Question

所以，这个程序有效，它运行，但我希望它是白痴证明！要做到这一点，我需要一些帮助.. 如果用户输入数字或根本没有输入任何数字，我不希望它转到下一行！

如果我按Enter键而没有写任何内容，它仍会转到下一行，变量navn在结尾处完全呈空。如果我写一个数字，它也会做同样的事情。如果答案不符合提示，我如何让它返回并再次循环尝试。非常感谢你:)）

import java.util.Scanner;

class Metoder {

    public static void main(String[] args) {
        String bosted; //Variable
        String navn; //Variable

        Scanner in = new Scanner(System.in);

        System.out.println("Skriv inn navn: "); //What shows up when you first start the program

        while (!in.hasNext("[A-Za-z]+")) { //Only allow letters A-Z
            in.next();
            System.out.println("Tall horer ikke hjemme i navn, prøv igjen!"); //Prints, "numbers dont belong in names, try again" if what the user entered is a number

        }
        System.out.println("Takk!"); //Says thank you if the user has entered letters

        navn = in.nextLine(); //Proceeds to next line

        System.out.println("Skriv inn bosted: "); //Next line, where the user is supposed to enter where he/she lives
        while (!in.hasNext("[A-Za-z]+")) { //Excactly the same loop as above
            in.next();
            System.out.println("Tall hører ikke hjemme i stedsnavn, prøv igjen!");
        }
        System.out.println("Takk!");

        bosted = in.nextLine();

        System.out.println("Hei, " + navn + "! Du er fra " + bosted + "."); //Prints out what the user has entered previously in a full sentence.

    }
}

Answer 1

请使用此代码：

(x,y,z)

Answer 2

您可以使用continue关键字重新启动循环。

while(!in.hasNext("[A-Za-z]+")) {
    try {
        String s = in.nextLine();
        if(s.isEmpty() || s.matches("^-?\\d+$")){
            throw new Exception("empty string or number detected");
        }
    } catch (Exception e){
        continue;
    }
}

在if条件下，我们检查输入的字符串是否为空或是整数，并且我们可以抛出一个异常，这将导致while循环继续请求输入，直到条件失败（即通过我们的检验）。

如果未满足参数，则重新启动while循环

2 个答案: