我已经编写了该成员资格类，如何创建找到给成员的特定ID的方法？

Question

Guys this is my membership class so far, i am struggling to create a method that finds the full members details that i have given just using a uniqueId finder. Please help.

public class Membership {

    private String FirstName;
    private String LastName;
    private int memberId;
    private String listOfMembers;
    private int uniqueId;
    private long phoneNumber;



    public Membership(String FirstName, String LastName, int uniqueId,
    long phoneNumber)
    {


       this.uniqueId = uniqueId;
       this.FirstName = FirstName;
       this.LastName = LastName;
       this.phoneNumber = phoneNumber;
    }


    public String getMember()
    {
        return FirstName + LastName;
    }
    public String getlistOfMembers()
    {
        return (FirstName + LastName);
    }
    public int getId()
    {
      return uniqueId;
    }

    public void MemberId (int Id)
    {
        System.out.println("Id" + Id);
    }

    public String getMemberDetails ()
    {
        System.out.println("Member Id: " + uniqueId);
        System.out.println("first name:  " + FirstName);
        System.out.println("LastName:  " + LastName);
        System.out.println("Member phone number:  " + phoneNumber);
        return listOfMembers;
    }


}

This is what i have done so far.

Answer 1

Issues:

• You've got user interface code where it doesn't belong. I would remove all System.out.println statements from this class and instead leave it in a UI class or main method (if very simple).
• In particular, getter methods should return field values, and should not have System.out.println statements
• I'm not sure why this class has a listOfMembers field, or why it's just a String. You look to be trying to combine Member and Membership together in one single class -- Don't do this.
• I'd name this class Member since it holds information for just a single Member.
• If I needed a Membership class, it would instead hold an ArrayList<Member>
• And it would have a public Member getMember(int id) method that would return the item in the list above that shares the id passed into the method. A simple for loop that iterated through the list, comparing id's would suffice.

Answer 2

To add on Hovercraft's answer with an example.

You have your class handling all the members, very basic implementation of it.

public class Membership {
    private final Map<Integer, Member> members = new HashMap<>();

    public void addMember (Integer uniqueId, Member member) {
        members.put (uniqueId, member);
    }

    public void getMember (Integer uniqueId) {
        return members.get (uniqueId);
    }

    ...
}

Then you have the Member s themselves like this, more fields can be added as you want them.

public class Member {
    private String firstName;
    private String lastName;

    public Member (String firstName, String lastName) {
        this.firstName = firstName;
        this.lastName = lastName;
    }

    public String getFirstName () {
        return firstName;
    }

    ...
}

This is a very basic, but strong, feature in OOP to use.

Again see Hovercraft's answer as it provides all the details. If they were to edit/remove I will update this one.

Map vs List

One minor thing is I'd vote against using an ArrayList<E> to store the Member s. If you add to the implementation that you can remove users the uniqueId will shift from user to user. Instead I would be for making sure that you are not adding to an existing user.

If you want to keep it simple and just get going, an ArrayList<E> works, do know the problem you might get in the feature, an uniqueId is not necessarily tied to a Member .

"I am quite new to java and have never come across "map" can you please explain what it is?"

"An object that maps keys to values. A map cannot contain duplicate keys; each key can map to at most one value." - From: Documentation .

Instead of working with direct indexes as you do in an Array :

• arr[5]; // here you get the value at index position 5.

Or like a List :

• list.get(5); // here you get the fifth element, it can be stored (almost) anywhere in the memory, before or after 4, doesn't matter, as 4 knows where 5 is.

And for a Map :

• map.get(5); // you get the object stored at 5, there might not be a 3 or 4 in the Map . You can store any Object s as anything. A String is another example of a common key .

Answer 3

I would suggest to use Map and use id as key of Map and store object of Membership as Value,thereby easy to retrieve and store also.

Something similar to this,

 Map<Integer,Membership> map = new HashMap<Integer,Membership>();
    Membership m = new Membership("First", "LastName", 1,1234567890);
    map.put(m.getId(), m);

To get member by id,

 System.out.println(map.get(id).getMemberDetails());