Develop Reference

How do you call a variable that was returned at the end of a function?

I created this function and want to call the returned result but, I'm not sure how to get the variable back. If possible I'd also like a different message to pop up if the user types n. Could anyone help?
def give_entertainment():
random_form_of_entertainment = random.choice(form_of_entertainment)
good_user_input = "y"
while good_user_input == "y":
user_input = input(f"We have chosen {random_form_of_entertainment} for your entertainment! Sound good? y/n: ")
if good_user_input != user_input:
random_form_of_entertainment = random.choice(form_of_entertainment)
continue
else:
print("Awesome! Glad we got that figured out. Your trip is all planned! ")
return random_form_of_entertainment

x = give_entertainment()
Should store the return of your function into x.
With:
print(x)
you should see what's stored in your x variable.

Call the method while assigning it to a variable.
some_var = your_method()
Now this some_var variable have the returning value.

I'd personally use a recursive function for this, I made it work with y/n/invalid_input cases. Also with this an invalid input won't update random_form_of_entertainment so the user has to say y or n for it to change. I hope this is what you're looking for!
def give_entertainment():
random_form_of_entertainment = random.choice(forms_of_entertainment)
good_user_input = "y"
bad_user_input = "n"
invalid_input = True
while invalid_input:
user_input = input(f'We have chosen {random_form_of_entertainment} for your entertainment! Sound good? y/n: ')
if user_input == good_user_input:
print("Awesome! Glad we got that figured out. Your trip is all planned for!")
return random_form_of_entertainment
elif user_input == bad_user_input:
print("Sorry to hear that, let me try again...")
return give_entertainment()
else:
print("I don't recognize that input, please try again.")
continue
chosen_result = give_entertainment()
print(f'You chose {chosen_result}!')

How do you stop a loop when a condition is met?

I'm a complete newbie to the coding trying my hands on python. While coding a simple program to calculate the age I ran into an error I cannot seem to fix no matter how hard I try. The code is pasted below. The program runs as far as I enter an year in the future, however when an year which has passed is used, the code loops again asking if the year is correct. Any help is appreciated.
from datetime import date
current_year = (date.today().year)
print('What is your age?')
myAge = input()
print('future_year?')
your_age_in_a_future_year= input()
future_age= int(myAge)+ (int(your_age_in_a_future_year)) - int(current_year)
def process ():
if future_age > 0:
print(future_age)
else:
print('do you really want to go back in time?, enter "yes" or "no"')
answer = input()
if answer =='yes':
print (future_age)
if answer == 'no':
print ('cya')
while answer != 'yes' or 'no':
print ('enter correct response')
process ()
process()

In this case, your function just needs to contain a while loop with an appropriate condition, and then there is no need to break out from it, or do a recursive call or anything.
def process():
if future_age > 0:
print(future_age)
else:
print('do you really want to go back in time?, enter "yes" or "no"')
answer = None
while answer not in ('yes', 'no'):
answer = input()
if answer == 'yes':
print(future_age)
elif answer == 'no':
print('cya')

try
...
if future_age > 0:
print(future_age)
return
else:
print('do you really want to go back in time?, enter "yes" or "no"')
...

break out of two loops without disturbing the if statements after it in python

I'm creating a simple login program. It asks you for a username, then the password. If you type the correct password, you're in, else, you get a genial request to repeat your password. here is the code:
while True:
turns= 5
turns= turns-1
print ("Username")
L_username= input ("")
print ("Authorising...")
time.sleep(2)
if(L_username)==("test#test.test"):
for x in range (5):
print ("Please enter the password")
passwordlogin= input("")
if passwordlogin == ("beta123"):
print ("Hello, developer.")
break
break
else:
print ("Incorrect. You have",turns," more turns")
now the thing is, it gives me an error message: incorrect syntax for else: print ("incorrect... I know this is meant an issue because I wrote two 'breaks', the latter out of the for loop... which means it will break the while True loop at the start whether or not I go into the loop... which means the 'else' statement is out of the loop (sorry, I am not the best explainer at these things)... which gives the error message. But I don't understand how I should fix this problem... can anyone help?
I hope you understood me!

This could be triviallized by using a function and returning from it once you reach that if statement.
def func():
while True:
turns= 5
turns= turns-1
print ("Username")
L_username= input ("")
print ("Authorising...")
time.sleep(2)
if(L_username)==("test#test.test"):
for x in range (5):
print ("Please enter the password")
passwordlogin= input("")
if passwordlogin == ("beta123"):
print ("Hello, developer.")
return
else:
print ("Incorrect. You have",turns," more turns")
However, if you insist on not having a function, you can basically have a flag in your code that you set to true inside that if block. Before continuing the outer loop, you should check if this flag is set to True, if it is, you can safely break.
while True:
success = False
turns= 5
turns= turns-1
print ("Username")
L_username= input ("")
print ("Authorising...")
time.sleep(2)
if(L_username)==("test#test.test"):
for x in range (5):
print ("Please enter the password")
passwordlogin= input("")
if passwordlogin == ("beta123"):
print ("Hello, developer.")
success = True
break
else:
print ("Incorrect. You have",turns," more turns")
if success:
break
Note should be taken for nested loops and such intense nesting for what is essentially a trivial problem. Please try to use a singular loop with your conditions.

