试图在python中做一个计数器
python 527
原文标题 :Trying to do a counter in python
我正在用 Python 创建一个猜数字游戏。我需要为三个试验做一个计数器。我做了下面的代码,它没有工作。
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()
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Pongpich Singhagumpon 评论
你的’count’变量只会增加’else’,你应该把它放在’else’之外。
我认为你会给用户 3 次猜测的机会,然后你需要从用户那里获得 3 次输入。所以你需要secretNum = input('Guess number 1 to 5: ')
在while循环中。
你想要的代码应该是这样的。
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
2年前