如何检查输入是否为 integer？ 当用户输入 7 程序结束时，我对 output 有问题

Question

/*
 * To change this license header, choose License Headers in Project Properties.
 * To change this template file, choose Tools | Templates
 * and open the template in the editor.
 */
package validatinginput;

import java.util.Scanner;

/**
 *
 * @author abdal
 */
public class ValidatingInput {

    /**
     * @param args the command line arguments
     */

    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
    // Declare variables
    int number = 0;
    int lowerbound = 5;
    int upperbound =10;
    //Prompt the user to enter an interger between 5 and 10
    System.out.println("Please enter a interger int between 5 and 10");
    while (!s.hasNextInt()) {
    System.out.println("Not an int");
    s.next(); 
    }
    number = s.nextInt();

    if (number<= lowerbound || number >= upperbound){ 
        System.out.println("out bound");
        number++;
    }else{
        System.out.println("you entered");
        s.next();

           }number = s.nextInt();


       } 
}

Answer 1

我认为这里有两个问题：

1. 如果用户输入了超出范围的 integer，那么您的程序将执行以下操作：它将调用number = s.nextInt() （等待更多输入）然后停止（从main返回）。 您需要另一个循环，直到用户输入有效的 integer。
2. 如果用户输入 integer，您可以使用number = s.nextInt()从扫描仪中读取。 如果 integer 在您调用s.next()和number = s.nextInt() *again*. Both calls look suspicious/wrong: you already have the integer to be read in variable *again*. Both calls look suspicious/wrong: you already have the integer to be read in variable number . There is no need to read any more information from the scanner. In fact, the calls to . There is no need to read any more information from the scanner. In fact, the calls to . There is no need to read any more information from the scanner. In fact, the calls to s.next() and s.nextInt()` 的调用将阻塞，直到您输入更多数据。

在第一个number = s.nextInt()行之后，您不应再调用s.nextXXX() 。