所以我对编码和Python非常基础。我曾经好多了,但是有一段时间没用过它,我正试图重新回到过去。我正在尝试执行一个If语句,其中输入被要求输入用户名和密码,如果正确则转到“定义”我认为它被调用,然后它执行 def access()<中的代码/强>
代码:
def main():
userName = ("u123")
userPass = ("p123")
userNameInput = input("Username: ")
userPassInput = input("Password: ")
if userPassInput == userPass and userNameInput == userName:
print("Access granted")
access()
else:
print("Access denied")
return main()
def access():
print("Welcome, " + userName)
access()
main()
但是,执行正确的输入时会出现此错误:
Username: u123
Password: p123
Access granted
Traceback (most recent call last):
File "C:/Users/Tom/Desktop/test.py", line 23, in <module>
main()
File "C:/Users/Tom/Desktop/test.py", line 11, in main
access()
UnboundLocalError: local variable 'access' referenced before assignment
>>>
任何帮助将不胜感激,谢谢。
答案 0 :(得分:1)
您的范围有误,您还需要在函数访问中添加一个参数:
def main():
userName = ("u123")
userPass = ("p123")
userNameInput = input("Username: ")
userPassInput = input("Password: ")
if userPassInput == userPass and userNameInput == userName:
print("Access granted")
#now you are able to use the function "access"
access(userName)
else:
print("Access denied")
return main()
#here, define the function in the same identation of the main function, not inside it
def access(userName):
print("Welcome, " + userName)
main()
答案 1 :(得分:1)
您尝试在声明之前调用access()
!这是因为定义access()
的代码是在声明之后放置的,所以你的代码本质上是试图找到一个不存在的函数(还)!这是一个scoping issue,很常见!
以下代码(通过将access()
的定义放在main()
之外)起作用:
def main():
userName = ("u123")
userPass = ("p123")
userNameInput = input("Username: ")
userPassInput = input("Password: ")
if userPassInput == userPass and userNameInput == userName:
print("Access granted")
access(userName)
else:
print("Access denied")
return main()
def access(userName):
print("Welcome, " + userName)
main()
我也不相信建议在函数中定义一个函数(对于这种类型的项目),而是可以创建一个包含各种函数的类:
class Login:
def __init__(self):
self.users = {}
self.userName = ("u123")
self.userPass = ("p123")
def addUser(self, uname, upass):
self.users[uname] = upass;
def login(self):
userNameInput = input("Username: ")
userPassInput = input("Password: ")
if userNameInput in self.users:
if self.users[userNameInput] == userPassInput:
print("Access granted")
self.access(userNameInput)
else:
print("Access denied")
return self.login()
else:
print("Access denied")
return self.login()
def access(self, username):
print("Welcome, "+username+"!")
def main():
mylogin = Login()
mylogin.addUser("u123","p123")
mylogin.login()
main()
展望未来,你必须实现某种安全性(上面的代码没有!所以一定不要将它用于重要的事情。不要使用用户和密码字典,这只是为了表演!)。
希望它有助于重燃你对Python的热爱!
答案 2 :(得分:0)
问题是你在调用它之后定义了这个函数,这样做:
def main():
userName = ("u123")
userPass = ("p123")
userNameInput = input("Username: ")
userPassInput = input("Password: ")
def access():
print("Welcome, " + userName)
if userPassInput == userPass and userNameInput == userName:
print("Access granted")
access()
else:
print("Access denied")
return main()
access()
main()
(假设你不只是混淆空间而实际上不想在另一个中定义一个函数)