自动和手动设置ID(PK)生成值

我想将用户持久保存到数据库中,并使用IDENTITY生成类型创建的User的ID(PK)的当前方案。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Create

Id(PK)在以下情况下进行操作:1)用户Id(PK)明确设置。2)如果未设置用户,Id(PK)则将自动分配它,并且它必须是唯一的。

请建议我一些可用的选项,以便我解决。谢谢。

回答:

您可以为此目的定义您的自定义ID生成器,如本SOAnswer中指出的那样

它的代码如下所示:

@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(){..}

  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.

}

}

}

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