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I am trying to make a calculator. I have the code that does the operations working, but I am trying to allow the user to keep the calculator running without having to rerun the code. i an trying to assign a Boolean to variable, but python keeps telling me there is a name error, which is the the variable is not defined. Can you please help me?
Thank you in advance.
I have tried to change the available name but it doesn't do anything.
The code stopped working when it got to the while run == true line.
else:
print('Invalid operator, please run code again')
run = True
while run == True:
print(' do you need another problem solved? y/n')
if input() == y:
run = True
elif input() == n:
run = False
I expected the code to ask me if I need another problem solved, but there is a name error.
If the error is complaining about y or n it's because you need to surround it with quotes
if input() == "y":
run = True
Also, run == True is not needed
while run:
does the trick
else:
print('Invalid operator, please run code again')
run = True
while run:
print(' do you need another problem solved? y/n')
inp=input()
if inp == 'y':
run = True
elif inp == 'n':
run = False
you have used input 2 times so your code will wait 2 times for input, use input() once.
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I have tried everything I could before asking this question... I am making a little text adventure game, really simple and the only code in it is basically just input(). But when they do something wrong, I want to add an else: and print("Thats not valid!") and go back to example = input(), but my else statement is printing and re-doing input with a correct statement, so its still executing else without any error, can anyone help? I don't know why its doing this and all the examples I have seen this is valid code, and my else statement just executes without checking if it was wrong...
```python
if prompt1 == "y":
print("Lets start")
else:
prompt1 = input()
It then executes input again...
If that's the exact code the indention is wrong.
The if and else of same condition should have same indention:
if condition:
do this
else:
do this
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When I run this easy program and want to exit it, I need to type exit 2 times (at line 7).
def registration():
def username_registration():
entering_username = True
while entering_username:
print("Type exit to leave the registration.")
usernam_from_registration = input("Enter username: ")
if usernam_from_registration == "exit":
break
lenght_usernam_from_registration = len(usernam_from_registration)
if lenght_usernam_from_registration > 15:
print("too long")
else:
return usernam_from_registration
username_registration()
print(username_registration())
registration()
Why is this and how can I make it so I only need to write it one time?
This is because you are calling the username_registration() function twice.
username_registration()
print(username_registration())
The first time you call it, nothing happens because you are not doing anything with the result.
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I am a beginner learning Python and wrote the following codes. To control the loop flow I tried using a variable "test", but the loop never stopped. Appreciate if any thoughts.
test=False
while test==False:
a=input("Please enter an even number to make test True: ")
if int(a)%2==0:
test==True
print("test is now True")
else:
print("Please try again!")
test=False
while test is False:
a=input("Please enter an even number to make test True: ")
if int(a) %2 == 0:
test = True
print("test is now True")
else:
print("Please try again!")
Fixed the test == true, should've been test = true :) have fun learning
heres the right code:
test=False
while test==False:
a=input("Please enter an even number to make test True: ")
if int(a)%2==0:
test=True
print("test is now True")
else:
print("Please try again!")
here's what you did wrong:
in the 5th line test==True the == checks if that's what the variable true stores, it doesn't set it; so this can be simply fixed by test=True
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I keep getting a syntax error on line 10 that says
Traceback (most recent call last):
File "main.py", line 5, in <module>
user = input("Choose a number")
File "<string>", line 1
python main.py
^
I can't figure it out. Can anyone help?
import random
number = random.randint(1, 20)
attempt = 5
def game():
global attempt
attempt -= 1
if attempt > 0:
if int(user) > number:
user = input("Choose a number")
print("Choose a smaller number\n " + (str(attempt) + " tries left"))
game()
elif int(user) < number:
print("Choose a larger number\n "+ (str(attempt) + " tries left"))
game()
elif int(user) == number:
print ("You guessed the number ")
else:
print("CONGRATULATIONS ")
game()
First, this is a run-time error, not a syntax error.
The problem is actually in the previous line: you've tried to use the value of user before it has one. You have to fix this before you can continue debugging your program. You need to get the user's first number before you can test it.
Problem A
I can't help but notice that your traceback message doesn't align with the code you have provided. It seems as though you already have the user variable assigned before defining and calling the game() function.
Solution A
If that is the case, try removing that line and apply the changes Prune recommended for you in the other answer. This will solve the code related problems in the program.
Problem B
However, the traceback also mentions that on line 1 you have python main.py. This is most likely caused by the way you are running the script.
Solution B
Try running it instead in the IDLE (distributed with Python) or Terminal / Command Prompt (depends on your OS) if not doing so already.
Last Resort
If the problem still arises, try running the snippet of code you included in the question by itself to see if it is functional ( add on Prune's revisions with it ).
Since the revised snippet should work, any errors that occur is definitely caused by how the code is ran. If no errors occur then look through your original source code for anything else that might be causing problems.
Good luck!
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Python Question: i need to run a program that asks for a password but if the wrong answer is input three times the user is thrown out of the program i can run it in a while loop but cant get it to quit if the wrong password is entered.
Thanks for your help
Adding an approximation of how I'd do it, in the absence of an example containing the problem. else on a for loop will only execute if you did not break out of the loop. Since you know the max number of times to run the loop is 3 you can just use a for loop instead of a while loop. break will still break you out early.
for _ in range(3):
if raw_input("Password:") == valid_passwd: # really should compare hashed values (as I shouldnt have passwords stored in the clear
print "you guessed correctly"
break
print "you guessed poorly"
else:
print "you have failed too many times, goodbye"
sys.exit(1)
# continue on your merry (they got the right password)
How about sys.exit()
>>> import sys
>>> guess = False
>>> if guess:
... pass
... else:
... sys.exit()
http://docs.python.org/3/library/sys.html