在函数内部调用函数？帮助修复[重复]答案

【问题标题】：Call a function inside a function? Help to fix [duplicate]在函数内部调用函数？帮助修复[重复]
【发布时间】：2017-08-07 07:47:48
【问题描述】：

我不确定如何修复我的代码，有人可以帮忙吗！它打印出这个 --> “NameError: free variable 'info' referenced before assignment in enclosure scope”，我不知道如何使 info 成为全局变量，我认为这是问题所在......请帮忙！

import time
import random

admincode = ["26725","79124","18042","17340"]
stulogin = ["NikWad","StanBan","ChrPang","JaiPat","JusChan","AkibSidd","VijSam"]
teachercode = ["KGV"]

def main():
    def accesscontrol():
        global teachercode, stulogin, admincode
        print("Enter: Student,Teacher or Admin")
        option = input("--> ")
        if option == "Student":
            info()
        elif option == "Teacher":
            print("Enter you teacher code(xxx)")
            option = input
            if option == teachercode:
                  print("Access Granted")
                  info()
            else:
                 print("Please input the correct code!")
                 accesscontrol()
        elif option == "Admin":
            print("Enter your admin code(xxxxx)")
            option = input("--> ")
            if option == admincode:
                print("access granted, my master!")
        else:
            accesscontrol()
    accesscontrol()

    def info():
        print("Hello, enter your information below")
        usname = input("Username: ")
        pwname = input("Password: ")
        done = False
        while not done:
            print("Is this the information correct?[Y/N]")
            option = input("--> ")
            if option == "Y":
                print("Information saved")
                print("Username :",usname,"\nPassword:",pwname)
                done = True
            else:
                main()
        return info()
    info()

main()

【问题讨论】：

发布完整的堆栈跟踪
这个问题也有其他错误@Nikhail

标签： python

【解决方案1】：

问题是您将accesscontrol 和info 定义为相对于main 的本地名称。因此，当您在 accesscontrol 中调用 info 时，它找不到它，因为它是“拥有”的名称，换句话说，是 main 的本地名称。

而不是像这样的功能：

def main():
    def accesscontrol():
        # ...
    def info():
        # ...
    # ...

像这样将它们移出main()：

def accesscontrol():
    # ...

def info():
    # ...

def main():
    # ...

因此保持main() 简单：

def main():
    accesscontrol()
    info()

【讨论】：

【解决方案2】：

您需要在调用之前定义info()。此外，您还给info() 打了一个不必要的电话，我已将其删除。

import time
import random

admincode = ["26725", "79124", "18042", "17340"]
stulogin = ["NikWad", "StanBan", "ChrPang", "JaiPat", "JusChan", "AkibSidd", "VijSam"]
teachercode = ["KGV"]


def main():

    def info():
        print("Hello, enter your information below")
        usname = input("Username: ")
        pwname = input("Password: ")
        done = False
        while not done:
            print("Is this the information correct?[Y/N]")
            option = input("--> ")
            if option == "Y":
                print("Information saved")
                print("Username :", usname, "\nPassword:", pwname)
                done = True
            else:
                main()
        return info()

    def accesscontrol():
        global teachercode, stulogin, admincode
        print("Enter: Student,Teacher or Admin")
        option = input("--> ")
        if option == "Student":
            info()
        elif option == "Teacher":
            print("Enter you teacher code(xxx)")
            option = input
            if option == teachercode:
                print("Access Granted")
                info()
            else:
                print("Please input the correct code!")
                accesscontrol()
        elif option == "Admin":
            print("Enter your admin code(xxxxx)")
            option = input("--> ")
            if option == admincode:
                print("access granted, my master!")
        else:
            accesscontrol()

    accesscontrol()

main()

【讨论】：

你是对的，这就是 OP 所拥有的。