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Java 如何显示实际输入而不是选项号?_Java_String_Random - Fatal编程技术网

Java 如何显示实际输入而不是选项号?

Java 如何显示实际输入而不是选项号?,java,string,random,Java,String,Random,我试图让它显示实际输入,而不是选项号 import java.util.*; public class RandomGenerator { public static void main(String[] args) { int length; Scanner input = new Scanner(System.in); System.out.println("How many options?"); //user input food

我试图让它显示实际输入,而不是选项号

import java.util.*;

public class RandomGenerator {

    public static void main(String[] args) {

       int length;
       Scanner input = new Scanner(System.in);
       System.out.println("How many options?"); //user input food options
       length = input.nextInt();

       String[] names = new String[length];
       for(int counter = 0; counter < length; counter++){
           System.out.println("Enter option #" + (counter+1) + ":");
           names[counter] = input.next();   
       }

       input.close();

       System.out.println("You are going to eat " + new Random().nextInt(names.length));
import java.util.*;
公共类随机发生器{
公共静态void main(字符串[]args){
整数长度;
扫描仪输入=新扫描仪(System.in);
System.out.println(“有多少个选项?”);//用户输入食物选项
length=input.nextInt();
字符串[]名称=新字符串[长度];
用于(int计数器=0;计数器<长度;计数器++){
System.out.println(“输入选项”#“+(计数器+1)+“:”;
名称[计数器]=输入。下一步();
}
input.close();
System.out.println(“你要吃”+new Random().nextInt(names.length));

您已经在此处生成了一个随机数:

new Random().nextInt(names.length)
您可以使用此随机数访问
names
数组中的元素

int randomNumber = new Random().nextInt(names.length);
String option = names[randomNumber]; // here is the important bit!
现在您可以打印
选项了

System.out.println("You are going to eat " + option);

那么问题出在哪里呢?您已经存储了输入,只需将其打印出来。请看这一行:
names[counter]=input.next();
。这就是访问数组元素的方式。将其应用于最后一行。