:
if guess == 0:
is_playing = False
your code...
else:
if guess == number:
is_playing = False
your code...
else:
your code...
Does not use any break etc. and It does breaks out of your loop as the loop will continue only while is_playing is True. This way you will break out of the loop when the guess is 0 (your simple exit way) or when the number is guessed correctly. Hope that helps.
I am not a fan of global variables but here it's your code with my solution implemented:
import random
def main() -> None:
attempts = 0
global is_playing
is_playing = True
while is_playing:
guess(attempts)
keep_playing()
def guess(attempts: int) -> None:
number = random.randint(1,100)
print(number)
is_guessing = True
while is_guessing:
attempts += 1
guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
if guess == 0:
is_guessing = False
print("\nYou quit too early.")
print("The number was ", number,sep='')
else:
if guess == number:
is_guessing = False
print("\nCongratulations! You guessed the right number!")
print("There were", attempts, "attempts")
else:
if guess < number:
print("Too low, try again.")
elif guess > number:
print("Too high, try again.")
def keep_playing() -> None:
keep_playing = input('Do you want to play again? Y/N ')
if keep_playing.lower() == 'n':
global is_playing
is_playing = False
main()
TIP:
instead
"There were", attempts, "attempts"
do: f'There were {attempts} attempts.'
Closed. This question needs debugging details. It is not currently accepting answers.
Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
Closed 1 year ago.
Improve this question
choice = int()
show_list= []
def main():
print("1. List all TV Shows")
print("2. Add a TV Show")
print("3. Delete a TV Show")
print("4. Exit")
choice = input("Please enter a number: ")
choice = int(choice)
if choice == 1:
print(show_list)
main()
elif choice == 2:
add_shows = int(input("Enter number of TV shows to add: "))
while True:2
for x in range(add_shows): #error
show = [input(("Please enter your TV Show: ").strip())]
show.append(show_list)
else:
break
main()
elif choice == 3:
remove_shows = int(input("Enter number of TV shows to delete: "))
while True:
for x in range(add_shows): #error
show = [input(("Please enter your TV Show: ").strip())]
show.append(show_list)
else:
break
main()
elif choice == 4:
print("Goodbye")
so this is the code i have so far. i am having an issue when i choose an option, do said option, and go back to select another, that it doesnt let you do it. im not sure what the error is in my for x in range() is but trinket(online python program) is running it as an error. also my list isnt working. i can do the add a show but it is not saving.
thank you any help in advance.
seems you are new to python. Your code has a lot of syntax and syntactical errors. Just take a look at the code, and try to figure out what was wrong.
show_list= []
choice = -1
while choice != 4:
print("1. List all TV Shows")
print("2. Add a TV Show")
print("3. Delete a TV Show")
print("4. Exit")
choice = input("Please enter a number: ")
choice = int(choice)
if choice == 1:
print(show_list)
elif choice == 2:
add_shows = int(input("Enter number of TV shows to add: "))
for x in range(add_shows):
show = input(("Please enter your TV Show: ").strip())
show_list.append(show)
elif choice == 3:
remove_shows = int(input("Enter number of TV shows to delete: "))
for x in range(remove_shows):
show = input(("Please enter your TV Show: ").strip())
show_list.remove(show)
elif choice == 4:
print("Goodbye")
If you have any question, feel free to ask in the comments.
PS: Don't get bothered about the downvotes. Happens.
I thought the logic of my while loop made sense, but it abruptly stops after the first loop.
choice=int(input("Enter choice:"))
if (choice=="" or (choice!=0 and choice!=1 and choice!=2)):
valid = False
while valid == False:
print("Invalid choice, please enter again")
choice=int(input("Enter choice:"))
return choice
if choice ==1:
valid=True
display_modules_average_scores()
menu()
elif choice ==2:
valid=True
display_modules_top_scorer()
menu()
elif choice==0:
exist=True
print("===============================================")
print("Thank you for using Students' Result System")
print("===============================================")
If I enter 5, it does:
print("Invalid choice, please enter again")
choice=int(input("Enter choice:"))
But if I enter 5 again, it stops the program. What am I doing wrong?
if I enter 5 again, it stops the program
Because you have a return statement that immediate ends the function you're running within.
You seem to be trying to create an infinite loop. You can start with testing exit and invalid conditions with this. Note:choice will never equal an empty string
while True:
choice=int(input("Enter choice (0 to exit):"))
if choice == 1:
pass # do something
elif choice == 2:
pass # do something else
elif choice == 0:
break
else:
print("Invalid choice, please enter again")
print("Thanks")
To exit the loop, you can use break, which executes code after the loop. Use return to end the function, as mentioned. There is a difference
If you're running this loop inside of the menu() function, you do not need to actually call the menu function again. That's the point of the while loop
By defining the function we can perform this task easily with no code duplication.
The Below code calls the function inputchoice() and then the inputchoice() will check the value entered by the user and if there the value is not valid then the inputchoice will call itself and the process continues untill the user enter correct input.
def inputchoice():
choice=int(input("Enter choice: "))
if (choice=="" or (choice!=0 and choice!=1 and choice!=2)):
print("Invalid choice!")
choice = inputchoice()
return choice
def menu():
choice = inputchoice()
print(choice)
if choice ==1:
valid=True
print("Do something if Valid = True")
elif choice ==2:
valid=True
print("Do something if Valid = True")
elif choice==0:
valid=True
print("Do something if Valid = True")
menu() #implementing menu function
I prefer making a dictionary with your functions, keeps the code clean in my eyes.
Consider this code here:
def choice1():
return 'choice1'
def choice2():
return 'choice2'
def defaultchoice():
return 'default'
choicedict = {
'1': choice1,
'2': choice2
}
while True:
choice = input("Enter choice (0 to exit):") # maintain as str to avoid error!
if choice == '0':
break
value = choicedict.get(choice, defaultchoice)()
print(value)
Single Function Code
def menu():
choice=int(input("Enter choice:"))
if (choice=="" or (choice!=0 and choice!=1 and choice!=2)):
print("Invalid choice, please enter again")
menu()
elif choice ==1:
print("Oh, its working")
menu()
elif choice ==2:
print("Oh, its working")
menu()
elif choice==0:
print("===============================================")
print("Thank you for using Students' Result System")
print("===============================================")
menu()
Hi i would use a while loop like this. It would seem from this assignment that we are from the same institution. This is what i use for my code, I hope this helps.
while True:
user_input = input("Enter choice: ")
if (user_input == "0"):
print("=====================================================")
print("Thank You for using Students' Result System")
print("=====================================================")
break
elif(user_input == "1"):
display_modules_average_scores()
elif(user_input == "2"):
display_modules_top_scorer()
else:
print("Invalid choice, please enter again")
I'm writing this program that is basically a limited calculator. I'm trying to make it so that if the user enters let's say "Power" instead of the number 1 for the desired mode, it prints out "Invalid selection". The same goes if they attempt to write "Quadratics" instead of 2 and so on for the rest.
#CALCULATOR
print("MY CALCULATOR")
print("1. Powers")
print("2. Quadratics")
print("3. Percents")
print("4. Basic Ops")
choice = int(input("Please enter your desired mode: "))
if choice == 1:
base = int(input("Enter the base: "))
exponent = int(input("Enter the exponent: "))
power = base**exponent
if choice == 2:
print("Please enter the values for A/B/C: ")
a = int(input("A: "))
b = int(input("B: "))
c = int(input("C: "))
I tried doing:
if choice not == 1:
print("Invalid Selection")
and
if choice not 1:
print("Invalid Selection")
but they don't seem to work. If you could please tell me what I am doing wrong. Thank you.
not is not a function. It is an operator.
The correct usage is to put it before an expression:
if not (choice == 1):
However in this case, it's much better to use != (not equal) instead:
if choice != 1:
For menu-driven programming, how is the best way to write the Quit function, so that the Quit terminates the program only in one response.
Here is my code, please edit if possible:
print("\nMenu\n(V)iew High Scores\n(P)lay Game\n(S)et Game Limits\n(Q)uit")
choose=input(">>> ")
choice=choose.lower()
while choice!="q":
if choice=="v":
highScore()
main()
elif choice=="s":
setLimit()
main()
elif choice=="p":
game()
main()
else:
print("Invalid choice, please choose again")
print("\n")
print("Thank you for playing,",name,end="")
print(".")
When the program first execute and press "q", it quits. But after pressing another function, going back to main and press q, it repeats the main function.
Thanks for your help.
Put the menu and parsing in a loop. When the user wants to quit, use break to break out of the loop.
source
name = 'Studboy'
while True:
print("\nMenu\n(V)iew High Scores\n(P)lay Game\n(S)et Game Limits\n(Q)uit")
choice = raw_input(">>> ").lower().rstrip()
if choice=="q":
break
elif choice=="v":
highScore()
elif choice=="s":
setLimit()
elif choice=="p":
game()
else:
print("Invalid choice, please choose again\n")
print("Thank you for playing,",name)
print(".")
def Menu:
while True:
print("1. Create Record\n2. View Record\n3. Update Record\n4. Delete Record\n5. Search Record\n6. Exit")
MenuChoice=int(input("Enter your choice: "))
Menu=[CreateRecord,ViewRecord,UpdateRecord,DeleteRecord,SearchRecord,Exit]
Menu[MenuChoice-1]()
You're only getting input from the user once, before entering the loop. So if the first time they enter q, then it will quit. However, if they don't, it will keep following the case for whatever was entered, since it's not equal to q, and therefore won't break out of the loop.
You could factor out this code into a function:
print("\nMenu\n(V)iew High Scores\n(P)lay Game\n(S)et Game Limits\n(Q)uit")
choose=input(">>> ")
choice=choose.lower()
And then call it both before entering the loop and then as the last thing the loop does before looping back around.
Edit in response to comment from OP:
The following code below, which implements the factoring out I had mentioned, works as I would expect in terms of quitting when q is typed.
It's been tweaked a bit from your version to work in Python 2.7 (raw_input vs. input), and also the name and end references were removed from the print so it would compile (I'm assuming those were defined elsewhere in your code). I also defined dummy versions of functions like game so that it would compile and reflect the calling behavior, which is what is being examined here.
def getChoice():
print("\nMenu\n(V)iew High Scores\n(P)lay Game\n(S)et Game Limits\n(Q)uit")
choose=raw_input(">>> ")
choice=choose.lower()
return choice
def game():
print "game"
def highScore():
print "highScore"
def main():
print "main"
def setLimit():
print "setLimit"
choice = getChoice()
while choice!="q":
if choice=="v":
highScore()
main()
elif choice=="s":
setLimit()
main()
elif choice=="p":
game()
main()
else:
print("Invalid choice, please choose again")
print("\n")
choice = getChoice()
print("Thank you for playing,")
This is a menu driven program for matrix addition and subtraction
def getchoice():
print('\n What do you want to perform:\n 1.Addition\n 2. Subtraction')
print('Choose between option 1,2 and 3')
cho = int(input('Enter your choice : '))
return cho
m = int(input('Enter the Number of row : '))
n = int(input('Enter the number of column : '))
matrix1 = []
matrix2 = []
print('Enter Value for 1st Matrix : ')
for i in range(m):
a = []
for j in range(n):
a.append(int(input()))
matrix1.append(a)
print('Enter Value for 2nd Matrix : ')
for i in range(m):
a = []
for j in range(n):
a.append(int(input()))
matrix2.append(a)
choice = getchoice()
while choice != 3:
matrix3 = []
if choice == 1:
for i in range(m):
a = []
for j in range(n):
a.append(matrix1[i][j] + matrix2[i][j])
matrix3.append(a)
for r in matrix3:
print(*r)
elif choice == 2:
for i in range(m):
a = []
for j in range(n):
a.append(matrix1[i][j] - matrix2[i][j])
matrix3.append(a)
for r in matrix3:
print(*r)
else:
print('Invalid Coice.Please Choose again.')
choice = getchoice()