You need to swap the else and break and change the break's indentation. If the user gets the correct username you want to test the password at most 5 times and then quit
while True:
turns= 5
print ("Username")
L_username= input ("")
print ("Authorising...")
time.sleep(2)
if(L_username)==("test#test.test"):
for x in range (5):
print ("Please enter the password")
passwordlogin= input("")
if passwordlogin == ("beta123"):
print ("Hello, developer.")
break
else:
turns -= 1
print ("Incorrect. You have",turns," more turns")
break
So in this case you run the for loop and then break

Well I have written your concept in my own way. Hope this works for you
turns = 5
while turns != 0:
username = input('Enter Username: ')
password = input(f'Enter password for {username}: ')
if username == 'test#test.test' and password == 'beta123':
print('Hello developer')
turns = 0 # Breaking while loop if user get access within number of "turns"
else:
turns = turns - 1 # Or turns -= 1
print('Incorrect username or password')
print(f'You have {turns} turns for gaining access')

User prompt to Continue or Clear a time duration calculator

I am very inexperienced with any programming language. I decided for my first for-fun/leaning project to be creating a calculator to add different durations of time (ex. add the length of several songs to find how long an album is).
So far the user input method is pretty basic, and I'll work to fix that once the basics work, so bear with me on that. The calculator is currently capable of adding two times together and printing the correct answer.
My main problem is creating a functional Continue/Clear user prompt. I have a loop asking if the user wants to continue, clear, or exit. User input works correctly, but I don't know how to actually "continue or clear".
My idea: Upon hitting continue, the previous total will become firstNumber, and the user will only be prompted for the second number (nextNumber) and will be given a new sum and receive the same continue/clear/exit prompt. If the user hits clear, it will start at the very beginning and the user will be prompted for a new firstNumber.
Any help is greatly appreciated.
Below is the code:
import datetime
# holds first entry
print("Enter the first number")
seconds_1 = int(input("Seconds: "))
minutes_1 = int(input("Minutes: "))
hours_1 = int(input("Hours: "))
# holds second entry
print("Enter another number")
seconds_2 = int(input("Seconds: "))
minutes_2 = int(input("Minutes: "))
hours_2 = int(input("Hours: "))
# calculation
duration_1 = datetime.timedelta(hours=hours_1, minutes=minutes_1, seconds=seconds_1)
duration_2 = datetime.timedelta(hours=hours_2, minutes=minutes_2, seconds=seconds_2)
total = duration_1 + duration_2
print(total)
# continue, clear, or exit
contClear = input("Continue: Y | Clear: N | Exit: X: ")
cont = False
while cont == False:
if contClear.upper() == "Y" or contClear.upper() == "YES":
print("Next")
cont = True
elif contClear.upper() == "N" or contClear.upper() == "NO":
print("Cleared")
cont = True
elif contClear.upper() =="X" or contClear.upper() == "EXIT":
print("Exiting")
cont = True
else:
print("Invalid Entry")
contClear = input("Continue: Y | Clear: N | Exit: X: ")
print("DONE")

python -- Crash when trying to deal with unexpected input

So, I'm just fooling around in python, and I have a little error. The script is supposed to ask for either a 1,2 or 3. My issue is that when the user puts in something other than 1,2 or 3, I get a crash. Like, if the user puts in 4, or ROTFLOLMFAO, it crashes.
EDIT: okay, switched it to int(input()). Still having issues
Here is the code
#IMPORTS
import time
#VARIABLES
current = 1
running = True
string = ""
next = 0
#FUNCTIONS
#MAIN GAME
print("THIS IS A GAME BY LIAM WALTERS. THE NAME OF THIS GAME IS BROTHER")
#while running == True:
if current == 1:
next = 0
time.sleep(0.5)
print("You wake up.")
time.sleep(0.5)
print("")
print("1) Go back to sleep")
print("2) Get out of bed")
print("3) Smash alarm clock")
while next == 0:
next = int(input())
if next == 1:
current = 2
elif next == 2:
current = 3
elif next == 3:
current = 4
else:
print("invalid input")
next = 0

Use raw_input() not input() the latter eval's the input as code.
Also maybe just build a ask function
def ask(question, choices):
print(question)
for k, v in choices.items():
print(str(k)+') '+str(v))
a = None
while a not in choices:
a = raw_input("Choose: ")
return a
untested though

since the input() gives you string value and next is an integer it may be the case that crash happened for you because of that conflict. Try next=int(input()) , i hope it will work for you :)