Question

Java控制台中精确判断输入数字用“什么抛异常”

Answer 1

import java.util.Scanner;
public class B {
public static void main(String args[]) throws Exception {
System.out.println("1.C# 2.Java 3.html");
boolean flag=true;//while循环是否继续，true继续循环，false停止循环
int num=0;//存放输入内容转成int之后的变量
while(flag){
Scanner scan = new Scanner(System.in);
System.out.println("请输入课程代号(1-3之间的数字)：");
String bh=scan.next();
try {
num = new B().checkInt(bh);
flag=false;//输入的内容转成int成功时，停止循环输入
}catch (ClassCastException e) {//输入为小数类型时
System.out.println("请输入整数");
} catch (NumberFormatException e) {//输入非数字类型时
System.out.println("请输入正确的数字");
}catch (ArithmeticException e) {//输入的数字不在1-3之间时
System.out.println("请输入1-3之间的数");
}
}
//以下输出你选择的课程
if(num==1){
System.out.println("你输入的课程是:1.C#");
}else if(num==2){
System.out.println("你输入的课程是:2.Java");
}else if(num==3){
System.out.println("你输入的课程是:3.html");
}
}

int checkInt(String bh)throws Exception{
try{
int num = Integer.parseInt(bh);//将输入的内容转换成int
if(num<1 || num>3){
throw new ArithmeticException();//转成int成功但不在1-3之间，手动抛出异常
}else{
return num;
}
}catch (NumberFormatException e) {//转换成int类型时失败
try{
Double.parseDouble(bh);//转成double类型
throw new ClassCastException();//转成double类型成功，手动抛出异常
}catch (NumberFormatException e2) {//转成double类型失败
throw new NumberFormatException();//手动抛出转型失败异常
}
}
}
}

//用到try...catch还有手动抛出异常做的

Answer 2

import java.util.InputMismatchException;
import java.util.Scanner;

public class Test {
private Scanner sc;

public void dispose() {
sc = new Scanner(System.in);
String[] course = { "C#", "Java", "html" };
System.out.print("请选择学习课程（1、C#\t2、Java\t3、html）:");
int courseNo = sc.nextInt();
switch (courseNo) {
case 1:
case 2:
case 3:
System.out.println("您选择了" + course[courseNo - 1]);
break;
default:
System.out.println("请输入1-3之间的数字。");
}
}

public static void main(String[] args) {
Test test = new Test();
try {
test.dispose();
} catch (InputMismatchException e) {
System.out.println("请输入数字!");
}
}
}