Question

我需要帮助。

我有一个完整的ArrayList，其中包含客户/客户的名称。我想要做的是编写一个代码，以便它询问我的客户端名称然后我输入它，如果它发现它添加一个电话号码给客户端。一旦添加，如果我想打印出列表，电话号码将在该特定客户端旁边。

我有3节课。 1是客户端类（带有变量 - 名称，地址，编号）和JobManager（整个测试程序，其中包括添加客户端和删除客户端等方法）。

这是我添加客户端的方式：

  public static void addClient() {

    System.out.println("Adding a client: \n");

    System.out.print("What is the name of the new client?: ");
    String name = keyboard.nextLine();

    System.out.println();

    if (name.length() == 0) {
        boolean invalid = true;

        while (invalid) {
            System.out.println("You cannot enter a blank name for the client. ");
            System.out.print("Please enter a valid name: "); 
            name = keyboard.nextLine();

            if (name.length() > 0) {
                invalid = false; 
            }
        }
    }

    System.out.print("What is the client's address?: ");
    String address = keyboard.nextLine();

    boolean isDuplicate = false;

    for (int i = 0; i < clientList.size(); i++) {
        String listName = clientList.get(i).getName();
        String listAddress = clientList.get(i).getAddress();

    if (listName.equalsIgnoreCase(name)
                && listAddress.equalsIgnoreCase(address)) {
            isDuplicate = true;

        if (listName.equalsIgnoreCase(name)) {
            isDuplicate = true;

        }
    }

    if (isDuplicate) {

        System.out.println();
        System.out.println("Error, that client already exists! \n");
        isDuplicate = false;

    } else {

        Client client = new Client(name, address);
        clientList.add(client);
        System.out.println();
        System.out.println("Client has been added. \n");

      }
    }
 }

这就是我的客户端类：

public class Client {

    private ArrayList<JobManager> jobManager;
    private ArrayList<TelephoneNumber> telephoneNumbers = new ArrayList<>();

    private String name;
    private String address;
    private static int number;

    public Client(String name,String address) {

        this.name = name;
        this.address = address;     
    }

    // public void createNumber() {
    //  TelephoneNumber teleNumber = new TelephoneNumber(number);
    //  telephoneNumbers.add(teleNumber);

    //}

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public String getAddress() {
        return address;
    }

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

    public static int getNumber() {
        return number;
    }

    public void setNumber(int number) {
        this.number = number;
    }

    public String toString() {
        String result;

        result = name + "\n" + address;

        result = (this.getName() + "\t" + this.getAddress());

        return result;
    }
}

先谢谢你的回答！

Answer 1

你可以在这里完成：

if (listName.equalsIgnoreCase(name)
            && listAddress.equalsIgnoreCase(address)) {
        isDuplicate = true;
        do {
            TelephoneNumber phoneNumber = ..;//ask user for phone number and keep repeating until user enters 0 as one user can have multiple numbers

            clientList.get(i).addNumber(phoneNumber);
        } while (!phoneNumber.getNumber().equals("0"))


    //if (listName.equalsIgnoreCase(name)) { //removed duplicate conditions
      //  isDuplicate = true;

    //}
}

在您的Client类中添加addNumber以将数字列表添加为：

public void addNumber(TelephoneNumber phoneNumber) {
    telephoneNumbers.add(phoneNumber);
}

Answer 2

如果您想先将电话号码添加到特定客户端，您必须在数据库中搜索该客户端：

public Client searchByName(String name){// find one client and return client;}

然后设置号码：

public void addNumber(int number){ Client client=searchByName(); client.setNumber(number);}

Answer 3

public void addNumberToClients(String clientName, TelephoneNumber number) {
    clientList.stream()
        .filter(client -> client.getName().equals(clientName))
        .forEach(client -> client.addNumber(number);
}

这会将数字添加到具有给定名称的所有客户端。要将其添加到一个，请将.forEach()更改为.findFirst().ifPresent()

然后将方法添加到Client

public void addNumber(TelephoneNumber number) {
    telephoneNumbers.add(number);
}

Answer 4

我已经从您的示例中开发了一小段代码示例，其中我很难enter code here编码电话号码，并且在主方法中我正在硬编码2，因为我在测试场景中输入用户输入为V3。 / p>

客户等级

public class Client {

private  String name;
private List<String> telephoneNumber;

Client(String name)
{
    this.name = name;
}

public List<String> getTelephoneNumber() {
    return telephoneNumber;
}

public void setTelephoneNumber(List<String> telephoneNumber) {
    this.telephoneNumber = telephoneNumber;
}

public String getName() {
    return name;
}

public void setName(String name) {
    this.name = name;
}

public boolean equals(Object obj)
{
    Client c = (Client)obj;
    return this.name.equals(c.name);
}

<强>识别TestClass

public class HelloWorld{
Scanner keyboard = new Scanner(System.in);



public static void main(String[] args) throws Exception {

    List<Client> clientList = new ArrayList<Client>();
    Client c1 = new Client("V1");
    Client c2 = new Client("V2");
    Client c3 = new Client("V3");
    Client c4 = new Client("V4");

    clientList.add(c1);
    clientList.add(c2);
    clientList.add(c3);
    clientList.add(c4);

    HelloWorld h = new HelloWorld();
    h.addTelephone(clientList);

    Client cTest = clientList.get(2);
    System.out.println(cTest.getTelephoneNumber());





}
public void addTelephone(List<Client> clientList)
{
    System.out.print("What is the name of the new client?: ");
    String name = keyboard.nextLine();
    Client c = new Client(name);
    if(clientList.contains(c))
    {//Client Already Exist
        int i =  clientList.indexOf(c);
        Client c1 = clientList.get(i);
        List<String> telNumber = new ArrayList<String>();
        getTelephoneNumbers(telNumber);
        c1.setTelephoneNumber(telNumber);
    }

}
}
public  void getTelephoneNumbers(List<String> telephone)
{
    boolean isExit = false;

    while(!isExit)
    {
        System.out.println("Enter Telephone Number  or 0 for exiting : ");
        String telp = keyboard.nextLine();
        if("0".equalsIgnoreCase(telp)) {
            System.out.println("Exiting");
            isExit = true;
        }
        else {
            telephone.add(telp);
        }
    }
}

您可以使用相同类型的addTelephone数字方法。

如何向ArrayList中的现有对象添加信息？

4 个答案: