java - Java通讯录程序

Question

我只是在做一个小程序。这是一个地址簿，有四个选项:

1. 插入新联系人
2. 按姓氏搜索联系人
3. 按姓名删除联系人
4. 显示所有联系人
5. 退出程序

只是想知道如何获取插入接触部件以及如何存储它。我对一个联系人进行了硬编码以对其进行测试。

这是我已经开始的代码

package addressbook;
import java.util.Scanner;

public class addressbooks
{


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

        //create a table to hold information

        String[][] addressbooks = new String[100][8];

        addressbooks[0][0]="Mobile Number";
        addressbooks[0][1]="First Name";
        addressbooks[0][2]="Last Name";
        addressbooks[0][3]="Address";
        addressbooks[0][4]="City";
        addressbooks[0][5]="County";
        addressbooks[0][7]="Telephone Number";

        //pre-populate address book for testing purposes and records

        addressbooks[1][0]="1";
        addressbooks[1][1]="David";
        addressbooks[1][2]="Needham";
        addressbooks[1][3]="Sraheens, Achill";
        addressbooks[1][4]="Galway";
        addressbooks[1][5]="Mayo";
        addressbooks[1][6]="086-1581077";
        addressbooks[1][7]="098-45368";

        addressbooks[2][0]="2";
        addressbooks[2][1]="Mc";
        addressbooks[2][2]="lovin";
        addressbooks[2][3]="Hawaii";
        addressbooks[2][4]="Hawaii";
        addressbooks[2][5]="Hawaii";
        addressbooks[2][6]="12345";
        addressbooks[2][7]="412-555-1234";

        //menu options
        System.out.print("Welcome to my Address book!");
        System.out.print("\n");
        System.out.print("\n1 - Insert a New Contact \n2 - Search Contact by Last Name \n3 - Delete Contact \n4 - Show All Contacts \n5 - Exit " );
        System.out.print("\n");
        System.out.print("\nChoose your option: ");

        int option = input.nextInt();

        if (option ==1)
        {
            System.out.print("\nPlease enter your First Name : ");
        }
        if (option ==2)
        {
        }

        if (option ==3)
        {
        }

        if (option ==4)
        {
        System.out.println(addressbooks[1][0]+
        "\t"+addressbooks[1][2]+ ", "+addressbooks[1][1]+
        "\n\t"+addressbooks[1][3]+
        "\n\t"+addressbooks[1][4]+ ", "+addressbooks[1][5]+ " "+addressbooks[1][6]+
        "\n\t"+addressbooks[1][7]);
        }

        if (option ==5)
        {
        }

    }
}

Answer 1

我会这样开始:

package model;

public class Person {
    private String firstName;
    private String lastName;
    // What else means something to your problem?  Birthday?
    // Constructors, getters (make them immutable), equals/hashCode
}

public class Address {
    private String street;
    private String city;
    private String county;
    private String postalCode;
    // Constructors, getters (make them immutable), equals/hashCode
}

public class AddressBook {
    private Map<Person, Address> contacts = new ConcurrentHashMap<Person, Address>();

    public void addContact(Person p, Address a) {
        this.contacts.put(p, a);
    }

    public void removeContact(Person p) {
        this.contacts.remove(p);
    }

    public Collection<Person> findAllContacts() {
        return new Collections.unmodifiableCollection(this.contacts.keySet());
    }

    public boolean hasContact(Person p) {
        return this.contacts.contains(p);
    }
    // etc.
}

我建议将所有有关基于文本 IO 的内容与问题的基本原理分开。如果你做对了，下一步就是编写 Web UI。如果你做得正确的话，大部分代码都可以重用。

考虑对应用进行分层:

view->services->persistence

模型类可以在所有三层中使用。

推荐 JDBC 的答案并没有错。如果您将服务和持久性类编写为接口(interface)，您会发现很容易将内存中的联系人 map 替换为使用 JDBC 的数据库版本:

package persistence;

public interface ContactDao {
    Collection<Contact> find();
    Contact find(Long id);
    Collection<Contact> find(String lastName);
    Collection<Contact> find(Address address);
    Long save(Contact c);
    void update(Contact c);
    void delete(Contact c);
}

做事的方法有很多种。我已经改变了主意:我引入了一个 Contact 类:

package model; 

public class Contact {
    private Person;
    private Address;
}