Question

我正在创建一个基本的电话簿程序，该程序可以执行操作（添加，更新，删除，查找和退出）。我有第一个选项作为添加条目，第二个选项作为删除条目。每次用户完成一个动作时，都会出现再次选择一个动作的选项。当用户第二次选择时，它返回到第一项而不是选择。例如; 1是添加新联系人，2是删除新联系人。用户选择1，添加新联系人被要求选择另一个选项，选择2，但是选项1的代码再次运行以添加新而不是删除。

print("Please select an option from the main menu :\n")

def print_menu():
    print("1 : Add New Entry")
    print("2 : Delete Entry")
    print("3 : Update Entry")
    print("4 : Lookup Number")
    print("5 : QUIT")
    print()

print_menu()

while 1 == 1:
    try:
        menuOption = int(input())
        break
    except:
        print("Invalid! Please choose an option between 1 and 5.")

while menuOption not in(1,2,3,4,5):
    print("Invalid! Please choose an option between 1 and 5.")
    try:
        menuOption = int(input())
        break
    except:
        print("Invalid! Please choose an option between 1 and 5.")

###the above works perfect to set menu and restrict entry

phonebook = {}

#while menuOption != 5:
    #menuOption = int(input("Enter your selection (1-5): "))

while 1 == 1 :

 if menuOption == 1: #
        print("\nNEW CONTACT")
        while True:
            name = input("Contact Name : ") 
            if name.replace(' ','').isalpha():
                break
            print('Please enter a valid name.')
        while True:
            try:
                number = int(input("Contact Number : "))
                if number:
                   break
            except:
                print("Please enter a valid number.")

        if number in phonebook:
            print("Contact already exists. Duplicates are not allowed.\n")
        else:
            phonebook[number] = name 
            print("Success! New contact has been added.\n")

            print("PLEASE SELECT AN OPTION BETWEEN 1 AND 5  \n")
            try:
                option = int(input())
            except:
                 print("Please enter a numeric value between 1 and 5  \n")        


    elif menuOption == 2: ##delete
        print("\nDELETE CONTACT") 
        name = input("Contact Name : ")
        if name in phonebook:
            del phonebook[name]
            print(name, "has been removed from your contacts.")
        else:
            print("Contact not found.")

        print("PLEASE SELECT AN OPTION BETWEEN 1 AND 5  \n")
        try:
            option = int(input())
        except:
            print("Please enter a numeric value between 1 and 5  \n")

Answer 1

欢迎来到堆栈，雨天！在查看/运行您的代码时，提示用户输入一到五个的消息比我预期的要多，还有可能还没有编码的其他随机错误。我建议定义更多功能（用于菜单选项）并更多地组织代码，这将使您的代码更易于阅读和遵循。

在下面（不完整或没有错误），我重新组织了代码，以便在调用main()时显示“电话簿”菜单选项，并且用户可以选择另一个选项。代替在各种功能之间使用长的“ else-if” / elif，在while函数中的一个main()语句中巧妙地组织了各种菜单例程，并将选项组织成5种不同的功能： add() / delete()等。（我将伪代码放入了update / lookup / exit fns）。我希望你觉得这有帮助。我确实发现，如果我在输入数字时输入了字符串，则会抛出错误消息。我插入了注释以提供帮助。

希望这会有所帮助

phonebook= [] def main(): print("\n\tPhone Book") #title # main menu print("\n\tMain Menu") print("\t1. Add a contact") print("\t2. Delete a contact") print("\t3. Update a contact") print("\t4. Look up") print("\t5. Quit") menuOption = int(input("\nPlease select one of the five options ")) while menuOption not in(1,2,3,4,5) : ## insert try/except error handling needed here to handle NaN ## print("\nPlease insert a numeric option between 1 and 5") menuOption =int(input()) while menuOption <=5: mOpt =menuOption if menuOption == 1: Add(mOpt) elif menuOption == 2: Delete(mOpt) elif menuOption == 3: Update(mOpt) elif menuOption == 4: LookUp(mOpt) elif menuOption == 5: ExitPhone(mOpt) else: print("Invalid input! Please insert a value between 1 and 5") # add contact def Add(mOpt): ##Option 1 add = "" contact = True print("\n\tADD CONTACT") while contact == True: if mOpt == 1: print("\nNEW CONTACT") while True: name = input("Contact Name : ") if name.replace(' ','').isalpha(): break print('Please enter a valid name.') while True: try: number = int(input("Contact Number : ")) if number: break except: print("Please enter a valid number.") if number in phonebook: print("Contact already exists. Duplicates are not allowed.\n") else: #item = name + number this won't be found in the delete function phonebook.append(name) phonebook.append(number) #print(phonebook) print("Success! New contact has been added.\n") add = input("Would you like to add another? Y (yes) or N (no)") add = add.lower() if add=="yes" or add=="y": contact = True else: contact = False main() # delete def Delete(mOpt): redel = "" delcontact = True print("\n\tDELETE CONTACT") while delcontact == True: if mOpt == 2: print("Enter Contact Name:") name = input("Contact Name : ") if name in phonebook: ## insert code here to find associated number ## and concatenate it if you have created an 'item' phonebook.remove(name) #removes name, leaves the number print(name, "has been removed from your contacts.") #print(phonebook) else: print("Contact not found.") redel = input("Would you like to delete another? Y (yes) or N (no)") redel = redel.lower() if redel =="yes" or redel =="y": delcontact = False else: delcontact = True main() def Update(mOpt): if mOpt == 3: print("\nUpdate function") main() def LookUp(mOpt): if mOpt == 4: print("\nLookUp function") main() def ExitPhone(mOpt): if mOpt == 5: print ("Exit?") main() main()

Answer 2

您的代码检查menuOption的值，但输入option。只需更改

option = int(input())

进入

menuOption = int(input())

基本选项菜单失败-需要帮助

2 个答案: