如何解决“惰性初始化角色集合失败”Hibernate异常

我有这样一个问题:

org.hibernate.LazyInitializationException:惰性初始化role: mvc3.model.Topic.comments集合失败，没有会话或会话已关闭

下面是模型:

@Entity
@Table(name = "T_TOPIC")
public class Topic {

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

    @ManyToOne
    @JoinColumn(name="USER_ID")
    private User author;

    @Enumerated(EnumType.STRING)    
    private Tag topicTag;

    private String name;
    private String text;

    @OneToMany(mappedBy = "topic", cascade = CascadeType.ALL)
    private Collection<Comment> comments = new LinkedHashSet<Comment>();

    ...

    public Collection<Comment> getComments() {
           return comments;
    }

}

调用model的控制器如下所示:

@Controller
@RequestMapping(value = "/topic")
public class TopicController {

    @Autowired
    private TopicService service;

    private static final Logger logger = LoggerFactory.getLogger(TopicController.class);


    @RequestMapping(value = "/details/{topicId}", method = RequestMethod.GET)
    public ModelAndView details(@PathVariable(value="topicId") int id)
    {

            Topic topicById = service.findTopicByID(id);
            Collection<Comment> commentList = topicById.getComments();

            Hashtable modelData = new Hashtable();
            modelData.put("topic", topicById);
            modelData.put("commentList", commentList);

            return new ModelAndView("/topic/details", modelData);

     }

}

jsp页面看起来如下所示:

<%@page import="com.epam.mvc3.helpers.Utils"%>
<%@ page language="java" contentType="text/html; charset=UTF-8" pageEncoding="UTF-8"%>
<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %>
<%@ page session="false" %>
<html>
<head>
      <title>View Topic</title>
</head>
<body>

<ul>
<c:forEach items="${commentList}" var="item">
<jsp:useBean id="item" type="mvc3.model.Comment"/>
<li>${item.getText()}</li>

</c:forEach>
</ul>
</body>
</html>

在查看jsp时，将引发异常。在c:forEach循环的行中

当前回答

我发现将@PersistenceContext声明为EXTENDED也解决了这个问题:

@PersistenceContext(type = PersistenceContextType.EXTENDED)

2015-06-24 10:24:29

其他回答

我发现将@PersistenceContext声明为EXTENDED也解决了这个问题:

@PersistenceContext(type = PersistenceContextType.EXTENDED)

2015-06-24 10:24:29

为了延迟加载集合，必须有一个活动会话。在web应用中，有两种方法可以做到这一点。您可以使用Open Session In View模式，其中使用拦截器在请求开始时打开会话，并在结束时关闭会话。风险在于你必须有可靠的异常处理，否则你可能会绑定所有的会话，你的应用程序可能会挂起。

The other way to handle this is to collect all the data you need in your controller, close your session, and then stuff the data into your model. I personally prefer this approach, as it seems a little closer to the spirit of the MVC pattern. Also if you get an error from the database this way you can handle it a lot better than if it happens in your view renderer. Your friend in this scenario is Hibernate.initialize(myTopic.getComments()). You will also have to reattach the object to the session, since you're creating a new transaction with every request. Use session.lock(myTopic,LockMode.NONE) for that.

2012-07-31 20:13:23

这是我最近遇到的问题，我用

<f:attribute name="collectionType" value="java.util.ArrayList" />

这里有更详细的描述，这挽救了我的一天。

2013-11-07 15:22:57

最好的解决方案之一是在应用程序中添加以下内容。属性文件: spring.jpa.properties.hibernate.enable_lazy_load_no_trans = true

2018-08-01 23:24:38

通过使用hibernate @Transactional注释，如果你从数据库中获得一个具有惰性获取属性的对象，你可以像这样简单地获取这些属性:

@Transactional
public void checkTicketSalePresence(UUID ticketUuid, UUID saleUuid) {
        Optional<Ticket> savedTicketOpt = ticketRepository.findById(ticketUuid);
        savedTicketOpt.ifPresent(ticket -> {
            Optional<Sale> saleOpt = ticket.getSales().stream().filter(sale -> sale.getUuid() == saleUuid).findFirst();
            assertThat(saleOpt).isPresent();
        });
}

这里，在Hibernate代理管理的事务中，调用ticket.getSales()执行另一个查询来获取sales，因为您显式地请求了它。

2018-07-25 09:53:53

如何解决“惰性初始化角色集合失败”Hibernate异常

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