After the input error of next numerical data type () occurs, it can still be input normally next time

next Numerical data type () After an input error occurs , You can still input normally next time

1 The scene

1.1 Content

next Numerical data () Method means nextInt()、nextDouble()、nextByte() Equal value receiving method

use try…catch Wrap the digital receiving method , When the input is a string , The program will not be interrupted , The user will be prompted to enter again , Call again nextInt() When the method is used , You will find that the user can't input directly , Directly into the dead circle , But there are always input prompt statements

1.2 Sample code

package Work;
import java.util.InputMismatchException;
import java.util.Scanner;
public class Test {
    
    public static void main(String[] args) {
    
       Scanner s=new Scanner(System.in);
       boolean flag=true;
        // When no exception occurs , Set it to false, Just exit the loop 
       int n=0;
       while(flag){
    
           System.out.println(" Please enter a number n:");
           try {
    
               n=s.nextInt();
               flag=false;
           } catch (InputMismatchException e) {
    
               System.out.println(" Incorrect input , Please re-enter !!!!");
           } catch (Exception e){
    

           }
       }
    }
}

1.3 The sample code runs

2 Why is this problem

use nextInt() Method to receive the data input by the user , Meeting First check whether there is data in the space , If it does not exist, the user will need to enter a data , After entering the data and pressing enter, the next statement will be executed , At this time, the user's data will be put into a space

When the user enters the type and nextInt() Approach is to The match is successful Of , It will take out the data stored in the space , Then put the value in the space Empty

When the user enters the type and nextInt() Methods The match didn't work when , At this time, the value in the space Will not be taken out , Next time you call nextInt() I still got this value when I was , Because the value can be obtained in the space, there is no need for users to enter the value , Also used try…catch Exception handling , This program will not be interrupted , So the program has been executed

3 How to solve this problem

3.1 emphasis

You have to put The error information in the space is absorbed and released ,

3.2 Solution

stay catch The code block says next() Method is used to receive error messages in space , So that the data in the space can be released

3.3 Example code for solving

package Work;
import java.util.InputMismatchException;
import java.util.Scanner;
public class Test {
    
    public static void main(String[] args) {
    
        Scanner s=new Scanner(System.in);
        boolean flag=true;
        // When no exception occurs , Set it to false, Just exit the loop 
        int n=0;
        while(flag){
    
            System.out.println(" Please enter a number n:");
            try {
    
                n=s.nextInt();
                flag=false;
            } catch (InputMismatchException e) {
    
                s.next();
                // Used to absorb the error information entered before 
                System.out.println(" Incorrect input , Please re-enter !!!!");
            } catch (Exception e){
    

            }
        }
        System.out.println(" The program ended normally !!!");
    }
}

3.4 Screenshot of solution example