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我已经使用 JPA(实现 Hibernate)有一段时间了，每次我需要创建实体时，我都会发现自己在解决诸如 AccessType、不可变属性、equals/hashCode 等问题。
所以我决定尝试找出每个问题的一般最佳实践，并将其写下来供个人使用。
然而，我不介意任何人对此发表评论或告诉我我错在哪里。

实体类

@Entity
@Table(name = "ROOM")
public class Room implements Serializable {

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue
    @Column(name = "room_id")
    private Integer id;

    @Column(name = "number") 
    private String number; //immutable

    @Column(name = "capacity")
    private Integer capacity;

    @ManyToOne(fetch = FetchType.LAZY, optional = false)
    @JoinColumn(name = "building_id")
    private Building building; //immutable

    Room() {
        // default constructor
    }

    public Room(Building building, String number) {
        // constructor with required field
        notNull(building, "Method called with null parameter (application)");
        notNull(number, "Method called with null parameter (name)");

        this.building = building;
        this.number = number;
    }

    @Override
    public boolean equals(final Object otherObj) {
        if ((otherObj == null) || !(otherObj instanceof Room)) {
            return false;
        }
        // a room can be uniquely identified by it's number and the building it belongs to; normally I would use a UUID in any case but this is just to illustrate the usage of getId()
        final Room other = (Room) otherObj;
        return new EqualsBuilder().append(getNumber(), other.getNumber())
                .append(getBuilding().getId(), other.getBuilding().getId())
                .isEquals();
        //this assumes that Building.id is annotated with @Access(value = AccessType.PROPERTY) 
    }

    public Building getBuilding() {
        return building;
    }


    public Integer getId() {
        return id;
    }

    public String getNumber() {
        return number;
    }

    @Override
    public int hashCode() {
        return new HashCodeBuilder().append(getNumber()).append(getBuilding().getId()).toHashCode();
    }

    public void setCapacity(Integer capacity) {
        this.capacity = capacity;
    }

    //no setters for number, building nor id

}

最佳答案

JPA 2.0 Specification指出:

• The entity class must have a no-arg constructor. It may have other constructors as well. The no-arg constructor must be public or protected.
• The entity class must a be top-level class. An enum or interface must not be designated as an entity.
• The entity class must not be final. No methods or persistent instance variables of the entity class may be final.
• If an entity instance is to be passed by value as a detached object (e.g., through a remote interface), the entity class must implement the Serializable interface.
• Both abstract and concrete classes can be entities. Entities may extend non-entity classes as well as entity classes, and non-entity classes may extend entity classes.

关于java - 创建完美的 JPA 实体，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/6033905/