如何实现无输入消息打印及每行一问的用户输入交互？

需求：程序需实现两项功能：1. 打印无需用户输入的提示消息（如欢迎语）；2. 每行输出一个交互问题，等待用户输入完成后再显示下一个问题。
现象：当前运行时多个提示语句挤在同一行，例如：Do you want to [E]ncrypt or [D]ecrypt? Do you want to use a [S]hift cipher or [V]igenere Cipher? Do you want to continue(Y/N)?

问题原因 #

程序开头多余的keyboard.nextLine()读取了欢迎语后的空行，导致第一次进入循环时，第一个operation = keyboard.nextLine()直接获取到空字符串，程序不会等待用户输入就执行下一个提示语句，最终造成多个提示挤在一行。

修复后的完整代码 #

public static void main(String[] args) {
    Cipher cipher; // 创建Cipher类型变量
       
    // 打印欢迎消息
    Scanner keyboard = new Scanner(System.in); // 创建Scanner对象用于用户输入
    System.out.println("Welcome to Nicholas Coleman's Cipher Program!\n"); 
    
    /*
     将所有逻辑放入while循环，支持用户多次重复操作
    */
    while (true) {
            
        /*
         提示用户选择加密或解密，再选择密码类型
         将输入分别存入operation和whichCipher变量
        */
            
        System.out.print("Do you want to [E]ncrypt or [D]ecrypt? ");
        String operation = keyboard.nextLine();
        System.out.print("Do you want to use a [S]hift cipher or [V]igenere Cipher? ");
        String whichCipher = keyboard.nextLine();
            
        /*
         判断要执行的操作
        */
            
        if(operation.equals("E")) { // 用户选择加密
            if(whichCipher.equals("S")) { // 用户选择ShiftCipher
                /*
                 提示输入密钥，创建ShiftCipher实例，调用encrypt方法加密明文
                */
                System.out.print("Please enter a number between 0 and 25 to use as a key: ");
                int key = keyboard.nextInt();
                keyboard.nextLine(); // 读取nextInt()留下的换行符
                System.out.print("Please enter the plaintext to be encrypted: ");
                String plaintext = keyboard.nextLine();
                cipher = new ShiftCipher(key);
                System.out.println("The corresponding ciphertext is: " + cipher.encrypt(plaintext));
                    
            } else if(whichCipher.equals("V")) { // 用户选择VigenereCipher
                /*
                 提示输入关键词，创建VigenereCipher实例，调用encrypt方法加密明文
                */
                System.out.print("Please enter a keyword: ");
                String keyword = keyboard.nextLine();
                System.out.print("Please enter the plaintext to be encrypted: ");
                String plaintext = keyboard.nextLine();
                cipher = new VigenereCipher(keyword);
                System.out.println("The corresponding ciphertext is: " + cipher.encrypt(plaintext));
                    
            }
                
        } else if (operation.equals("D")) { // 用户选择解密
            if(whichCipher.equals("S")) { // 用户选择ShiftCipher
                /*
                 提示输入密钥，创建ShiftCipher实例，调用decrypt方法解密密文
                */
                System.out.print("Please enter a number between 0 and 25 to use as a key: ");
                int key = keyboard.nextInt();
                keyboard.nextLine(); // 读取nextInt()留下的换行符
                System.out.print("Please enter the ciphertext to be decrypted: ");
                String ciphertext = keyboard.nextLine();
                cipher = new ShiftCipher(key);
                System.out.println("The corresponding plaintext is: " + cipher.decrypt(ciphertext));
                    
            } else if(whichCipher.equals("V")) { // 用户选择VigenereCipher
                /*
                 提示输入关键词，创建VigenereCipher实例，调用decrypt方法解密密文
                */
                System.out.print("Please enter a keyword: ");
                String keyword = keyboard.nextLine();
                System.out.print("Please enter the ciphertext to be decrypted: ");
                String ciphertext = keyboard.nextLine();
                cipher = new VigenereCipher(keyword);
                System.out.println("The corresponding plaintext is: " + cipher.decrypt(ciphertext));

            }    
        }
        // 询问用户是否继续
        System.out.print("Do you want to continue(Y/N)? ");
        String cont = keyboard.nextLine();
            
        // 如果用户输入Y，继续循环
        if(cont.equals("Y")) {
            continue;
                
        }
        // 如果用户输入N，显示感谢信息并终止循环
        else if(cont.equals("N")) {
            System.out.println("Thank you for using Nicholas Coleman's cipher program.");
            break;

        }
    }
}

关键修改说明 #

1. 删除了程序开头多余的String nextLine = keyboard.nextLine();，避免提前读取空行导致后续输入步骤被跳过。
2. 修正了Vigenere解密部分的两处语义错误：
   • 将提示文本中的plaintext to be decrypted改为ciphertext to be decrypted，符合解密操作的输入逻辑。
   • 将输出结果中的corresponding ciphertext改为corresponding plaintext，匹配解密操作的输出结果。
3. 保留了nextInt()后的keyboard.nextLine()，用于清除输入缓冲区中的换行符，避免后续nextLine()读取到空字符串。

内容的提问来源于stack exchange，提问作者Pooh_86