我有一个包含多对一关系的jpa持久化对象模型:一个Account有多个transaction。一个事务有一个帐户。

下面是一段代码:

@Entity
public class Transaction {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @ManyToOne(cascade = {CascadeType.ALL},fetch= FetchType.EAGER)
    private Account fromAccount;
....

@Entity
public class Account {
    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;
    @OneToMany(cascade = {CascadeType.ALL},fetch= FetchType.EAGER, mappedBy = "fromAccount")
    private Set<Transaction> transactions;

我能够创建Account对象,向其添加事务,并正确地持久化Account对象。但是,当我创建一个事务,使用现有的已经持久化的帐户,并持久化的事务,我得到一个异常:

导致:org.hibernate.PersistentObjectException:传递给persist: com.paulsanwald.Account的分离实体 org.hibernate.event.internal.DefaultPersistEventListener.onPersist (DefaultPersistEventListener.java: 141)

因此,我能够持久化一个包含事务的Account,但不能持久化一个具有Account的Transaction。我认为这是因为帐户可能没有附加,但这段代码仍然给了我相同的异常:

if (account.getId()!=null) {
    account = entityManager.merge(account);
}
Transaction transaction = new Transaction(account,"other stuff");
 // the below fails with a "detached entity" message. why?
entityManager.persist(transaction);

如何正确地保存与已经持久化的帐户对象相关联的事务?


当前回答

使用合并是有风险和棘手的,所以在您的情况下,这是一种肮脏的变通方法。您至少需要记住,当您将一个实体对象传递给merge时,它将停止附加到事务,而是返回一个新的、现在已附加的实体。这意味着如果任何人仍然拥有旧的实体对象,那么对它的更改将被无声地忽略并在提交时丢弃。

You are not showing the complete code here, so I cannot double-check your transaction pattern. One way to get to a situation like this is if you don't have a transaction active when executing the merge and persist. In that case persistence provider is expected to open a new transaction for every JPA operation you perform and immediately commit and close it before the call returns. If this is the case, the merge would be run in a first transaction and then after the merge method returns, the transaction is completed and closed and the returned entity is now detached. The persist below it would then open a second transaction, and trying to refer to an entity that is detached, giving an exception. Always wrap your code inside a transaction unless you know very well what you are doing.

使用容器管理的事务,它看起来像这样。注意:这假设方法在会话bean中,并通过本地或远程接口调用。

@TransactionAttribute(TransactionAttributeType.REQUIRED)
public void storeAccount(Account account) {
    ...

    if (account.getId()!=null) {
        account = entityManager.merge(account);
    }

    Transaction transaction = new Transaction(account,"other stuff");

    entityManager.persist(account);
}

其他回答

@OneToMany(mappedBy = "xxxx", cascade={CascadeType.MERGE, CascadeType.PERSIST, CascadeType.REMOVE})

为我工作。

可能在这种情况下,您使用merge逻辑获得了帐户对象,而persist用于持久化新对象,如果层次结构有一个已经持久化的对象,它将报错。在这种情况下,应该使用saveOrUpdate,而不是持久化。

在您的实体定义中,您没有为联接到事务的帐户指定@JoinColumn。你会想要这样的东西:

@Entity
public class Transaction {
    @ManyToOne(cascade = {CascadeType.ALL},fetch= FetchType.EAGER)
    @JoinColumn(name = "accountId", referencedColumnName = "id")
    private Account fromAccount;
}

编辑:嗯,我想如果您在类上使用@Table注释,那么这将是有用的。哈。:)

我基于Spring Data jpa的回答是:我只是在外部方法中添加了一个@Transactional注释。

为什么它有效

由于没有活动的Hibernate Session上下文,子实体立即被分离。提供一个Spring (Data JPA)事务可以确保存在Hibernate会话。

参考:

https://vladmihalcea.com/a-beginners-guide-to-jpa-hibernate-entity-state-transitions/

这是我的药。

Below is my Entity. Mark that the id is annotated with @GeneratedValue(strategy = GenerationType.AUTO), which means that the id would be generated by the Hibernate. Don't set it when entity object is created. As that will be auto generated by the Hibernate. Mind you if the entity id field is not marked with @GeneratedValue then not assigning the id a value manually is also a crime, which will be greeted with IdentifierGenerationException: ids for this class must be manually assigned before calling save()

@Entity
@Data
@NamedQuery(name = "SimpleObject.findAll", query="Select s FROM SimpleObject s")
public class SimpleObject {

    @Id
    @GeneratedValue(strategy = GenerationType.AUTO)
    private Long id;

    @Column
    private String key;

    @Column
    private String value;

}

这是我的主类。

public class SimpleObjectMain {

    public static void main(String[] args) {

        System.out.println("Hello Hello From SimpleObjectMain");

        SimpleObject simpleObject = new SimpleObject();
        simpleObject.setId(420L); // Not right, when id is a generated value then no need to set this.
        simpleObject.setKey("Friend");
        simpleObject.setValue("Bani");

        EntityManager entityManager = EntityManagerUtil.getEntityManager();
        entityManager.getTransaction().begin();
        entityManager.persist(simpleObject);
        entityManager.getTransaction().commit();

        List<SimpleObject> simpleObjectList = entityManager.createNamedQuery("SimpleObject.findAll").getResultList();
        for(SimpleObject simple : simpleObjectList){
            System.out.println(simple);
        }

        entityManager.close();
        
    }
}

我想救它的时候,它把它扔出去了

PersistentObjectException: detached entity passed to persist.

我所需要修复的是删除主方法中simpleObject的id设置行。