我是python的新手，并且可以进行网络安全。当然，模块之一是Python。
我正在尝试尝试编写没有教程的数字猜谜游戏。有人可以看一下这段代码，然后告诉我为什么代码似乎没有遵循if语句。
谢谢。
编辑：错误：我正在尝试创建代码，通过检查播放器适合的范围来判断玩家选择了哪个“难度”。问题是，它正在打印每个“困难”选项，而不仅仅是它所适合的选项。
编辑2：将标题更改为更具体，以便对他人有所帮助。
print("Welcome to the number guesser game!")
tutorial = input("Would you like a tutorial? (Y/N): ")
while tutorial != "Y" or "y" or "N" or "n":
input("Input not recognised, would you like a tutorial? Y (for yes) or N (for no)")
if tutorial == "N" or "n":
print("Let's get right into it!")
if tutorial == "Y" or "y":
print("In this game you will pick a number, the number you pick will be added to a random number between 1 and 20 /nYou will then attempt to guess the number generated, with a hint given every guess")
break
difficulty = input("Please choose a random number \nGuide:\nEasy 1-10\nMedium 1-100\nHard 1-1,000\nExtreme Numbers above 1,000-10,000")
if difficulty >="1" <="10":
print("You have chosen easy! You must be a noob!")
if difficulty >"10" <="100":
print("You have chosen medium! Good choice!")
if difficulty >"100" <="1000":
print("You have chosen hard! Pfft, good luck")
if difficulty >"1000" <="10000":
print("You have chosen extreme! You must be nuts")
if difficulty <="0" >"10000":
difficulty = input("Nice try, pick again: ")
else:
difficulty = input("Input not recognised, please input a number greater than 0: ")```
那是因为首先，如果输入不是y，那么它将说无法识别。这是因为，如果它是y，它将检查no语句，然后它会中断。这意味着，它将同时说两个。
然后，这变得很奇怪，因为您正在使用str和str做> =。它会做出奇怪的反应。因此，您应该将int与int进行比较。
首先，在第一个语句中使用和。然后，使用elif。并在if语句中也使用break。并做int(input())转换为int。然后，您可以再次使用elif。并删除引号。
尝试这个：
print("Welcome to the number guesser game!")
tutorial = input("Would you like a tutorial? (Y/N): ")
while True:
if tutorial != "Y" and tutorial !="y" and tutorial !="N" and tutorial !="n":
tutorial = input("Input not recognised, would you like a tutorial? Y (for yes) or N (for no)")
if tutorial == "N" or tutorial =="n":
print("Let's get right into it!")
break
elif tutorial == "Y" or tutorial == "y":
print("In this game you will pick a number, the number you pick will be added to a random number between 1 and 20 /nYou will then attempt to guess the number generated, with a hint given every guess")
break
difficulty = int(input("Please choose a random number \nGuide:\nEasy 1-10\nMedium 1-100\nHard 1-1,000\nExtreme Numbers above 1,000-10,000"))
if difficulty >=1 and difficulty <=10:
print("You have chosen easy! You must be a noob!")
elif difficulty >10 and difficulty<=100:
print("You have chosen medium! Good choice!")
elif difficulty >100 and difficulty<=1000:
print("You have chosen hard! Pfft, good luck")
elif difficulty >1000 and difficulty<=10000:
print("You have chosen extreme! You must be nuts")
elif difficulty <=0 and difficulty >10000:
difficulty = input("Nice try, pick again: ")
else:
difficulty = input("Input not recognised, please input a number greater than 0: ")
对于while循环，将其打开，while True:然后使用if语句进行检查。然后您可以检查y或n。
本文收集自互联网，转载请注明来源。
如有侵权，请联系[email protected] 删除。
我来说两句