输入'done'时Java终止数组

Question

更新<！/强> 我现在已经修复了原始问题，但结果是显示最年轻和最老的成员时出错并抛出以下错误。有什么建议？感谢

线程“main”中的异常java.lang.IndexOutOfBoundsException：索引：4，大小：4     at java.util.ArrayList.rangeCheck（ArrayList.java:653）     at java.util.ArrayList.get（ArrayList.java:429）     在NameAge.main（NameAge.java:46）

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

public class NameAge {
    public static void main(String[] args) {
    Scanner input = new Scanner(System.in);

    final int MAX_VALUE = 4;

    ArrayList<String> nameList = new ArrayList<String>();
    ArrayList<Integer> ageList = new ArrayList<Integer>();

    Integer[] ages = new Integer[10];

    for (int i = 0; i < MAX_VALUE; i++) {

        System.out.print("Enter a name: ");
        String currentLine = input.next();

        if (currentLine.equals("DONE")) {
        break;
        }

        nameList.add(currentLine);

        System.out.print("Now enter an age for " + currentLine + ": ");
        ageList.add(input.nextInt());

      }

        System.out.print("\n");
        for(int i = 0; i < MAX_VALUE; i++) {

            System.out.println("Name: " + nameList.get(i) + "   Age: " + ageList.get(i));


        }

        // DISPLAY YOUNGEST AND OLDEST OF ARRAY, PRODUCING ERRORS
        int smallest = ageList.get(0);
        int largest = ageList.get(0);

        String oldest = nameList.get(0);
        String youngest = nameList.get(0);

        for (int i = 0; i < ages.length; i++) {
            if(ageList.get(i) > largest) {
                largest = ageList.get(i);
                oldest = nameList.get(i);

            }
             else if(ageList.get(i) < smallest) {
                smallest = ageList.get(i);
                youngest = nameList.get(i);
            }

        }
            System.out.println("\nThe youngest person is " + youngest + " who is " + smallest + " years old");
            System.out.println("The oldest person is " + oldest + " who is " + largest + " years old");

        }

    }

Answer 1

正如评论所示，我会在这里使用两个正则表达式，一个用于名称，一个用于年龄。另外，我会将扫描仪的输入作为字符串。目前，您正在调用Scanner.nextInt()，我认为这对于非数字数据会失败。如果发生这种情况，您的正常应用程序逻辑将检查输入是否永远不会被命中。相反，将它们都读为字符串，然后使用正则表达式来验证输入的格式。

这样的事情应该有效：

String age = "35";
String name = "Tim Biegeleisen";

if (name.matches(".*[^a-zA-Z].*")) {
    System.out.println("Not a valid name");
}
if (age.matches(".*[^0-9].*")) {
    System.out.println("Not a valid age");
}

Answer 2

正如所指出的那样，正则表达式是完成验证的最简单方法。您可能还想查看java.util.regex.Pattern（https://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html）的javadoc。

name.matches("\\D+") // for anything other than numbers with length > 1

name.matches("\\d+") // for numbers of length 1 and more