java - JPA递归延迟加载失败

Question

我有一个名为 Menu 的实体有一个 parent和多个children我使用 JPA 加载。一切正常，当 children设置为@OneToMany(mappedBy = "parent", fetch = FetchType.EAGER) ，但如果我将其设置为 FetchType.LAZY ，我收到错误:

org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: com.mydom.myapp.domain.Menu.children, could not initialize proxy - no Session

我使用Hibernate 4.3.1.Final、JPA、Wildfly 10.1.0作为服务器、MySQL作为数据库。

我已经尝试过 Hibernate.initialize(menu.getChildren()) 的技巧以及调用 size() 方法，如 menu.getChildren().size() 中所示，但在这两种情况下我都会导致相同的错误。

我也尝试设置 @Fetch(FetchMode.JOIN)这是有效的，但它非常类似于使用 FetchType.Eager ，因此没有用。两者@Fetch(FetchMode.SELECT)和@Fetch(FetchMode.SUBSELECT)行不通。

我必须做什么才能延迟加载menu.getChildren() ？

所以，下面我有三个类来处理这个问题。方法MenuManagerController.buildNodeStructureFromMenu(TreeNode root, List<Menu> menus)执行递归工作。这是延迟加载发生的地方。

菜单.java:

@Entity
@NamedQueries({
    @NamedQuery(name = Menu.FIND_MENU_ROOT,
                query = "SELECT m FROM Menu m WHERE m.parent.id IS NULL")
})
public class Menu implements Serializable {
    private static final String PREFIX = "Menu.";
    public static final String FIND_MENU_ROOT = PREFIX + "findMenuRoot";

    @Id
    @GeneratedValue
    private long id;
    @ManyToOne
    private Languages languageType;
    @Column(length = 100)
    private String nameDisplay;
    private String nameAddress;
    private int position;
    @ManyToOne
    private Menu parent;
    @OneToMany(mappedBy = "parent", fetch = FetchType.LAZY)
    private List<Menu> children;

    // Getters, setters, etc.
}

MenuManagerController.java:

@Named
@ViewScoped
public class MenuManagerController implements Serializable {
    @Inject
    private MenuService menuService;
    private Languages selectedLanguage;
    private List<Languages> languages;
    private String menuName;
    private TreeNode rootMenu;
    private TreeNode selectedMenu;

    @PostConstruct
    private void init() {
        // Find root menu.
        Menu menu = menuService.findRootMenu();
        // Create root node.
        this.rootMenu = new DefaultTreeNode(menu, null);
        // Lazy load children.
        Hibernate.initialize(menu.getChildren());
        // Build Node from Menu.
        buildNodeStructureFromMenu(this.rootMenu, menu.getChildren());
    }

    private void buildNodeStructureFromMenu(TreeNode root, List<Menu> menus) {
        App.printWithTrace("Add all the " + menus.size() + " children to \"" + root.getData().toString() + "\".");

        for(Menu menu : menus) {
            App.printWithTrace("Create node for \"" + menu.getNameDisplay() + "\".");
            // Create child node of root node.
            TreeNode child = new DefaultTreeNode(menu, root);
            // Lazy load children.
            Hibernate.initialize(menu.getChildren());
            // Create all children of this menu.
            List<Menu> children = menu.getChildren();
            // Recursion.
            buildNodeStructureFromMenu(child, children);
        }
    }
}

MenuService.java:

@Stateless
public class MenuService {
    @PersistenceContext(unitName = "MyappPU")
    private EntityManager em;

    public enum Layer {
        All,
        FirstOnly
    }

    public Menu findRootMenu() {
        return this.em.createNamedQuery(Menu.FIND_MENU_ROOT, Menu.class)
                      .getSingleResult();
    }

    // More methods, etc.
}

Answer 1

异常表明您在事务关闭后尝试获取延迟加载的关系。

我建议你看看我的文章:https://arnoldgalovics.com/lazyinitializationexception-demystified/

但是，我的文章是针对基于 Spring 的配置编写的，您可以轻松地根据您的情况进行调整。