使用Spring Roo项目并获得org.hibernate.lazyInitializationException：无法初始化代理-没有会话

Question

所涉及的项目：

• memdrill-data
• 数据-使用

memdrill-data是一个Spring Roo管理的项目。这里没有web组件。我只是：

persistence setup --provider HIBERNATE --database HYPERSONIC_PERSISTENT

@RooJavaBean
@RooToString
@RooEntity
public class Item {

    @NotNull
    @Column(unique = true)
    @Size(min = 1, max = 200)
    private String itemId;

    @NotNull
    @Lob
    @OneToOne(cascade = javax.persistence.CascadeType.ALL, fetch = FetchType.LAZY)
    private LobString content;
}

@RooJavaBean
@RooEntity
public class LobString {

    @Lob
    @Basic(fetch=FetchType.LAZY)
    private String data;

    public LobString() {
        super();
    }

    public LobString(String data) {
        this.data = data;
    }
}

<?xml version="1.0" encoding="UTF-8" standalone="no"?>
<beans xmlns="http://www.springframework.org/schema/beans" xmlns:aop="http://www.springframework.org/schema/aop" xmlns:context="http://www.springframework.org/schema/context" xmlns:jee="http://www.springframework.org/schema/jee" xmlns:tx="http://www.springframework.org/schema/tx" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://www.springframework.org/schema/aop http://www.springframework.org/schema/aop/spring-aop-3.0.xsd         http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.0.xsd         http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.0.xsd         http://www.springframework.org/schema/jee http://www.springframework.org/schema/jee/spring-jee-3.0.xsd         http://www.springframework.org/schema/tx http://www.springframework.org/schema/tx/spring-tx-3.0.xsd">

<context:property-placeholder location="classpath*:META-INF/spring/*.properties"/>

<context:spring-configured/>

<context:component-scan base-package="com.memdrill.data">
    <context:exclude-filter expression=".*_Roo_.*" type="regex"/>
    <context:exclude-filter expression="org.springframework.stereotype.Controller" type="annotation"/>
</context:component-scan>
<bean class="org.apache.commons.dbcp.BasicDataSource" destroy-method="close" id="dataSource">
    <property name="driverClassName" value="${database.driverClassName}"/>
    <property name="url" value="${database.url}"/>
    <property name="username" value="${database.username}"/>
    <property name="password" value="${database.password}"/>
    <property name="testOnBorrow" value="true"/>
    <property name="testOnReturn" value="true"/>
    <property name="testWhileIdle" value="true"/>
    <property name="timeBetweenEvictionRunsMillis" value="1800000"/>
    <property name="numTestsPerEvictionRun" value="3"/>
    <property name="minEvictableIdleTimeMillis" value="1800000"/>
</bean>
<bean class="org.springframework.orm.jpa.JpaTransactionManager" id="transactionManager">
    <property name="entityManagerFactory" ref="entityManagerFactory"/>
</bean>
<tx:annotation-driven mode="aspectj" transaction-manager="transactionManager"/>
<bean class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean" id="entityManagerFactory">
    <property name="persistenceUnitName" value="persistenceUnit"/>
    <property name="dataSource" ref="dataSource"/>
</bean>

public static void main(String[] args) {
    ClassPathXmlApplicationContext appContext = new ClassPathXmlApplicationContext(
            "classpath*:META-INF/spring/applicationContext*.xml");

    Item item = new Item();
    item.setItemId("abc123");
    item.setContent(new LobString("this is the content of the item"));

    item.persist();

    System.out.println("persist - OK");

    List<Item> items = Item.findAllItems();

    Item i = items.get(0);
    System.out.println("i.getId() = " + i.getId());
    System.out.println("i.getItemId() = " + i.getItemId());
    System.out.println("i.getContent() = " + i.getContent());
    System.out.println("i.getContent().getId() = " + i.getContent().getId());
    System.out.println("i.getContent().getData() = " + i.getContent().getData());

    appContext.close();
}

public static List<Item> Item.findAllItems() {
    return entityManager().createQuery("SELECT o FROM Item o", Item.class).getResultList();
}

