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(){..}
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.
}
}
}
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