while循环中的If语句与用户inpu

Question

这段代码以前可以工作，但现在我遇到了while0<；=int（relationship）>；3代码块的问题。用户应该能够选择与信件收件人的关系类型，这将在信件中填充问候语和告别语。你知道吗

如果在while循环中保留If/elif语句，sal/val字符串就不会填充到字母中。你知道吗

如果使If/elif语句与while循环齐平，则仅当用户第一次键入1、2或3时，才会填充字母的字符串。如果用户键入1、2或3之外的数字，程序将继续，但sal/val字符串将不会填充。你知道吗

最初，我只有一组if/elif语句，但我创建了两组语句（这是多余的，也没有解决问题），以查看这是否能解决问题。我也把压痕弄得一团糟。我也尝试过为每种类型的关系创建函数，但我不断收到称呼的名称错误。你知道吗

如果用户第一次按指示操作，我可以让程序工作；如果用户在第一次输入错误后被提示输入1、2或3，我可以让程序运行。但我不能让sal/val在两种情况下都填充。我觉得这个问题一定和缩进有关，但是当我改变缩进级别时，问题就没有解决。你知道吗

# Function for an infinite nested dictionary to add content to a personalized letter 
def lkng_glss_lttr_bot():             

    lgl_num = 0 
    while command == 'compose': 

        store_lkng_glss_lttr = {}
        new_lgl = {} 
        new_lgl_num = len(store_lkng_glss_lttr) + 1 
        store_lkng_glss_lttr[new_lgl_num] = new_lgl

        address_acquaintence = ['Dear', 'Cordially,']
        address_friend = ['Dearest', 'With warmest regards,']
        address_beloved = ['My Darling', 'With all my love and affection,']
        salutation = ''
        valediction = ''

        recipient = input("\nWhat is the recipient’s name?\n") 
        new_lgl['recipient'] = recipient

        email = input("\nWhat is the recipient's email?\n")
        new_lgl['email'] = email


        print("\nWhat is your relationship to %s?" % recipient)
        relationship = int(input("""Type 1 for Acquaintence
    Type 2 for Friend
    Type 3 for Beloved\n"""))

        while 0 <= int(relationship) > 3:
            relationship = input("Please type 1, 2, or 3.\n")

            if relationship == 1:
                for salutation in address_acquaintence:
                    salutation = address_acquaintence[0]                
                for valediction in address_acquaintence:
                    valediction = address_acquaintence[1]

            elif relationship == 2:    
                for salutation in address_friend:
                    salutation = address_friend[0]        
                for valediction in address_friend:
                    valediction = address_friend[1]

            elif relationship == 3:
                for salutation in address_beloved:
                    salutation = address_beloved[0]      
                for valediction in address_beloved:
                    valediction = address_beloved[1]

        if relationship == 1:
            for salutation in address_acquaintence:
                salutation = address_acquaintence[0]                
            for valediction in address_acquaintence:
                valediction = address_acquaintence[1]

        elif relationship == 2:    
            for salutation in address_friend:
                salutation = address_friend[0]        
            for valediction in address_friend:
                valediction = address_friend[1]

        elif relationship == 3:
            for salutation in address_beloved:
                salutation = address_beloved[0]      
            for valediction in address_beloved:
                valediction = address_beloved[1]



        print("\nPlease inquire about %s's well being." % recipient)
        inquiry = input("You can ask about the weather, or what %s has been doing to pass the time in the interim between your last exchange:\n" % recipient)

        body = input("\nPlease write a few sentences to inform %s of the recent happenings in your life:\n" % recipient)

        final_sentiment = input("\nPlease close the letter by expressing the importance of your friendship and your desire to spend time with %s in the future:\n" % recipient ) 

        sender = input("\nWhat is your name?\n")

        postscript = input("\nPlease write a short postscript to %s:\n" % recipient) 


        # Concatenate greeting and recipient
        part_1 = salutation + " " + recipient + "," + "\n"

        # Concatenate inquiry and body
        part_2 = inquiry + " " + body + "\n"

        # Concatenate part_1, part_2, final_sentiment, closing, sender, & postscript in reverse order for a non-tradional letter format
        non_trdtnl_lttr = "P.S. " + postscript + "\n" + sender + "\n" + valediction + "\n" + final_sentiment + "\n" + part_2 + part_1

        # Using the non-traditional letter format, reverse the order of the entire letter using an extended slice to create a Looking Glass Letter & store in dictionary
        lkng_glss_lttr = non_trdtnl_lttr[::-1]
        new_lgl['lkng_glss_lttr'] = lkng_glss_lttr


        # Notify sender that their letter contents have been added to the Bot
        print(store_lkng_glss_lttr)
        print("\nYour letter to %s has been composed by the Looking Glass Letter Bot:\n" % recipient + lkng_glss_lttr)
        print("\nYour Looking Glass Letter to %s will be emailed immediately." % recipient)


        # Ask user to add another client 
        print("\nWould you like to compose another letter? Y/N")
        sender_response = input().lower().strip() 


        if sender_response == 'y' or sender_response == 'yes': 
            continue 

        else:
            print("\nWe hope you enjoyed using the Looking Glass Letter Bot.\nWe look forward to serving your Looking Glass Letter composition needs in the future.")
            break 

输出应该类似于：

,boB raeD

!doog m'I ?uoy era woH

!!uoy ssim I

,yllaidroC

G

.em llaC .S.P

但是，如果用户键入除1、2或3之外的内容，并且必须再次提示，那么亲爱的、真诚的用户将不会填充。你知道吗

Answer 1

在循环中，不能将input的返回值转换为整数。你知道吗