Python Int对象不可调用

请帮忙！我不明白这里的错误。为什么在键入0、1或2以外的数字时会出现错误：“int”对象不可调用？相反，它应该打印“您输入的号码不正确，请再试一次”，然后返回询问问题

第二个问题：我如何更改代码，即使我键入字母字符，它也不会给我值错误并继续重新提问？谢谢大家!

def player_action():        
    player_action = int(input("Enter 0 to stay, 1 to go Up, or 2 to go Down: "))

    if player_action == 0:
        print ("Thank You, you chose to stay")


    if player_action == 1:
        print ("Thank You, you chose to go up")

    if player_action == 2:
        print ("Thank You, you chose to go down")

    else:
        print ("You have entered an incorrect number, please try again")
        player_action()

player_action()

---

只需将变量名

player\u action

更改为函数的不同名称，即：

def player_action():
    user_input = int(input("Enter 0 to stay, 1 to go Up, or 2 to go Down: "))
    if user_input == 0:
        print ("Thank You, you chose to stay")
    elif user_input == 1:
        print ("Thank You, you chose to go up")
    elif user_input == 2:
        print ("Thank You, you chose to go down")
    else:
        print ("You have entered an incorrect number, please try again")
        player_action()

player_action()

---

Pedro已经回答了您问题的第一个答案，但对于第二个答案，try-Exception语句应该可以解决这个问题：

编辑：是的，对不起，我把事情搞砸了。。。有更好的答案，但我认为我应该花时间来解决这个问题

def player_action():
    try:
        player_action_input = int(input("Enter 0 to stay, 1 to go Up, or 2 to go Down: "))
    except ValueError:
        print("Non valid value") # or somehting akin
        player_action()
    if player_action_input == 0:
        print ("Thank You, you chose to stay")
    elif player_action_input == 1:
        print ("Thank You, you chose to go up")
    elif player_action_input == 2:
        print ("Thank You, you chose to go down")
    else:
        print ("You have entered an incorrect number, please try again")
            player_action()

player_action()

---

您应该将变量名更改为@Pedro Lobito suggest，使用@Craig建议的

while

循环，还可以包括

try…除了
语句，但不使用@polarisfox64的方式，因为他将其放置在了错误的位置

以下是完整版本供您参考：

def player_action():    
    while True:   
        try:
            user_input = int(input("Enter 0 to stay, 1 to go Up, or 2 to go Down: "))
        except ValueError:
            print('not a number')
            continue

        if user_input == 0:
            print ("Thank You, you chose to stay")          

        if user_input == 1:
            print ("Thank You, you chose to go up")

        if user_input == 2:
            print ("Thank You, you chose to go down")

        else:
            print ("You have entered an incorrect number, please try again")
            continue
        break

player_action()

您的变量名会影响函数名。您试图调用函数
player\u action（）
，但实际上正在调用变量
player\u action
，该变量是一个int。函数和变量的名称不要相同！而且，没有理由让这个函数递归。只需在while
循环中输入一个
，直到您得到一个有效的输入，然后对该输入进行处理。哦，好的，谢谢！对不起，你能告诉我如何用while函数重写这个吗？