Question

我之前已经问过这个问题但是被告知我包含了太多不必要的代码，所以我现在要用更少的代码来询问，希望我所包含的更多是正常的。

我想让一名成员离开团队，如果他们愿意的话。通过这样做，系统将从系统中删除所有细节。我的代码收到错误。有人可以告诉我我做错了什么以及如何实现这一目标？

我想根据用户输入和成员的需要，添加成员并删除成员以便一直更新。我希望这是有道理的！

以下是我的代码：

all_users = []

class Team(object):
    members = []  # create an empty list to store data
    user_id = 1

    def __init__(self, first, last, address):
        self.user_id = User.user_id
        self.first = first
        self.last = last
        self.address = address
        self.email = first + '.' + last + '@python.com'
        Team.user_id += 1

    @staticmethod
    def remove_member():
        print()
        print("We're sorry to see you go , please fill out the following information to continue")
        print()
        first_name = input("What's your first name?\n")
        second_name = input("What's your surname?\n")
        address = input("Where do you live?\n")
        unique_id = input("Finally, what is your User ID?\n")
        unique_id = int(unique_id)
        for i in enumerate(all_users):
            if Team.user_id == unique_id:
                all_users.remove[i]

def main():
    user_1 = Team('chris', 'eel', 'london')
    user_2 = Team('carl', 'jack', 'chelsea')

    continue_program = True
    while continue_program:
        print("1. View all members")
        print("2. Want to join the team?")
        print("3. Need to leave the team?")
        print("4. Quit")
        try:
            choice = int(input("Please pick one of the above options "))

            if choice == 1:
                Team.all_members()
            elif choice == 2:
                Team.add_member()
            elif choice == 3:
                Team.remove_member()
            elif choice == 4:
                continue_program = False
                print()
                print("Come back soon! ")
                print()
            else:
                print("Invalid choice, please enter a number between 1-3")
                main()
        except ValueError:
           print()
           print("Please try again, enter a number between 1 - 3")
           print()


if __name__ == "__main__":
    main()

Answer 1

让我们关注删除代码：

for i in enumerate(all_users):
    if Team.user_id == unique_id:
        all_users.remove[i]

enumerate返回两个值：索引和索引处的对象。在您的情况下，i是这两个值中的tuple。 .remove是一个函数，而不是集合，因此.remove[i]将失败。即使这样，它也是错误的工具：它会扫描列表i并删除它。你只想删除。最后，一旦更改了列表，就需要停止枚举。

所以，要清理它：

for i, user in enumerate(all_users):
    if user.user_id == unique_id:
        del all_users[i]
        break

如何从python类

1 个答案: