我如何在对象数组上显示所有数据

Question

我想创建一种在我制作的对象数组上显示所有数据的方法。 我设法通过这里的人们的帮助来创建特定的搜索。 我可以重用他的代码吗？ 怎么样？ 我要创建的方法在最后一部分。 我如何在对象数组上显示所有数据

import java.util.Dictionary;
import java.util.Hashtable;
import java.util.Scanner;

public class TestMouse {

    private static Dictionary<String, Mouse> mouseList = new Hashtable<String, Mouse>(); 

    public static void main(String[] args) {

        menu();
    }

    @SuppressWarnings("resource")
    public static void menu() {
        Scanner input = new Scanner(System.in);

        int choice;
        do {
            System.out.println("CMPE 325 Student Record Holder System");
            System.out.println("--------------------------------");
            System.out.println("1.Add Student");
            System.out.println("2.View Records");
            System.out.println("3.Update Students");
            System.out.println("4.Get Average");
            System.out.println("5.Exit");
            System.out.println();
            System.out.println("--------------------------------");
            System.out.print("Enter your choice: ");
            choice = input.nextInt();

            switch (choice) {

            case 1:
                record();
                break;
            case 2:
                display();
                // display();
                break;
            case 3:
                menu();
                // update();
                // break();
            case 4:
                menu();
                // average();
                // break();
            case 5:
                break;
            default:
                continue;
            }

        } while (choice != 5);
        System.out.println();
        System.out.println("Good-Bye");

    }

    // end of menu(); method.
    // ------------------------------------------------------------
    @SuppressWarnings({ "resource", "unused" })
    public static void record() {

        Scanner Input = new Scanner(System.in);

        int total;

        System.out.println("How many students you are going to input? ");
        total = Input.nextInt();

        Mouse[] keyboard = new Mouse[total];

        for (int index = 0; index < total; index++) {
            String space;

            keyboard[index] = new Mouse();

            System.out.printf("Student[%d]", index + 1);

            System.out.println();

            System.out.println("");
            space = Input.nextLine();

            System.out.println("ID Number: ");
            keyboard[index].setId(Input.nextLine());

            System.out.println("First Name: ");
            keyboard[index].setFirstName(Input.nextLine());

            System.out.println("Middle Name: ");
            keyboard[index].setMiddleName(Input.nextLine());

            System.out.println("Last Name: ");
            keyboard[index].setLastName(Input.nextLine());

            System.out.println("Degree: ");
            keyboard[index].setDegName(Input.nextLine());

            System.out.println("Year Level: ");
            keyboard[index].setYear(Input.nextInt());

            //Save current object
            mouseList.put(keyboard[index].getId(), keyboard[index]);
        }

        for (int index2 = 0; index2 < total; index2++) {
            System.out.printf("Student[%d]", index2 + 1);
            System.out.println();
            System.out.println("ID Number:" + keyboard[index2].getId());
            System.out.println("Name:" + keyboard[index2].getFirstName() + " "
                    + keyboard[index2].getMiddleName() + " "
                    + keyboard[index2].getLastName());
            System.out.println("Degree:" + keyboard[index2].getDegName());
            System.out.println("Year Level:" + keyboard[index2].getYear());
        }
    }

    public static void specific() {
        String id = "";
        System.out.println("Enter an Id Number");
        Scanner Input = new Scanner(System.in);
        id = Input.nextLine();

        Mouse mouse = mouseList.get(id);
        if (mouse != null) {
            System.out.println();
            System.out.println("ID Number:" + mouse.getId());
            System.out.println("Name:" + mouse.getFirstName() + " "
                    + mouse.getMiddleName() + " "
                    + mouse.getLastName());
            System.out.println("Degree:" + mouse.getDegName());
            System.out.println("Year Level:" + mouse.getYear());
        }
    }
    @SuppressWarnings("resource")
    public static void display(){
        Scanner input = new Scanner(System.in);

        int choice;


            System.out.println("--------------------------------");
            System.out.println("1.View List");
            System.out.println("2.View Specific Record");
            System.out.println("3.Exit");
            System.out.println();
            System.out.println("--------------------------------");
            System.out.print("Enter your choice: ");
            choice = input.nextInt();

            switch (choice) {

            case 1:
                displayall();
                break;
            case 2:
                specific();
                break;
            case 3:
                menu();
                break;

            }
}

 this part is the one im talking about. i hope u can help me
 public static void displayall(){

        String id = "";
        System.out.println("Enter an Id Number");
        Scanner Input = new Scanner(System.in);
        id = Input.nextLine();

        Mouse mouse = mouseList.get(id);
        if (mouse != null) {
            System.out.println();
            System.out.println("ID Number:" + mouse.getId());
            System.out.println("Name:" + mouse.getFirstName() + " "
                    + mouse.getMiddleName() + " "
                    + mouse.getLastName());
            System.out.println("Degree:" + mouse.getDegName());
            System.out.println("Year Level:" + mouse.getYear());
        }
    }
}             

Answer 1

因为您使用的是旧版Dictionary接口 ，所以这有点烦人，但是可以这样做。

您要做的是创建元素的Enumeration ，然后对其进行迭代。 通过调用Dictionary#elements() ，或更具体地说，通过mouseList.elements()提供实例。

然后，按照Enumeration规定对其进行迭代。

如果Mouse也覆盖toString() ，实际上会容易得多，因为这是打印这类复杂对象的最佳方法。

最后，您在该方法中进行了许多不必要的调用-您不需要关心ID或类似的东西。 您也不需要检查null ，因为如果它不在元素集中，则不会在Enumeration 。

如果您想更现代一点...那么请使用Map代替。 Hashtable已经使用了Map界面，因此您不需要更改具体对象（尽管我会强烈推荐）。 遍历Map的元素将成为对Map#entrySet()的调用，需要进行更多工作。