Question

我正在尝试制作一个简单的python脚本来诱骗我的朋友认为得到病毒除非他写密码p3nis47但是每当我尝试运行它时我会在第1,4和9行得到错误.17和4只是方法，我不知道为什么我会收到错误，在9我只是减去1到计数，以确保我只给我的朋友3次尝试输入“密码”。对不起，如果它真的很明显，我刚开始学习python。

count = 4
def ask():
    answer = input("do you wan't a virus ")
    respond(answer)
def respond(response):
    if(response == "p3nis47"):
        print("congrats!!! you don't have a virus")
    else:
        count = count - 1
        if(count == 0):
            print("trololololololololololol")
        print(answer,"is not a vailid answer")
        print("you have ",count," attempts remaining")
        ask()
ask()

Answer 1

使用输入错误运行程序会出错

$scope.Ctrl.type = "" + $scope.Ctrl.type + "";

那是因为你正在使用全局变量，但是你需要让python知道它。将Traceback (most recent call last): File "c.py", line 15, in <module> ask() File "c.py", line 4, in ask respond(answer) File "c.py", line 9, in respond count = count - 1 UnboundLocalError: local variable 'count' referenced before assignment添加到函数的开头。然后还有另一个错误

global count

这也很简单。你只是意外地使用了错误的变量名。具有两个更改的工作脚本是

Traceback (most recent call last):
  File "c.py", line 16, in <module>
    ask()
  File "c.py", line 4, in ask
    respond(answer)
  File "c.py", line 13, in respond
    print(answer,"is not a vailid answer")
NameError: name 'answer' is not defined

所以，我运行了修复程序......但现在我已经感染了病毒！

每当我尝试使用方法时出错

1 个答案: