星球大战名称生成器代码

时间:2014-02-16 00:09:52

标签: java

我必须制作一个代码,使一个星球大战的名称生成器,它要求名称,但只需要从名称中的某些字母提出星球大战的名称,我的程序不会编译任何人都可以看到我做错了什么:

import java.util.Scanner;

public class StarWarsName {
    public static void main(String[] args) {
        System.out.printf("Enter your first name: ");
        firstname = input.nextLine();
        first = first.substring(0,3);

        System.out.printf("Enter your last name: ");
        lastname = input.nextLine();
        last = last.substring(0,2);

        System.out.printf("Enter your mother's maiden name: ");
        mothersname = input.nextLine();
        mother = mother.substring(0,2);

        System.out.printf("Enter the name of the city in which you were born: ");
        cityname = input.nextLine();
        city = city.substring(0,3);

        StarWarsName = first +" "+ last +" "+ mother +" "+ city + " of " + last +" "+$
        System.out.println("May the force be with you, " + StarWarsName + "May the fo$
    }

}

2 个答案:

答案 0 :(得分:2)

从这一行开始:

firstname = input.nextLine();

提示:

1)firstname在哪里宣布?

2)阅读编译错误消息!

3)当你向别人询问有关编译错误的问题时,你需要说明编译错误是什么,以及编辑错误发生的位置。

考虑这一行:

StarWarsName = first +" "+ last +" "+ mother +" "+ city + " of " + last +" "+$

4)最后的$不是有效的Java。它看起来像一个复制和粘贴错误。如果Java程序中的一行太长而无法放在您的(预期)显示设备上,那么将它拆分为常规练习; e.g。

    StarWarsName = first +" "+ last +" "+ mother +" "+ city + " of " + 
            last + " " + and + " " + the + " " + rest;

答案 1 :(得分:0)

您有几个未声明的变量,例如firstname。我认为还有一些逻辑错误,因为你不使用firstname和其他人。