如果輸入錯誤而未在列表中獲得錯誤值，則再次要求用戶輸入

Question

為什么elif b:= 0 and b!= 1: print (b)不起作用？ 用戶必須輸入 1 或 0 才能確定 b。 如果用戶不寫 1 或 0，我希望程序再次詢問他“玩家是計算機 (0) 還是人類 (1)？” 沒有 list_2 附加初始錯誤數字。

list_2 = []
players = int(input("number of players: ")) 
   
for i in range (players):
    a = input("name of player: ")
    b = int(input ("is player a Computer (0) or a human (1)?"))

    if b == 0:
        list_2.append([a] + [True])
      
    elif b == 1:
        list_2.append([a] + [False])

    elif b!= 0 and b!= 1:
        print (b)
         
print (list_2)

Answer 1

將該部分輸入包裝在一個while True循環中。 當輸入正確時打破循環，否則繼續循環以獲得更多輸入。

while True:
    b = int(input ("is player a Computer (0) or a human (1)?"))
    if b == 0:
        # player is a computer ...
        # do computer stuff
        break
    elif b == 1:
        # player is a human ...
        # do human stuff
        break
    else:
        print("That was not 1 or 0.  Please try again")

Answer 2

因為 elif 不是循環，編譯器只是檢查條件，然后繼續執行語句。

您可以通過在第一個 if 條件之前添加一個 ***while 循環 *** 來解決這個問題，如下所示：

for i in range (players): a = input("玩家名稱：") b = int(input ("玩家是計算機 (0) 還是人類 (1)？"))

while (b!=0 and b!=1): # Checking if b is not equal to both '0' and '1'
    b = int(input ("is player a Computer (0) or a human (1)?"))
    
if b == 0:
    list_2.append([a] + [True])
  
elif b == 1:
    list_2.append([a] + [False])

elif b!= 0 and b!= 1:
    print (b)
     

Answer 3

您還可以嘗試拆分代碼並創建一個新的checkInput function，如下所示。

 list_2 = [] players = int(input("number of players: ")) def checkInput(value): if value == 0: list_2.append([a] + [True]) return True elif value == 1: list_2.append([a] + [False]) return True else: return False for i in range (players): check = False; a = input("name of player: ") while check == False: b = int(input ("is player a Computer (0) or a human (1)?")) check = checkInput(b) print('Please enter correct input') print (list_2)