Question

print "Password?"

main = 1.0
tries = 1
Pass = raw_input()
if Pass == "hi": print "Granted access."
elif Pass == "swag": print "really?"
else:
        print "Wrong, try again."
        print "Wrong", tries, "try."
        tries = tries + 1

print "Password?"
Pass = raw_input()
if Pass == "hi": print "Granted access."
else:
        print "Wrong,", tries, """tries.
try again?""" #test # line 18
        tries = tries + 1
answer = raw_input()
if answer == "Yes": print "Password?"
 Pass = raw_input() # line 22
 if Pass == "hi": print "WOW SUCH PASS"
        else:
                print "Wrong,", tries, "tries...aborting"      
tries = tries + 1
else: # line 27
        print "Okay."

检查此代码。

我正在第18行尝试询问用户是否想再次尝试接听密码。如果是的话，请求密码并检查我的“通行证”。如果用户不想回答密码，则应将其重定向到代码行27。

但已经在第22行出现了问题。

我已经尝试了几种方法来格式化这段代码，而且我很新。有人想帮忙解决这个问题吗？

编辑：对于误导性的标题感到抱歉，我们不确定该怎么做。

Answer 1

你必须让你的缩进正确。第21行和第22行似乎是奇数，以及24上的挂起else。您可能希望暂时避免内联if cond: pass构造，直到您掌握Python的缩进。

我可能会也可能没有正确地对您的块进行分组，但是这里的代码应该看起来：

if answer == "Yes":
    print "Password?"
    Pass = raw_input()
    if Pass == "hi":
        print "WOW SUCH PASS"
    else:
        print "Wrong,", tries, "tries...aborting"      
    tries = tries + 1
else:
    print "Okay."

Answer 2

评论扩展：python要求您正确缩进代码。每个街区有4个空间。有关要求列表，请参阅http://www.python.org/dev/peps/pep-0008/。

print "Password?"                                                                                                                                                                                            

main = 1.0 
tries = 1 
Pass = raw_input()
if Pass == "hi":
    print "Granted access."
elif Pass == "swag":
    print "really?"
else:
    print "Wrong, try again."
    print "Wrong", tries, "try."
    tries = tries + 1 

print "Password?"
Pass = raw_input()
if Pass == "hi":
    print "Granted access."
else:
    print "Wrong,", tries, """tries.
try again?"""  # test # line 18
    tries = tries + 1 
answer = raw_input()
if answer == "Yes":
    print "Password?"
    Pass = raw_input()  # line 22
    if Pass == "hi":
        print "WOW SUCH PASS"
    else:
        print "Wrong,", tries, "tries...aborting"
        tries = tries + 1 
else:  # line 27
    print "Okay."

Answer 3

我认为这是代码的正确缩进版本。它在语法方面是正确的，但可能不是在行为方面 - 这是你要调查的。您应该阅读注释中建议的Python教程和http://www.python.org/dev/peps/pep-0008/。

print "Password?"

main = 1.0
tries = 1
Pass = raw_input()

if Pass == "hi":
    print "Granted access."
elif Pass == "swag":
    print "really?"
else:
    print "Wrong, try again."
    print "Wrong", tries, "try."
    tries = tries + 1

print "Password?"
Pass = raw_input()

if Pass == "hi":
    print "Granted access."
else:
    print "Wrong,", tries, "tries.try again?"
    tries = tries + 1

answer = raw_input()
if answer == "Yes":
    print "Password?"
    Pass = raw_input()

    if Pass == "hi":
        print "WOW SUCH PASS"
    else:
        print "Wrong,", tries, "tries...aborting"      
        tries = tries + 1
else:
        print "Okay."

Answer 4

自由改写：

users = {
    'hi': 'hi',
    'swag': None,
    'nsa': 'hail_the_fatherland'
}

def validate_user():
    for tries in range(3):
        user = raw_input('User: ')
        pwd  = raw_input('Password: ')
        if user in users and users[user] == pwd:
            print('Access Granted')
            return True
    print('Go away, you l33t h4ck3r')
    return False

def main():
    if validate_user():
        print('Direct me, master!')

if __name__=="__main__":
    main()

使用变量时的优先级错误

4 个答案: