Question

我真的很擅长开发，我正在从Codecademy重新编写Battleship游戏，成为2名玩家。我真的很难通过这个，但到目前为止，我觉得这是一个很好的练习。我想保持良好的OOP和DRY原则。不过我有一个问题。我试图传递用户输入来创建对象，我不得不创建两个类似的定义来实现我的目标。我只写了部分程序，并且我一直在测试。要查看我的问题，请查看下面的get_name1和get_name2：

 from random import randint


 class Person(object):

     def __init__(self, name, turn):
         self.name = name
         self.turn = turn


     def get_name1():
         while 1:
             name = input("What is the name of Player 1? ")
             if name.isalpha() == False:
                 print("\nPlease share your name with me.\n")
             else:
                 print("\nNice to meet you %s. It will be fun to play Battleship!\n" % name)
                 return name
                 break

     def get_name2():
         while 1:
             name = input("What is the name of Player 2? ")
             if name.isalpha() == False:
                 print("\nPlease share your name with me.\n")
             else:
                 print("\nNice to meet you %s. It will be fun to play Battleship!\n" % name)
                 return name
                 break



 Player1 = Person(Person.get_name1(), 1)
 Player2 = Person(Person.get_name2(), 2)

 print("Welcome to Battleship, %s!" % Player1.name)
 print("You will take turn %s.\n" % Player1.turn)
 print("Welcome to Battleship, %s!" % Player2.name)
 print("You will take turn %s.\n" % Player2.turn)

有没有办法将get_name1和get_name2合并到一个函数中，同时保持＆＃34的唯一输入行;播放器1的名称是什么？＆＃34;和＃34;播放器2的名称是什么？＆＃34;并仍然将唯一输入传递给两个不同的类对象？

Answer 1

最好，使用classmethod创建一个名字的人：

class Person(object):
    def __init__(self, name, turn):
        self.name = name
        self.turn = turn

    @classmethod
    def create(cls, turn):
        while True:
            name = input("What is the name of Player %d? " % turn)
            if name.isalpha():
                break;
            print("\nPlease share your name with me.\n")
        print("\nNice to meet you %s. It will be fun to play Battleship!\n" % name)
        return cls(name, turn)

Player1 = Person.create(1)
Player2 = Person.create(2)

print("Welcome to Battleship, %s!" % Player1.name)
print("You will take turn %s.\n" % Player1.turn)

Answer 2

def get_name(name_str):
         while 1:
             name = input("What is the name of %s? " % name_str)
             if name.isalpha() == False:
                 print("\nPlease share your name with me.\n")
             else:
                 print("\nNice to meet you %s. It will be fun to play Battleship!\n" % name)
                 return name
                 break

Player1 = Person(Person.get_name('Player 1'), 1)
Player2 = Person(Person.get_name('Player 2'), 2)

Answer 3

所以两个功能之间的唯一区别是球员号码？然后将播放器编号作为参数传递给函数，并使用字符串格式化提示。

def get_name(player_number):
    prompt = 'What is the name of Player {}? '.format(player_number)
    while True:
        name = input(prompt)
        # rest of code goes here..

通常，当您发现两个函数几乎完全相同时，您会查找不同的函数，将其作为变量，并将变量作为参数传递。那么你只有一个函数可以被参数修改。

用Python编写DRY代码

3 个答案: