如何正确使用试抓？

Question

我正在尝试处理如何使用try-catch。我知道它将“尝试”主代码，如果它不工作，它将捕获它并执行不同的操作。我还希望不断提示用户输入正确的值。

我一直得到输入不匹配异常错误，即使我设置我的捕获在其块中。

澄清一下：当我向用户询问他们计划停留多长时间，以及他们希望呆在哪一层楼的INT时，try-catch就会出现。因此，我想处理的错误包括非整数，以及它们是否超出“hotel”的范围。

这是我的密码：

public class Hotel{

   public static void main(String[] args) throws IOException {
      int choice = 0;
      String guestName = " ";
      int stayTime = 0;
      int floorPref = 0;

      System.out.println("Welcome to the Hotel California.");
      Scanner sc = new Scanner(System.in);


      Room[][] hotel = new Room[8][20];         


      for(int i = 0; i< hotel.length; i++){
         for(int j = 0; j<hotel[i].length;j++){
            hotel[i][j] = new Room(0,false,"none",0.00,0);

            int roomNum = ((i+1) * 100) + (j + 1);
            hotel[i][j].setRoom(roomNum);

            int roomCheck = hotel[i][j].getRoomNumber();

            if(roomCheck > 500){
               hotel[i][j].setPrice(350.00);   
            }

            else if(roomCheck < 500){
               hotel[i][j].setPrice(200.00);
            } 
         }
      }

       // Guest check-in interface.

      do{

         System.out.println("What business have you today?");
         System.out.println("1. Guest Registration");
         System.out.println("2. Guest Checkout");
         System.out.println("3. Show me occupied rooms");
         System.out.println("4. Exit");

         choice = sc.nextInt();

         if(choice == 1){  


            System.out.println("Tell us about yourself.");

            System.out.println("Please input your name:");

            guestName = sc.next();

            System.out.print("How long are you planning to stay?");

            try{
               stayTime = sc.nextInt();
            }
            catch(InputMismatchException e){
               System.out.println("Please input a valid integer.");
               stayTime = sc.nextInt();
            }

            System.out.println("Great. What floor would you like to be on? Enter a number 1-8, 0 for no preference.");

            floorPref = sc.nextInt();

            System.out.println("The following rooms are available based on your floor preference (floors 1-8, 0 for no preference: ");

         }    
         if(floorPref > 0){

            for(int i = 0; i < hotel[floorPref].length; i++){
               if(hotel[floorPref][(i)].getOccupation() == false){


                  System.out.print("Rooms " +  hotel[floorPref-1][i].getRoomNumber() + ", ");


               }
            }

            System.out.println("Are available today.");
         }


         else if(floorPref == 0){
            for(int i = 0; i < hotel.length; i++){
               for(int j = 0; j < hotel[i].length; j++){
                  System.out.print("Room " +  hotel[i][j].getRoomNumber() + ", ");

               }
            }

            System.out.println("Is available.");

         }


      }while(choice != 4);


   }
}

Answer 1

下面是小编的一段代码进行划分。如果用户输入零，那么它将去抓，你也可以输入数字。请注意，代码将只捕获一次。我已经用了治疗方法。这只是根据您的要求的一个例子。

import java.io.DataInputStream;
import java.io.IOException;


public class test1 {
    public static void main(String args[]) throws NumberFormatException, IOException
    {
        DataInputStream d=new DataInputStream(System.in);int x;
        try {
             x=Integer.parseInt(d.readLine());
            int z=8;
            System.out.println(z/x);
        } catch (Exception e)
        {
            System.out.println("can not divide by zero");
            x=Integer.parseInt(d.readLine());
        }
    }

}

Answer 2

试着把这个放在代码的顶部

import java.util.InputMismatchException;
import java.util.Scanner;
import java.io.IOException;

在您的publicstaticvoid抛出IOException maindeletethrows IOException中，它只剩下publicstaticvoid main希望它能工作

Answer 3

您现在使用的try-catch块有缺陷，因为一旦进入catch块，用户所要做的就是输入非整数的内容，以使整个程序崩溃。

相反，要获取stayTime和从扫描仪提取的所有其他int，请创建一个单独的函数，在用户输入int之前阻塞该函数：

private static int parseIntFromScanner(Scanner sc) {
    while(true) {
        try {
            int toReturn = sc.nextInt();
            return toReturn;
        } catch(InputMismatchException ime) {
            //Continue back to the top of the loop, ask again for the integer
        }
    }
}