Python：无法在while循环中调用函数

Question

刚从这里开始soz。我一直试图通过输入一个数字来创建一个带有几个选项的菜单（def logged() :)，并且它会跳转到该功能的目的。但是，我似乎无法使用while语句放入if语句来调用指定函数，而是在日志函数应该永久保持while循环运行时跳转到menu()函数。

当我在logged()的菜单中输入相应的数字时，它应该调用该特定功能，但它只是跳回第一个菜单。我似乎无法让这两个菜单永远循环，而不会让它们来回跳动。那么，我究竟如何让两个while循环分别永远循环而不是彼此？

def menu(): 
    mode = input("""Choose options:\n 
    a) Test1 Calls logged() function 
    b) Test2 
    Enter the letter to select mode\n 
    > """) 
    return mode 

def test1(): 
    print("Test1") 
    logged() 

def test2(): 
    print("Test2") 

def logged(): #Logged menu is supposed to run through a while loop and not break out when reached. 
    print("----------------------------------------------------------------------\n") 
    print("Welcome user. ") 
    modea = input("""Below are the options you can choose:\n 
    1) Function1 
    2) Function2 
    3) Function3 
    4) Exit 
    \n 
    Enter the corresponding number 
    > """).strip() 
    return modea 

def funct1(): #EXAMPLE FUNCTIONS 
    print("Welcome to funct1") 


def funct2(): 
    print("Welcome to funct2") 


def funct3(): 
    print("Welcome to funct3") 

#Main routine 
validintro = True 
while validintro: 
    name = input("Hello user, what is your name?: ") 
    if len(name) < 1: 
     print("Please enter a name: ") 
    elif len(name) > 30: 
     print("Please enter a name no more than 30 characters: ") 
    else: 
     validintro = False 
     print("Welcome to the test program {}.".format(name)) 

#The main routine 
while True: 
    chosen_option = menu() #a custom variable is created that puts the menu function into the while loop 

    if chosen_option in ["a", "A"]: 
     test1() 

    if chosen_option in ["b", "B"]: 
     test2() 

    else: 
     print("""That was not a valid option, please try again:\n """)  

while True: 
    option = logged() 
    if option == "1": 
     funct1() 

    elif option == "2": 
     funct2() 

    elif option == "3": 
     funct3() 

    elif option == "4": 
     break 
    else: 
     print("That was not a valid option, please try again: ") 

print("Goodbye") 

Answer 1

好的，所以你犯了一些错误（很明显），没什么大不了的，每个人都得开始学习。

最大的问题是你进入菜单循环（你有第二个while循环），但从不做任何事情来退出它。我还评论了其他一些变化。我不是100％确定你在某些地方做什么...但是...

我认为this is what you were going for though，我评论了这些变化。有一些奇怪的东西，我只是留下，因为我认为这是意图。

def menu(): 
    mode = input("""Choose options:\n 
    a) Test1 Calls logged() function 
    b) Test2 
    Enter the letter to select mode\n 
    > """) 
    return mode 

def test1(): 
    print("Test1") 
    logged() 

def test2(): 
    print("Test2") 

def logged(): #Logged menu is supposed to run through a while loop and not break out when reached. 
    print("----------------------------------------------------------------------\n") 
    print("Welcome user. ") 
    modea = input("""Below are the options you can choose:\n 
    1) Function1 
    2) Function2 
    3) Function3 
    4) Exit 
    \n 
    Enter the corresponding number 
    > """).strip() 
    return modea 

def funct1(): #EXAMPLE FUNCTIONS 
    print("Welcome to funct1") 


def funct2(): 
    print("Welcome to funct2") 


def funct3(): 
    print("Welcome to funct3") 

#Main routine 
validintro = False # I like it this way 
while not validintro: 
    name = input("Hello user, what is your name?: ") 
    if len(name) < 1: 
     print("Please enter a name: ") 
    elif len(name) > 30: 
     print("Please enter a name no more than 30 characters: ") 
    else: 
     validintro = True 
     print("Welcome to the test program {}.".format(name)) 

#The main routine 
validintro = False # need a way out 
while not validintro: 
    chosen_option = menu() #a custom variable is created that puts the menu function into the while loop 
    validintro = True # start thinking we're okay 
    if chosen_option in ["a", "A"]: 
     test1() # you're calling this, which calls the logged thing, but you do nothing with it 
     # I just left it because I figured that's what you wanted 

    elif chosen_option in ["b", "B"]: # You want an elif here 
     test2() 

    else: 
     print("""That was not a valid option, please try again:\n """) 
     validintro = False # proven otherwise 

validintro = False 
while not validintro: 
    validintro = True 
    option = logged() 
    print(option) 
    if option == "1": 
     funct1() 

    elif option == "2": 
     funct2() 

    elif option == "3": 
     funct3() 

    elif option == "4": 
     break 
    else: 
     print("That was not a valid option, please try again: ") 
     validintro = False 

print("Goodbye") 

Answer 2

的问题是，你的代码犯规跟着你想要的流动路径，尝试上面的代码，看看是否是你想要的，我会考虑的一点点，试图解释我做了什么（现在我只是创建了一个函数whileloop（）并将其添加到正确的位置）。

def whileloop(): 
    while True: 
    option = logged() 
    if option == "1": 
     funct1() 

    elif option == "2": 
     funct2() 

    elif option == "3": 
     funct3() 

    elif option == "4": 
     break 
    else: 
     print("That was not a valid option, please try again: ") 

print("Goodbye") 
def menu(): 
    mode = input("""Choose options:\n 
    a) Test1 Calls logged() function 
    b) Test2 
    Enter the letter to select mode\n 
    > """) 
    return mode 

def test1(): 
    print("Test1") 
    whileloop() 

def test2(): 
    print("Test2") 
    whileloop() 

def logged(): #Logged menu is supposed to run through a while loop and not break out when reached. 
    print("----------------------------------------------------------------------\n") 
    print("Welcome user. ") 
    modea = input("""Below are the options you can choose:\n 
    1) Function1 
    2) Function2 
    3) Function3 
    4) Exit 
    \n 
    Enter the corresponding number 
    > """).strip() 
    return modea 

def funct1(): #EXAMPLE FUNCTIONS 
    print("Welcome to funct1") 


def funct2(): 
    print("Welcome to funct2") 


def funct3(): 
    print("Welcome to funct3") 

#Main routine 
validintro = True 
while validintro: 
    name = input("Hello user, what is your name?: ") 
    if len(name) < 1: 
     print("Please enter a name: ") 
    elif len(name) > 30: 
     print("Please enter a name no more than 30 characters: ") 
    else: 
     validintro = False 
     print("Welcome to the test program {}.".format(name)) 

#The main routine 
while True: 
    chosen_option = menu() #a custom variable is created that puts the menu function into the while loop 

    if chosen_option in ["a", "A"]: 
     test1() 

    if chosen_option in ["b", "B"]: 
     test2() 

    else: 
     print("""That was not a valid option, please try again:\n """) 

我觉得发生了什么事。我列出你的代码正在经历的流程，你可以用简单的方法理解它。

1. 进入循环while validintro;
2. 进入while True环（chosen_option = menu()）
3. 进入menu()，并呼吁test1()
4. 进入test1()，并呼吁logged()
5. 进入logged()和现在的事情。当您在While True循环中调用test1()函数时，您的执行流程将返回到。
6. 您输入if chosen_option in ['b', 'B']。
7. 因为它不在b，B你会激活你的else声明并打印你的错误信息。之后，循环重新开始。