Python打印这两个“if”场景，尽管1显然不适用

Question

这是我的密码：

varsec =raw_input("enter secret number between 1 and 100(the system won't know):")

if int(varsec.strip()) >=100: print("Cheater caught! >:D")

if int(varsec.strip()) <=1: print("Cheater caught! >:D")

import random

varGuess = random.randint(2,99)

if 2 <= int(varsec.strip()) <= 99 : print(varGuess,
                                          ("If this number is less than your secret number, type <, if more then type >, if it IS your secret number then type =."))


VarConfirm = raw_input(":")


if (VarConfirm.strip()) is "<": from random import randint
print(randint(varGuess + 1,99))

if (VarConfirm.strip()) is ">": from random import randint
print(randint(2,varGuess - 1))

if (VarConfirm.strip()) is "=": print("YES, the program has successfully guessed your number ^^")    

如果我键入“<；”对于VarConfirm，它打印两个数字（1个介于varGuess +1和99之间，另一个介于2和varGuess -1之间）

我希望它只打印介于varGuess + 1和99之间的数字

另外，当我键入"<"或"="时，会出现NameError: 'randint' is not defined.

Answer 1

我不知道复制这里的代码是否会弄乱缩进，但是缩进表明if语句中“：”后面的部分只算作条件部分，然后print语句只是普通的python行，完全独立于if语句

另外，不需要在if语句中间导入这样的内容。只需在程序的开头做import语句，然后再使用它。所以你的代码应该是

from random import randint

# [.... previous steps ....]

if (VarConfirm.strip()) == "<":
    print randint(varGuess + 1,99)

if (VarConfirm.strip()) == ">":
    print randint(2,varGuess-1)

if (VarConfirm.strip()) == "=":
    print "Yes, bla bla"