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我是 python 新手，我应该创建一个游戏，其中输入只能在 1 和 3 的范围内。（玩家 1、2、3）如果用户输入超过 3，则输出应该是错误的，如果它是错误的在字符串中。

def makeTurn(player0):

    ChoosePlayer= (raw_input ("Who do you want to ask? (1-3)"))

    if ChoosePlayer > 4:
        print "Sorry! Error! Please Try Again!"
        ChoosePlayer= (raw_input("Who do you want to ask? (1-3)"))

    if ChoosePlayer.isdigit()== False:
        print "Sorry! Integers Only"
        ChoosePlayer = (raw_input("Who do you want to ask? (1-3)"))
    else:
        print "player 0 has chosen player " + ChoosePlayer + "!"
        ChooseCard= raw_input("What rank are you seeking from player " + ChoosePlayer +"?")

我是这样做的，但问题是我的代码似乎有问题。如果输入是 1，它仍然说“错误请重试”我很困惑！

4

3 回答 3

1

raw_input返回一个字符串。因此，您正在尝试做"1" > 4. 您需要使用将其转换为整数int

如果要捕获输入是否为数字，请执行以下操作：

while True:
    try:
        ChoosePlayer = int(raw_input(...))
        break
    except ValueError:
        print ("Numbers only please!")

请注意，现在它是一个整数，下面的连接将失败。在这里，您应该使用.format()

 print "player 0 has chosen player {}!".format(ChoosePlayer)
于 2013-07-27T22:43:11.317 回答
1

您可能需要将 ChoosePlayer 转换为 int，例如：

ChoosePlayerInt = int(ChoosePlayer)

否则，至少在 pypy 1.9 中，ChoosePlayer 作为 unicode 对象返回。

于 2013-07-27T22:45:27.630 回答
-1

您必须使用方法将您的值转换为 int int()：

 def makeTurn(player0):
    ChoosePlayer= (raw_input ("Who do you want to ask? (1-3)"))

    if int(ChoosePlayer) not in  [1,2,3]:
        print "Sorry! Error! Please Try Again!"
        ChoosePlayer= (raw_input("Who do you want to ask? (1-3)"))

    if ChoosePlayer.isdigit()== False:
        print "Sorry! Integers Only"
        ChoosePlayer = (raw_input("Who do you want to ask? (1-3)"))
    else:
        print "player 0 has chosen player " + ChoosePlayer + "!"
        ChooseCard= raw_input("What rank are you seeking from player " + ChoosePlayer +"?")
于 2013-07-27T22:44:08.337 回答