Question

我写了一个菜单功能，该功能一直循环到用户按下3退出。但是，当用户通过按1或2询问其他选项时，我希望它打印一些内容。这是我的代码：

def Menu():
    return("\n"
          + "Menu:\n-------------"
          + "\n1 - Does This"
          + "\n2 - Does That"
          + "\n3 - Quit"
          + "\n")
    if option == 1:
        return "\nThis"
    if option == 2:
        return "\nThat"

option = None
while True:
    print(Menu())
    option = int(input("Please choose an option: "))
    if option == 3:
        print("\nBye!")
        break
    if option < 1 or option > 3:
        print("\nIncorrect input!")

代替按1时获得“ This”或按2时获得“ That”，它只是再次循环菜单，直到我按3退出为止。

这是我当前的输出：

Menu:
-------------
1 - Does This
2 - Does That
3 - Quit

Please choose an option: 1

Menu:
-------------
1 - Does This
2 - Does That
3 - Quit

Please choose an option: 2

Menu:
-------------
1 - Does This
2 - Does That
3 - Quit

Please choose an option: 3

Bye!

我想要的结果只是让它打印“ This”或“ That”，然后再次循环菜单直到退出。

Answer 1

目前，当您选择1或2时，while循环将再次执行。您需要在用户输入有效值的地方实现代码。

只需澄清一下，因为您将返回Menu()函数，所以这些行永远不会被击中，

if option == 1:
    return "\nThis"
if option == 2:
    return "\nThat"

编辑

这是经过改进且可以使用的版本，

input_string = """Menu:-------------
1 - Does This"
2 - Does That"
3 - Quit
Please choose an option:"""

def Menu(option):
    if option == "1":
        return "\nThis"
    if option == "2":
        return "\nThat"

while True:
    option = input(input_string)
    if option in {"1","2"}:
        print(Menu(option))

    elif option == "3":
        print("\nBye!")
        break
    else:
        print("\nIncorrect input!")

Answer 2

您忘了实际编码呼叫以打印文本。试一试：

def Menu():
return("\n"
      + "Menu:\n-------------"
      + "\n1 - Does This"
      + "\n2 - Does That"
      + "\n3 - Quit"
      + "\n")


option = None
while True:
    print(Menu())
    option = int(input("Please choose an option: "))
    if option == 3:
        print("\nBye!")
        break
    elif option == 1:
        print ("\nThis")
    elif option == 2:
        print ("\nThat")
    elif option < 1 or option > 3:
        print("\nIncorrect input!")

Answer 3

您需要一些搬迁

deleteLocationItem(id) {
  this.setState({
   items: this.state.items.filter(item => item.id !== id)
  });
}

函数不返回任何值-Python

3 个答案: