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Python 如何在用户输入后验证列表中的正确猜测?_Python_Python 3.x - Fatal编程技术网

Python 如何在用户输入后验证列表中的正确猜测?

Python 如何在用户输入后验证列表中的正确猜测?,python,python-3.x,Python,Python 3.x,我不知道用户输入将如何满足编译器 我试过:“gussing!=猜测列表[1]或…….gussing!=猜测列表[5]”但失败了 guessing_list = ["New York", "Mumbai", "Brooklyn", "Tokyo", "Peris"] gussing = "" gussing_counting = 0 gussing_limits = 4 gussing_range = False while gussing != guessing_list[1:5] and n

我不知道用户输入将如何满足编译器

我试过:“gussing!=猜测列表[1]或…….gussing!=猜测列表[5]”但失败了

guessing_list = ["New York", "Mumbai", "Brooklyn", "Tokyo", "Peris"]
gussing = ""
gussing_counting = 0
gussing_limits = 4
gussing_range = False

while gussing != guessing_list[1:5] and not(gussing_range):
    if gussing_counting < gussing_limits:
        gussing = str(input("Enter Your dream city: "))
        gussing_counting += 1
    else:
        gussing_range = True
if gussing_range:
    print("You're out or range!")
else:
    print("You got your dream city!")'
猜测清单=[“纽约”、“孟买”、“布鲁克林”、“东京”、“佩里斯”] gussing=“” gussing_计数=0 gussing_极限=4 gussing_范围=假 边唱边唱!=猜测列表[1:5]与否(gussing_范围): 如果浇注计数<浇注极限: gussing=str(输入(“进入你的梦想城市:”) gussing_计数+=1 其他: gussing_范围=真 如果是gussing_范围: 打印(“您超出或超出范围!”) 其他: 打印(“你有你的梦想之城!”)
我希望如果用户在给定范围内输入“Brooklyn”,那么输出必须显示“You got your dream city!”

要检查序列中是否存在值(如列表或字符串),应使用
in
关键字。我想你要找的是:

while gussing not in guessing_list and not gussing_range :
请注意,您的
while
循环可以简化为:

while gussing not in guessing_list and gussing_counting < gussing_limits:
    gussing = str(input("Enter Your dream city: "))
    gussing_counting += 1
当不在猜测列表和计数中时
要检查序列中是否存在值(如列表或字符串),应使用
in
关键字。我想你要找的是:

while gussing not in guessing_list and not gussing_range :
请注意,您的
while
循环可以简化为:

while gussing not in guessing_list and gussing_counting < gussing_limits:
    gussing = str(input("Enter Your dream city: "))
    gussing_counting += 1
当不在猜测列表和计数中时