这是我得到的：

persist - OK
i.getId() = 1
i.getItemId() = abc123
2014-07-05 13:23:30,732 [main] ERROR org.hibernate.LazyInitializationException - could not initialize proxy - no Session
org.hibernate.LazyInitializationException: could not initialize proxy - no Session
    at org.hibernate.proxy.AbstractLazyInitializer.initialize(AbstractLazyInitializer.java:167)
    at org.hibernate.proxy.AbstractLazyInitializer.getImplementation(AbstractLazyInitializer.java:215)
    at org.hibernate.proxy.pojo.javassist.JavassistLazyInitializer.invoke(JavassistLazyInitializer.java:190)
    at com.memdrill.data.util.LobString_$$_javassist_0.toString(LobString_$$_javassist_0.java)
    at java.lang.String.valueOf(String.java:2826)
    at java.lang.StringBuilder.append(StringBuilder.java:115)
    at com.memdrill.prototypes.datause.Main.main(Main.java:34)

因此，当我试图访问懒惰加载的数据以填充LobString POJO时，没有会话在进行，会话崩溃。

我最接近（在我的代码中，即，不是我所依赖的库中）放入LobString访问代码（例如，item.getContent().getId（））的方法是插入findAllItems()，或者创建一个类似于这样的新方法来说明这一点：

html" target="_blank">public static List<Item> findAllItemsWithRefs() {
    List<Item> items = entityManager().createQuery("SELECT o FROM Item o", Item.class).getResultList();
    for(Item item : items) {
        System.out.println("item.getContent().getId() = " + item.getContent().getId());
    }
    return items;
}

正如其他SO文章所建议的那样，我尝试将@Transactional置于该方法之上，但仍然得到了LazyInitializationException。但即使这真的有效...为什么我需要启动一个事务来从数据库中读取数据？事务难道不是为了在编写时保持数据完整性吗？

在我看来，解决这个问题的一个方法是以某种方式指定一个不同于“从项目o中选择o”的查询...一个还获取LobString数据的...但这似乎不太正确，即使它能在这种特殊情况下飞行。

当我访问它还没有加载的数据时，我希望Hibernate开始一个新的会话...但很明显，它并不像我期望的那样起作用。

顺便说一句，如果我将以下log4j.propeties添加到类路径：

log4j.rootLogger=error, stdout

log4j.appender.stdout=org.apache.log4j.ConsoleAppender
log4j.appender.stdout.layout=org.apache.log4j.PatternLayout

# Print the date in ISO 8601 format
log4j.appender.stdout.layout.ConversionPattern=%d [%t] %-5p %c - %m%n

log4j.logger.org.hibernate.SQL=DEBUG
log4j.logger.org.hibernate.type=TRACE

我得到：

persist - OK
2014-07-05 16:17:35,707 [main] DEBUG org.hibernate.SQL - select item0_.id as id1_, item0_.content as content1_, item0_.item_id as item2_1_, item0_.version as version1_ from item item0_
2014-07-05 16:17:35,711 [main] TRACE org.hibernate.type.descriptor.sql.BasicExtractor - found [1] as column [id1_]
2014-07-05 16:17:35,713 [main] TRACE org.hibernate.type.descriptor.sql.BasicExtractor - found [1] as column [content1_]
2014-07-05 16:17:35,713 [main] TRACE org.hibernate.type.descriptor.sql.BasicExtractor - found [abc123] as column [item2_1_]
2014-07-05 16:17:35,713 [main] TRACE org.hibernate.type.descriptor.sql.BasicExtractor - found [0] as column [version1_]
i.getId() = 1
i.getItemId() = abc123
2014-07-05 16:17:35,717 [main] ERROR org.hibernate.LazyInitializationException - could not initialize proxy - no Session

因此，正如“Found 1 as column[content1_]”一行所示，我至少在包含LobString数据的表中有行的外键······但是如果没有获得LazyInitializationException，我就无法访问该id。如果我能找到它，也许我可以通过身份证找到龙虾串...但是i.getContent（）和i.getContent（）.getID（）给出了例外情况。外键应该与LobString表的主键id匹配，即i.getContent().getID（）。但是似乎没有办法从i...如日志所示，实际上是从数据库中提取的。

        // Invoke method on actual Query object.
        try {
            Object retVal = method.invoke(this.target, args);
            return (retVal == this.target ? proxy : retVal);
        }
        catch (InvocationTargetException ex) {
            throw ex.getTargetException();
        }
        finally {
            if (method.getName().equals("getResultList") || method.getName().equals("getSingleResult") ||
                    method.getName().equals("executeUpdate")) {
                EntityManagerFactoryUtils.closeEntityManager(this.em);
            }
        }

