Question

我正在使用Spring Roo基于表（在MS SQL Server上创建）生成实体。以下是我的sql脚本：

CREATE TABLE [dbo].[JMSMessage](
     [MessageID] [nvarchar](100) NOT NULL,
     [Destination] [varchar](50) NOT 
     [Status] [varchar](15) NULL,
CONSTRAINT [PK_JMSMessage] PRIMARY KEY CLUSTERED 
(
    [MessageID] ASC
) WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF,ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]

Spring Roo生成的Java实体

@Entity
@Table(schema = "dbo",name = "JMSMessage")
@Configurable
public class Jmsmessage {

    @Column(name = "Destination", length = 50)
    @NotNull
    private String destination;

    @Column(name = "Status", length = 15)
    private String status;

    public String getDestination() {
        return destination;
    }

    public void setDestination(String destination) {
        this.destination = destination;
    }

    public String getStatus() {
        return status;
    }

    public void setStatus(String status) {
        this.status = status;
    }

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    @Column(name = "MessageID", length = 100)
    private String messageId;

    public String getMessageId() {
        return this.messageId;
    }

    public void setMessageId(String id) {
        this.messageId = id;
    }

    @PersistenceContext
    transient EntityManager entityManager;

    public static final EntityManager entityManager() {
        EntityManager em = new Jmsmessage().entityManager;
        if (em == null) throw new IllegalStateException("Entity manager has not been injected (is the Spring Aspects JAR configured as an AJC/AJDT aspects library?)");
        return em;
    }

    public static Jmsmessage findJmsmessage(String messageId) {
        if (messageId == null || messageId.length() == 0) return null;
        return entityManager().find(Jmsmessage.class, messageId);
    }

    @Transactional
    public void persist() {
        if (this.entityManager == null) this.entityManager = entityManager();
        this.entityManager.persist(this);
    }

    @Transactional
    public void remove() {
        if (this.entityManager == null) this.entityManager = entityManager();
        if (this.entityManager.contains(this)) {
            this.entityManager.remove(this);
        } else {
            Jmsmessage attached = Jmsmessage.findJmsmessage(this.messageId);
            this.entityManager.remove(attached);
        }
    }

    @Transactional
    public void flush() {
        if (this.entityManager == null) this.entityManager = entityManager();
        this.entityManager.flush();
    }

    @Transactional
    public void clear() {
        if (this.entityManager == null) this.entityManager = entityManager();
        this.entityManager.clear();
    }

    @Transactional
    public Jmsmessage merge() {
        if (this.entityManager == null) this.entityManager = entityManager();
        Jmsmessage merged = this.entityManager.merge(this);
        this.entityManager.flush();
        return merged;
    }

    public String toString() {
        return ReflectionToStringBuilder.toString(this, ToStringStyle.SHORT_PREFIX_STYLE);
    }
}

然后，当我运行以下代码来创建实体实例并尝试持久化时，程序抛出一个PersistentObjectException，说“传递给持久化的分离实体”。

Jmsmessage message = new Jmsmessage();
message.setMessageId("m1");
message.setDestination("queue1");
message.setStatus("status1");
message.persist(); // If i use message.merge(), it will throws another Sql exception saying "Cannot insert the value NULL into column 'MessageID'"

我如何能够坚持不懈地工作，有什么建议吗？提前谢谢！

Answer 1

您不应为messageId设置值，因为您的策略为@GeneratedValue(strategy = GenerationType.AUTO)

另一方面，我注意到您使用了roo提供的活动记录模式。我建议您将此更改为DAO模式，因为这样的代码在Java / Spring / Hibernate世界中非常罕见：）

Answer 2

您拥有ID生成策略自动，但您已在实体中分配了一些ID值。意思是ORM引擎会将您的新实体视为已分离的实体，因为它获得了id并且它会像尝试保存分离的实体一样抛出异常..如果您尝试合并实体以克服问题，ORM引擎将检查您提供给它的ID实体，它不能匹配任何行与您的实体匹配，因为没有这样的行与主键。
错误的底线是，您将ORM引擎与ID生成混淆。

希望这很有用！

Hibernate＆amp; Spring Roo：org.hibernate.PersistentObjectException：传递给persist的分离实体

2 个答案: