Question

我有一个ArrayList来保存联系人。创建时，用户会为每个联系人分配ID。如何引用ID来打印联系人的详细信息。例如： Contact ID = 1。我想从1拨打arrayList并打印Jone Smith，123 West St等。

以下是我创建ArrayList：

的方法

主类：

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


  public class ContactList {

   public static void main(String[] args) {

    ArrayList<Contact> contacts = new ArrayList<Contact>();

    Scanner input1 = new Scanner(System.in);
    int type = 0;
    while(type != 4){
    System.out.println("Please select an option:");
    System.out.println("Add a Personal Contact: Enter 1");
    System.out.println("Add a Business Contact: Enter 2");
    System.out.println("Display Contacts List: Enter 3");
    System.out.println("4 to quit");

    type = input1.nextInt();

    if(type == 4){
        System.out.println("Goodbye");
        break;
    }
if (type==1 || type==2){

     Contact contact = null;
    //ArrayList<Contact> contacts = new ArrayList<Contact>();
    Scanner input = new Scanner(System.in);
    System.out.println("Please enter ContactId : ");
    String contactId = input.nextLine();
    System.out.println("Please enter First Name : ");
    String firstName = input.nextLine();
    System.out.println("Please enter Last Name : ");
    String lastName = input.nextLine();
    System.out.println("Please enter Address : ");
    String address = input.nextLine();
    System.out.println("Please enter Phone Number : ");
    String phoneNumber = input.nextLine();
    System.out.println("Please enter Email Address : ");
    String emailAddress = input.nextLine();

    if(type == 1){
       System.out.println("Please enter Birthday: ");
       String dateofBirth = input.nextLine();
       Contact pcontact = new PersonalContact(contactId, firstName, lastName, address, phoneNumber, emailAddress, dateofBirth);
       contacts.add(pcontact);

       for (Contact showcontact: contacts){
           System.out.println(showcontact.displayContact());}
    }

    else if(type == 2){
        System.out.println("Please enter Job Title: ");
        String jobTitle = input.nextLine();
        System.out.println("Please enter Organization: ");
        String organization = input.nextLine();
        Contact bcontact = new BusinessContact(contactId, firstName, lastName, address, phoneNumber, emailAddress, jobTitle, organization);
        contacts.add(bcontact);

        for (Contact showcontact: contacts){
           System.out.println(showcontact.displayContact());}
    }

}
    if(type == 3){
        //System.out.println(contacts);
        for (Contact showcontact: contacts){
            System.out.println(showcontact.displayFullName());}
        }
    }     
}  
}

父类：

public abstract class Contact {

String contactId;
String firstName;
String lastName;
String address;
String phoneNumber;
String emailAddress;

public Contact(String contactId,String firstName,String lastName, String address, String phoneNumber, String emailAddress)
{
    this.contactId = contactId;
    this.firstName = firstName;
    this.lastName = lastName;
    this.address = address;
    this.phoneNumber = phoneNumber;
    this.emailAddress = emailAddress;
}
public void setContactId(String input){
    this.contactId = input;
}
public String getContactId(){
    return contactId;
}

public void setFirstName(String input){
    this.firstName = input;
}
public String getFirstName(){
    return firstName;
}

public void setLastName(String input){
    this.lastName = input;
}
public String getLastName(){
    return lastName;
}

public void setAddress(String input){
    this.address = input;
}
public String getAddress(){
    return address;
}

public void setPhoneNumber(String input){
    this.phoneNumber = input;
}
public String getPhoneNumber(){
    return phoneNumber;
}

public void setEmailAddress(String input){
    this.emailAddress = input;
}
public String getEmailAddress(){
    return emailAddress;
}

@Override
public String toString(){
   return ("ContactID: " + this.getContactId() + "\nFirst Name: " + this.getFirstName() + "\nLast Name: " + this.getLastName() + "\nAddress: " + this.getAddress() + "\nPhone Number: " + this.getPhoneNumber() + "\nEmail Address " + this.getEmailAddress());
}

public String displayFullName(){
    System.out.println("Contact List:");
    return ("ContactID: " + this.getContactId() + "\nFirst Name: " + this.getFirstName() + "\nLast Name: " + this.getLastName());
}

public String displayContact(){
    return ("ContactID: " + this.getContactId() + "\nFirst Name: " + this.getFirstName() + "\nLast Name: " + this.getLastName() + "\nAddress :" + this.getAddress() + "\nPhone Number :" + this.getPhoneNumber() + "\nEmail Address " + this.getEmailAddress());
}

}

子类是个人和商业，他们只是覆盖一些父类，所以不知道他们对这个问题感兴趣。如果需要，我可以发布它们。

Answer 1

我认为在这里使用Map可能更容易。这是一个由键值对组成的数据结构。您可以将您的ID用作密钥，将联系人用作值。一个例子：

Map<String, Contact> contacts = new HashMap<String, Contact>();
contacts.put("anId", aContact);
contacts.put("anotherId", anotherContact);
...
contacts.get("anId"); // returns aContact

迭代它进行打印可以这样做：

for(String contactId : contacts.keySet(){
    System.out.println(contacts.get(contactId).displayFullName());}
}

Answer 2

我还建议您使用Map<String, Contact>，但万一您不想这样做，以下是如何使用您的代码：

System.out.println("Please enter ID of contact: ");
String soughtId = input.nextLine();

for (Contact showcontact: contacts)
{
   if (showcontact.getContactId().equals(soughtId))
           System.out.println(showcontact.displayContact());
}

如何根据输入的字符串从arraylist打印出特定项目

2 个答案: