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Java 姓名首字母_Java_Char - Fatal编程技术网

Java 姓名首字母

Java 姓名首字母,java,char,Java,Char,我正在编写这段代码,它允许用户输入一个可以包含空格和连字符的名称 Scanner scanner = new Scanner(System.in); System.out.print("Enter name: "); String input = scanner.nextLine(); String result = ""; char firstChar = input.charAt(0); char fUpper = Character.toU

我正在编写这段代码,它允许用户输入一个可以包含空格和连字符的名称

     Scanner scanner = new Scanner(System.in);
    System.out.print("Enter name: ");
    String input = scanner.nextLine();
    String result = "";
    char firstChar = input.charAt(0);
    char fUpper = Character.toUpperCase(firstChar);
    result = result + fUpper;
    for (int i = 1; i < input.length(); i++) {
        char currentChar = input.charAt(i);
        char cUpper = Character.toUpperCase(currentChar);
        char previousChar = input.charAt(i - 1);
        char pUpper = Character.toUpperCase(previousChar);
        if (pUpper == ' ')  {
            result = result + "." + cUpper;
        }
    }
    System.out.println("Initials Are: " + result);
    }
Scanner Scanner=新的扫描仪(System.in);
系统输出打印(“输入名称:”);
字符串输入=scanner.nextLine();
字符串结果=”;
char firstChar=input.charAt(0);
char fUpper=Character.toUpperCase(firstChar);
结果=结果+福珀;
对于(int i=1;i

我遇到的问题是,代码无法识别连字符,并将其包含在if语句的结尾首字母输出中,如果要检查空格,还需要检查连字符:

if(pUpper == ' ' || pUpper == '-')
编辑:

要在输出中包含连字符,只需执行以下操作:

    if(pUpper == ' ')
    {
        result = result + "." + cUpper;
    }
    else if(pUpper == '-')
    {
        result = result + "-" + cUpper;
    }

现在可以识别连字符了,谢谢,但我如何将其包含在最终输出中?因此,如果名称中有连字符,我希望它打印出连字符,而不是DOT。我刚刚用一个代码段更新了答案,该代码段将在输出中包含连字符。