好奇为什么这条线对于函数的其余部分正常工作如此重要

Question

最初，当我编写函数时，我没有在set_list函数中包含print(UserInput.users)，但是，我认为列表向用户显示他们能够创建哪些列表会很有帮助。删除它后，该功能无法正常工作：

class UserInput:
    users=[]
    def __init__(self, name,lista,listb,listc,listd):
        self.name=""
        self.lista=lista
        self.listb=listb
        self.listc=listc
        self.listd=listd
    @classmethod
    def create_new_users(cls):
        print("how many users do you want to create")
        x=int(input())
        for _ in range(x):       
            print("assign the users names")
            name = input()
            if name == '' or name.lower()  == 'none':
                raise RuntimeError("name cannot be None or empty")              
            user=cls(name,"","","","")      
            cls.users.append(name)
        return(name)
    @classmethod  
    def show_users(cls):
        print(UserInput.users)
    def set_lists():
        print("Do you want to create lists")
        decision = input()
        print( "select the user you intend on adding lists for")
        #why does the function require the code print(UserInput.users) on this line to   function properly?
        choice = input()
        for elem in UserInput.users:
            if choice == elem:
                if decision == "yes":   
                    print("how many lists would you like to create?(up to 4)")
                    decision2= int(input())
                    if decision2 == 1:
                        print("what would you like the list to be named?")
                        self.lista=input()
                        print("you have created 1 list, with the name:"+ self.lista)                                         
                    elif decision2 == 2:
                        print("what would you like your lists to be?")
                        self.lista,self.listb=input().split(",")
                        print("You have created 2 lists with the names," + self.lista ,self.listb)
                    elif decision2 == 3:
                        print("what name would you like your lists to be?")
                        self.lista,self.listb,self.listc = input().split(",")
                        print("you have created 3 lists with the names," +self.lista,self.listb,self.listc)
                    elif decision2 == 4:
                        print("what name would you lists to be?")
                        self.lista,self.listb,self.listc,self.listd = input().split(",")
                        print("you have created 4 lists with the names of," +self.lista,self.listb,self.listc,self.listd)
                    else:
                        print("quitting")
                        return

            else:
                print("not in users list")
                break

我的问题：当我从中选择一个名称时，print(UserInput.users)包含在函数中时，它将识别列表，但如果未包含，它会跳到底部的其他名称：为什么？

Answer 1

搜索代码会发现至少一个混合制表符和空格的实例。这可能是问题所在，但我对此表示怀疑。相反，我怀疑这是一个逻辑错误，与您删除的行无关。

    for elem in UserInput.users:
        if choice == elem:
            stuuuuuuuuuff...
        else:
            print("not in users list")
            break

如果choice不是UserInput.users的第一个元素，这将在第一次迭代时命中else子句并中断。我怀疑您只想在choice根本不在列表中时才这样做。与其循环，我推荐使用 in 运算符：

if choice in UserInput.users:
    stuuuuuuuuuuff...
else:
    print("Not in user list.")