為什么`input`不顯示在python上？

Question

我正在做一個簡單的游戲，有2個玩家和20個棍棒。 每個玩家可以選擇1-3個桿，而玩家選擇最后一個桿將輸掉比賽。

def stix(num):
    for _ in range(5): print('|  '* num)
    print
stix(20)
game_over = 0
while game_over !=0:
    players={}
    for i in range(2):
        players[i] = int(input('Player %d: Please pick stick(s) up to 3' %i))
        if players > 3 or players<0:
            print('Please pick between 1 - 3 stick(s)')
        else:
            stix-=players
            if stix <= 0:
                print('Player[i] lost')
                break
            else:
                print('There is %d stick(s) left' %stix)
                print(stix-players[i])

因此，功能stix顯示20條，僅此而已。 它不要求please pick stick(s) up to 3 。 我在這里想念什么？

*我正在使用python 2.6

提前致謝！

Answer 1

您根本不會進入while循環：

game_over = 0
while game_over !=0: # Evaluated to false the first time so it's skipped.
    # code

因此，在這種情況下，錯誤與input()無關

Answer 2

您的while情況是“ game_over”不等於0，而是直接位於其上方的行將其設置為0。

我認為您想要的是：

game_over = True
while not game_over:
   ...

Answer 3

您的問題是當game_over 不為 0時，您的while循環將運行，但是在前面的一行中，您將game_over設置為0。

game_over = 0
while game_over !=0:
    ...

因此，將game_over更改為1，您的代碼將起作用！

def stix(num):
    for _ in range(5): print('|  '* num)
    print
stix(20)
game_over = 1
while game_over !=0:
    players={}
    for i in range(2):
        players[i] = int(input('Player %d: Please pick stick(s) up to 3' %i))
        if players > 3 or players<0:
            print('Please pick between 1 - 3 stick(s)')
        else:
            stix-=players
            if stix <= 0:
                print('Player[i] lost')
                break
            else:
                print('There is %d stick(s) left' %stix)
                print(stix-players[i])