Question

我有一个整数数组列表，它将由用户填充，我想接受输入直到用户想要。我该怎么办？

我已经尝试过了，但是不满意我。 s是用户的最终令牌。

for(int i=0;input.nextInt()!='s';i++)
{   int a=input.nextInt();
    number.add(a);
}

Answer 1

使用nextInt()不会解决您的问题，因为一旦用户输入非整数，它将生成一个异常，因此将永远不会执行您的循环条件。另外，由于您的迭代次数未知，因此您应该只使用while循环而不是使用for循环。我已经修改了您的代码。请参阅下面的代码和代码注释。

import java.util.ArrayList;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        ArrayList<Integer> number = new ArrayList<>();
        String endToken = "s"; // this will be the holder of your end token
        while (true) { // iterate while user doesn't input the end token
            try {
                System.out.println("Enter an integer or enter 's' to exit!"); // ask the user for input
                String userInput = input.nextLine(); // get the user input
                if (userInput.equals(endToken)) { // check if user input is end token
                    break; // exit the loop
                }
                int a = Integer.parseInt(userInput); // convert the input into an int
                number.add(a); // add the input to the list
             // if an error inputs a String but not the end token then an exception will occur
            } catch (Exception e) {
                System.out.println("Invalid Input!"); // tell the user that the input is invalid and ask the input again
            }
        }

        input.close(); // don't forget to close the scanner

        // loop below is just used to print the user inputs
        System.out.println("Your inputs are: ");
        for (int i = 0; i < number.size(); i++) {
            System.out.println(i);
        }
    }

}

Answer 2

您可以使用Regex解析传递的输入，解决方案将排除浮点数和字符串

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输入

public static void main(String[] args) {
    // TODO Auto-generated method stub

    Scanner s = new Scanner(System.in);
    ArrayList<Integer> userNumber = new ArrayList<>();
    Pattern p = Pattern.compile("\\d+\\d");
    Matcher m = null ;

    System.out.println("Enter your number list use chanracter 'NO' or 'no' as the terminator: ");
    String input = s.nextLine();
    String[] data = input.split(",");
    for(String word : data) {
        if(word.equals("NO") || word.equals("no")) {
            break;
        }
        //m = p.matcher(word).matches();
        if(p.matcher(word).matches()) {
            m = p.matcher(word);
            m.find();
            int n = Integer.valueOf((m.group()));
            userNumber.add(n);
        }else {
            System.out.println("Data entered " + word + " is not a number : ");
        }

    }
    System.out.println("Numbers entered are : " + userNumber);

}

输出

12,12.1,123,"asd",123,"NO","NO"

如何为用户使用标志变量来结束输入

2 个答案: