Question

当我输入值“Adam J”作为我的名字时，输出以下内容：

Please enter student first name: Adam J
Please enter student surname: Please select Subject Unit:

我希望它拒绝这个值，就像输入一个数字一样。理想情况下，如果它有空格，我不希望它接受它，所以它会说“请输入有效的名字”。

 case 1:
                System.out.print("Please enter student first name: ");
                firstName = scan.next(); 

                while(!firstName.matches("[-a-zA-Z]*"))
                {
                    System.out.print("Please enter a valid first name: ");
                    firstName = scan.next(); 
                }
                firstCap = firstName.substring(0,1).toUpperCase() + firstName.substring(1);

    System.out.print("Please enter student surname: ");
        lastName = scan.next();

        while(!lastName.matches("[-a-zA-Z]*"))
        {
            System.out.print("Please enter a valid surname: ");
            lastName = scan.next();
        }
        surCap = lastName.substring(0,1).toUpperCase() + lastName.substring(1);

        System.out.print("Please select Subject Unit: "); 
        unit = scan.next();

我很困惑为什么：“请输入学生姓：请选择主题单位：”也在同一行。

你能帮忙吗？感谢

Answer 1

next()方法从流中检索下一个标记，默认情况下由空格分隔。 next()对名字的调用仅消耗Adam。下一次调用next()会消耗J，让您处于第三次提示状态。

您应该拨打nextLine()来检索整行，以便Adam J一次性使用firstName。

Answer 2

case 1:
                System.out.println("Please enter student first name: ");
                firstName = scan.nextLine(); 

                while(!firstName.matches("[a-zA-Z]+"))
                {
                    System.out.println("Please enter a valid first name: ");
                    firstName = scan.nextLine(); 
                }
                firstCap = firstName.substring(0,1).toUpperCase() + firstName.substring(1);

    System.out.println("Please enter student surname: ");
        lastName = scan.nextLine();

        while(!lastName.matches("[a-zA-Z]+"))
        {
            System.out.println("Please enter a valid surname: ");
            lastName = scan.nextLine();
        }
        surCap = lastName.substring(0,1).toUpperCase() + lastName.substring(1);

        System.out.println("Please select Subject Unit: "); 
        unit = scan.nextLine();

我假设额外 - 你的正则表达式是无意的。另外，你的正则表达式应该检查[a-zA-Z] +而不是[a-zA-Z] *，因为空名对我来说似乎没有用。

Answer 3

尝试nextLine()代替next()。

输入字母+空格+字母后，程序会变得有趣

3 个答案: