NameError：变量未定义[关闭]答案

【问题标题】：NameError: variable is not defined [closed]NameError：变量未定义[关闭]
【发布时间】：2021-06-05 11:29:30
【问题描述】：

需要帮助。我创建了一个空列表来附加名称输入。但是，我不断收到错误“NameError: name 'players' is not defined。如果我将 player = [] 定义为全局变量而不是主函数，它将起作用。仅供参考，我不允许使用全局变量.

def existingPlayers(name):
    '''
    Check if player name already exists
    '''
    for player in players:
        if player[0].lower() == name.lower():
            return True
    return False

def getPlayerNames():
    '''
    Get the names of the players. Check if the number of players and 
    player names are valid
    '''
    while True:
        name = input("Enter player name or <ENTER> to exit (min 2, max 4 players):")
        if len(name) == 0:
            if len(players) < 2:
                print(f"Minimum 2 players. Currently, {len(players)} player")
                continue
            else:
                break
        if existingPlayers(name):
            print("Repeated name. Choose another name")
            continue
        players.append([name])
        if len(players) == 4:
            break
def main():
    players = []
    getPlayerNames()
main()

【问题讨论】：

你的问题到底是什么？
嗯，这些函数仍然可以在一个类中，你可以使用类变量，或者在你的函数中添加另一个参数来接受来自 main 的播放器列表。
我的问题是如何附加到我的空列表中，主函数中的玩家 = []
你可以使用和定义一个 Players 类吗？这将是避免全局变量的好方法。
要么使用类对象，要么让玩家列表成为函数的参数。

标签： python nameerror

【解决方案1】：

这是你想要的吗？

def existingPlayers(name, players):
    '''
    Check if player name already exists
    '''
    for player in players:
        if player[0].lower() == name.lower():
            return True
    return False

def getPlayerNames(players):
    '''
    Get the names of the players. Check if the number of players and 
    player names are valid
    '''
    while True:
        name = input("Enter player name or <ENTER> to exit (min 2, max 4 players):")
        if len(name) == 0:
            if len(players) < 2:
                print(f"Minimum 2 players. Currently, {len(players)} player")
                continue
            else:
                break
        if existingPlayers(name, players):
            print("Repeated name. Choose another name")
            continue
        players.append([name])
        if len(players) == 4:
            break
def main():
    players = []
    getPlayerNames(players)
main()

【讨论】：

它可以工作，但现在代码接受重复的名称，这是不应该的
好的，修复了再测试一下