猜猜數字遊戲給出的答案太快

Question

我正在學習Python的介紹，並遇到了一個問題。

請看看我的腳本：

import random 
secretNumber = random.randint(1,20) 
print (' I am thinking of a number between 1 and 20.') 

#ask the player to guess 6 times. 
for guessesTaken in range (1,7): 
    print ('take a guess.') 
    guess = int(input()) 

    if guess < secretNumber: 
     print ('your guess is too low') 
    elif guess > secretNumber: 
     print ('your guess is too high') 
    else: 
     break 

    if guess == secretNumber: 
     print ('good job! you guessed my number in ' + str(guessesTaken) + ' guesses!') 
    else: 
     print ('nope. the number i was thinking of was ' + str(secretNumber)) 

我試圖做的猜測非數字遊戲。當我運行該腳本，它問我：

1. 猜數字

2. 當我在數字輸入，它會自動給我它存儲的，而不是給我6次嘗試我的號碼。

如果我在任何步驟都不清楚，請告訴我，下次我會盡全力回答更清晰的問題！

Answer 1

以if guess == secretNumber:開頭的最後一部分是循環的一部分，但您不希望在循環中給出最終結果。相反，把它在循環後（外的話），這意味着你應該減少這4行代碼的縮進：

import random 
secretNumber = random.randint(1,20) 
print (' I am thinking of a number between 1 and 20.') 

#ask the player to guess 6 times. 
for guessesTaken in range (1,7): 
    print ('take a guess.') 
    guess = int(input()) 

    if guess < secretNumber: 
     print ('your guess is too low') 
    elif guess > secretNumber: 
     print ('your guess is too high') 
    else: 
     break 

if guess == secretNumber: 
    print ('good job! you guessed my number in ' + str(guessesTaken) + ' guesses!') 
else: 
    print ('nope. the number i was thinking of was ' + str(secretNumber)) 

Answer 2

最簡單的解決方法是，上面已經爲說明，只需將最後的if語句跳出循環。你使用的語法（除了那個錯誤的縮進）非常好，這是python的一個很好的開始。然而，在python 3中，你不需要在循環內部的if語句中使用其他語言。從根本上說，因爲如果滿足ELSE條件，你不需要做任何事情，你可以把它拿出來。