私はカスタム入力を受け入れるカウントダウンを望みます

Question

私は自分のコードを手に入れようと努力しています。

print ("Would You like To Start A Countdown? Y/N (CASE SENSITIVE)") 
countdownyn = input (':') 
if countdownyn == ('Y'): 
    print ("Please Enter Your Designated Time To Countdown From") 
x = input (':') 
def countdown(x) : 
    while x> 0: 
     print (x) 
     print ("") 
     time.sleep(1) 
     x = x1 
    if x ==0: 
     print("BLAST OFF!") 
countdown(x) 

Answer 1

私は表示されませんが、ここで支援するための修正がされているいくつかの問題。多分。 Python 3を使用している場合は、raw_inputを入力に変更します。

import time 

def countdown(x) : 
    while x > 0: 
     print (x) 
     print ("") 
     time.sleep(1) 
     x = x - 1 
    print("BLAST OFF!") 

countdownyn = raw_input ('Would You like To Start A Countdown? Y/N (CASE SENSITIVE): ') 
if countdownyn == ('Y'): 
    x = raw_input ('Please Enter Your Designated Time To Countdown From: ') 
    countdown(int(x)) 

必要なポーズで出力：

Would You like To Start A Countdown? Y/N (CASE SENSITIVE): Y 
Please Enter Your Designated Time To Countdown From: 10 
10 

9 

8 

7 

6 

5 

4 

3 

2 

1 

BLAST OFF! 

**** **** EDIT

あなたはさらに行く得るために...：

import time 

def countdown(ticks) : 
    for tick in range(ticks, 0, -1): 
     print (tick) 
     print ("") 
     time.sleep(1) 
    print("BLAST OFF!") 



while True: 
    countdownyn = raw_input('Would You like To Start A Countdown? Y/N (CASE SENSITIVE): ') 
    if countdownyn == 'Y': 
     x = raw_input('Please Enter Your Designated Time To Countdown From: ') 
     try: 
      countdown(int(x)) 
     except ValueError: 
      print("Please enter a valid integer") 
      continue 
    elif countdownyn == 'N': 
     print("Goodbye!") 
     break 
    else: 
     print("Please enter only Y or N") 

Answer 2

あなたは決してx1に何もしません。おそらく、あなたはx-1を意味：

while x> 0: 
    print (x) 
    print ("") 
    time.sleep(1) 
    x = x-1 #here