'Trying to do a counter in python
I'm creating a number guessing game in Python. I need to do a counter for three trials. I did the code below and it didn't work.
secretNum = input('Guess number 1 to 5: ') # the guessed number
random_number = random.randrange(1, 6) # the secret number
ssent = random.choice(sent)
def numba():
count = 0
while count < 3:
if count < 3:
if int(secretNum) == random_number:
print(f'''Yaay you guessed right
The secret number is {random_number}''')
elif int(secretNum) < random_number:
print(f'''HINT
Guessed number is less than secret number {random_number}''')
elif int(secretNum) > random_number:
print(f'''HINT
Guessed number is more than secret number {random_number}''')
else:
print('so close yet so far!!')
count +=1
break
return
numba()
Solution 1:[1]
your 'count' variable will increase only 'else', you should put it outside 'else'.
I think you will give the user 3 chances of guesses, then you will need to get the inputs from the user 3 times. So you need
secretNum = input('Guess number 1 to 5: ')
in the while loop.
The code you want should be something like this.
def numba():
count = 0
random_number = random.randrange(1, 6)
while count < 3:
userGuess = input('Guess number 1 to 5: ')
# if count < 3: # you don't need this, you already have while loop.
if int(userGuess) == random_number:
print(f'''Yaay you guessed right
The secret number is {random_number}''')
# User guessed it right! Stop the while loop.
break
elif int(userGuess) < random_number:
print(f'''HINT
Guessed number is less than secret number {random_number}''')
elif int(userGuess) > random_number:
print(f'''HINT
Guessed number is more than secret number {random_number}''')
else:
print('so close yet so far!!')
count +=1
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | Pongpich Singhagumpon |