首先，如果使用“GetResultList”，则EntityManager将被关闭。我假设EntityManager是一个JPA结构，它映射到Hibernate的会话，但不详细说明。

二...如果我在调试时钻得足够深，我可以看到什么应该懒洋洋地加载数据“这是项目的内容”...所以我实际上把数据加载到内存里了...1）因为它是懒惰的，所以不应该2）即使它在那里，我也不能访问它，因为我得到了LazyInitializationException。

在调试时，我可以看到数据加载在以下两个地方：

懒惰加载的数据到此为止...我到底做错了什么？

我似乎也懒洋洋地加载了LobString的规范。但如果是这样的话，而我确实记忆中有这些内容...为什么我不能访问它？

编辑3：

    EntityManager em = Item.entityManager();
    List<Item> items = em.createQuery("SELECT o FROM Item o JOIN FETCH o.content", Item.class).getResultList();

2014-07-06 21:44:56,862 [main] INFO  org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean - Building JPA container EntityManagerFactory for persistence unit 'persistenceUnit'
in LobString default constructor
in LobString (String data) constructor
2014-07-06 21:44:57,891 [main] DEBUG org.springframework.orm.jpa.JpaTransactionManager - Creating new transaction with name [com.memdrill.data.entity.Item.persist]: PROPAGATION_REQUIRED,ISOLATION_DEFAULT; ''
2014-07-06 21:44:57,944 [main] DEBUG org.springframework.orm.jpa.JpaTransactionManager - Opened new EntityManager [org.hibernate.ejb.EntityManagerImpl@779d9c0d] for JPA transaction
2014-07-06 21:44:57,948 [main] DEBUG org.springframework.orm.jpa.JpaTransactionManager - Not exposing JPA transaction [org.hibernate.ejb.EntityManagerImpl@779d9c0d] as JDBC transaction because JpaDialect [org.springframework.orm.jpa.DefaultJpaDialect@d7e60a1] does not support JDBC Connection retrieval
2014-07-06 21:44:57,999 [main] DEBUG org.springframework.orm.jpa.JpaTransactionManager - Initiating transaction commit
2014-07-06 21:44:58,000 [main] DEBUG org.springframework.orm.jpa.JpaTransactionManager - Committing JPA transaction on EntityManager [org.hibernate.ejb.EntityManagerImpl@779d9c0d]
2014-07-06 21:44:58,002 [main] DEBUG org.springframework.orm.jpa.JpaTransactionManager - Closing JPA EntityManager [org.hibernate.ejb.EntityManagerImpl@779d9c0d] after transaction
2014-07-06 21:44:58,002 [main] DEBUG org.springframework.orm.jpa.EntityManagerFactoryUtils - Closing JPA EntityManager
persist - OK - in Main of memdrill-data not data-use
2014-07-06 21:44:58,008 [main] DEBUG org.springframework.orm.jpa.SharedEntityManagerCreator$SharedEntityManagerInvocationHandler - Creating new EntityManager for shared EntityManager invocation
in LobString default constructor
2014-07-06 21:44:58,133 [main] DEBUG org.springframework.orm.jpa.EntityManagerFactoryUtils - Closing JPA EntityManager
i.getId() = 1
i.getItemId() = abc123
i.getContent() = com.memdrill.data.util.LobString@6fe30af
i.getContent().getId() = 1
i.getContent().getData() = this is the content of the item
2014-07-06 21:44:58,134 [main] INFO  org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean - Closing JPA EntityManagerFactory for persistence unit 'persistenceUnit'

这几乎是我想要的。我只想要龙虾串的ID。这个想法是，这将是一个REST调用，我只想返回项“基本”数据和LobString的ID，它们可以用于单独的请求。

知道什么样的JPQL查询可以实现这一点吗？（如有任何有助于理解幕后发生的事情的评论，将不胜感激）-谢谢

Answer 1

要解决这个问题，可以在非静态方法上使用@transactional注释。

示例：

@RooJavaBean
@RooToString
@RooEntity
public class Item {

    @NotNull
    @Column(unique = true)
    @Size(min = 1, max = 200)
    private String itemId;

    @NotNull
    @Lob
    @OneToOne(cascade = javax.persistence.CascadeType.ALL, fetch = FetchType.LAZY)
    private LobString content;

    @Transactional(readOnly=true)
    LobString loadContent(){
       getContent().getId();
       return getContent();
    }
}

然后，使用LoadContent()方法获取LobString。