Question

我是python中的新手，我应该创建一个输入，其输入只能在1和3的范围内。（播放器1,2,3）如果用户输入超过3或输出应该是错误的错误，如果它在字符串中。

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。如果输入为1，它仍然显示“错误，请再试一次”我很困惑！

Answer 1

raw_input返回一个字符串。因此，您正在尝试"1" > 4。您需要使用int

将其转换为整数

如果要捕获输入是否为数字，请执行：

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)

Answer 2

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会以unicode对象的形式返回。

Answer 3

您必须使用方法int()将值转换为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 +"?")

在python中只允许最大整数输入且没有字母表

3 个答案: