设置Id（PK）生成值auto和manual

我想将用户持久化到数据库以及使用IDENTITY生成类型创建的用户ID（PK）的当前场景。 例如

@Entity @Table(name = "USER_PROFILES", uniqueConstraints = @UniqueConstraint(columnNames = "USERNAME")) public class UserProfiles implements java.io.Serializable { private Long id; private String username; private String password; public UserProfiles() { } @Id @GeneratedValue(strategy=GenerationType.IDENTITY) @Column(name = "ID", unique = true, nullable = false, precision = 20, scale = 0) public Long getId() { return this.id; } public void setId(Long id) { this.id = id; } @Column(name = "USERNAME", unique = true, nullable = false, length = 32) public String getUsername() { return this.username; } public void setUsername(String username) { this.username = username; } @Column(name = "PASSWORD", nullable = false, length = 32) public String getPassword() { return this.password; } public void setPassword(String password) { this.password = password; } } 

但我想在以下场景中Create Id(PK) ：1）用户明确设置Id(PK) 。 2）如果用户未设置Id(PK) ，则会自动分配，并且必须是唯一的。

请建议我一些可用的选项，以便我可以解决它。 谢谢。

Answer 1

您可以为此目的定义自定义ID生成器，如本SO答案中所述

以下是其代码的外观： –

 @Id @Basic(optional = false) @GeneratedValue(strategy=GenerationType.IDENTITY, generator="IdOrGenerated") @GenericGenerator(name="IdOrGenerated",strategy="....UseIdOrGenerate") @Column(name = "ID", unique = true, nullable = false, precision = 20, scale = 0) public Long getId(){..} 

和

  public class UseIdOrGenerate extends IdentityGenerator { @Override public Serializable generate(SessionImplementor session, Object obj) throws HibernateException { if (obj == null) throw new HibernateException(new NullPointerException()) ; if ((((EntityWithId) obj).getId()) == null) {//id is null it means generate ID Serializable id = super.generate(session, obj) ; return id; } else { return ((EntityWithId) obj).getId();//id is not null so using assigned id. } } }