Question

所以我是新手，我一直在努力遵循DRY原则。

所以我有一个班级

class Player(object):
    lifecondition = True
    hitpoints = 100
    def __init__(self, name):
        self.name = name
    def death(self):
        self.lifecondition = False

列表

playerlist = ["Player1", "Player2", "Player3", "Player4", "Player5", 
"Player6", "Player7", "Player8", "Player9", "Player10"]

和一个功能

def addplayers(alist):
    for item in alist:
        item = Player("%s" % item)
        print "%s has %s HP of life and feels pretty much alive" %  (item.name, item.hitpoints, )

我通过函数

向对象添加对象

addplayers(playerlist)

到目前为止一切都很好，但如果我做了 print Player1.name 我收到Player1没有定义。我错过了什么？

Answer 1

你可能应该这样做：

def addplayers(alist):
    players = []
    for item in alist:
        # NOTE reuse the name item is not a good idea here. You can use player instead.
        item = Player("%s" % item)
        players.append(item)
        print "%s has %s HP of life and feels pretty much alive" %  (item.name, item.hitpoints, )
    return players

并用

调用它

players = addplayers(playerlist)
print(players[0])

最后一点不是关于你的问题，但最好将你的班级定义为

class Player(object):
    def __init__(self, name):
        self.name = name
        self.lifecondition = True
        self.hitpoints = 100
    def death(self):
        self.lifecondition = False

因为您可能希望每个玩家都有不同的hitpoints。请参阅有关类变量和实例变量here的Python文档。

函数通过列表将对象添加到类中 - 调用对象

1 个答案: