日期:2014-05-20  浏览次数:20946 次

猜数游戏输入字母会错误
代码如下

import java.util.Random ;
import java.util.Scanner ;

public class Kaikouzhong{

public static void main(String[] args){
int n = 0 , m = 99 ;
Random r = new Random();
int z = r.nextInt(100);
int s = z + 1 ;
System.out.println( "猜数游戏\n");
System.out.println( "请输入一个1-99之间的数");
Scanner sc = new Scanner(System.in);
int i = sc.nextInt();
while (i != s){
while (i>99||i<1){
System.out.println( "输入有误,请输入一个1-99之间的数");
i = sc.nextInt();
}
System.out.println("输入的数为"+i);
if (i>s){
System.out.println(n+" 到 "+i);
m = i ;
System.out.println( "请输入一个"+n+"到"+m+"之间的数");
i = sc.nextInt();
}else{
System.out.println(i+" 到 "+m);
n = i ;
System.out.println( "请输入一个"+n+"到"+m+"之间的数");
i = sc.nextInt();
}
while (i>m||i<n){
System.out.println( "输入有误,请输入一个"+n+"到"+m+"之间的数");
i = sc.nextInt();
}
}
System.out.println("中奖号码为"+s);
System.out.println("中奖了");
}
}


怎么可以限制输入字母返回重新输入呢、




------解决方案--------------------
for example
Java code
public static void main(String[] args){
    int n = 1, m = 99 ;
    Scanner sc = new Scanner(System.in);
    Random r = new Random();
    int s = r.nextInt(99) + 1;
    System.out.println( "猜数游戏\n");
    while (true) {
        System.out.printf( "请输入一个%d-%d之间的数\n", n, m);
        String str = sc.nextLine();
        int i = 0;
        try {
            i = Integer.parseInt(str);
            if (i<n || i>m) {
                throw new Exception("out of range.");
            }
        } catch (Exception e) {
            System.out.println("输入有误,请重新输入");
            continue;
        } 

        if (i < s) {
            System.out.printf("输入的数是%d,比中奖号码小,再猜\n", i);
            m = i;
        } else if (i > s) {
            System.out.printf("输入的数是%d,比中奖号码大,再猜\n", i);
            n = i;
        } else {
            System.out.println("输入的数是%d,中奖号码是%d,中奖了\n", i, s);
            System.out.println("是否继续?请输入[quit]退出");
            str = sc.nextLine();
            if ("quit".equals(str)) {
                break;
            }
            n = 1;
            m = 99;
            s = r.nextInt(99) + 1;
        }
    }
}

------解决方案--------------------
Java code

public class Kaikouzhong {
    public static void main(String[] args) {
        int n = 0, m = 99;
        Random r = new Random();
        int z = r.nextInt(100);
        int s = z + 1;
        System.out.println("猜数游戏\n");
        System.out.println("请输入一个1-99之间的数");
        Scanner sc = new Scanner(System.in);
        int i=0;
        try{
            i = sc.nextInt();
            while (i != s) {
                while (i > 99 || i < 1) {
                    System.out.println("输入有误,请输入一个1-99之间的数");
                    i = sc.nextInt();
                }
                System.out.println("输入的数为" + i);
                if (i > s) {
                    System.out.println(n + " 到 " + i);
                    m = i;
                    System.out.println("请输入一个" + n + "到" + m + "之间的数");
                    i = sc.nextInt();
                } else {
                    System.out.println(i + " 到 " + m);
                    n = i;
                    System.out.println("请输入一个" + n + "到" + m + "之间的数");
                    i = sc.nextInt();
                }
                while (i > m || i < n) {
                    System.out.println("输入有误,请输入一个" + n + "到" + m + "之间的数");
                    i = sc.nextInt();
                }
            }
            System.out.println("中奖号码为" + s);
            System.out.println("中奖了");
        }catch(Exception e){
            System.out.println("==只能输入数字!游戏结束==");
        }
        
    }

}