接線問題。 org.hibernate.LazyInitializationException：無法初始化懶洋洋地收集

Question

我捕捉一些異常試圖getUserRole（）

org.apache.jasper.JasperException: An exception occurred processing JSP page /WEB-INF/jsp/index.jsp at line 19 

16: } 
17: */ 
18: User user = (User)request.getAttribute("list"); 
19: out.print(user.getLogin() + " : " + user.getPassword() + " : " + user.getUserRole()); 
20: %> 

org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: com.myproject.User.userRole, could not initialize proxy - no Session 
    org.hibernate.collection.internal.AbstractPersistentCollection.throwLazyInitializationException(AbstractPersistentCollection.java:575) 
    org.hibernate.collection.internal.AbstractPersistentCollection.withTemporarySessionIfNeeded(AbstractPersistentCollection.java:214) 
    org.hibernate.collection.internal.AbstractPersistentCollection.initialize(AbstractPersistentCollection.java:554) 
    org.hibernate.collection.internal.AbstractPersistentCollection.read(AbstractPersistentCollection.java:142) 
    org.hibernate.collection.internal.PersistentSet.toString(PersistentSet.java:316) 
    java.lang.String.valueOf(String.java:2847) 
    java.lang.StringBuilder.append(StringBuilder.java:128) 
    org.apache.jsp.WEB_002dINF.jsp.index_jsp._jspService(index_jsp.java:90) 
    org.apache.jasper.runtime.HttpJspBase.service(HttpJspBase.java:70) 
    javax.servlet.http.HttpServlet.service(HttpServlet.java:725) 
    org.apache.jasper.servlet.JspServletWrapper.service(JspServletWrapper.java:432) 
    org.apache.jasper.servlet.JspServlet.serviceJspFile(JspServlet.java:396) 
    org.apache.jasper.servlet.JspServlet.service(JspServlet.java:340) 

這裏我註解類時：

@Entity 
@Table(name="users") 
public class User implements Serializable{ 

    private static final long serialVersionUID = 1L; 

    private String login; 
    private String password; 
    private String email; 
    private boolean enabled; 
    private Set<UserRole> userRole = new HashSet<UserRole>(0); 

    public User(){} 
    public User(String login, String password, String email, boolean enabled){ 
     this.login = login; 
     this.password = password; 
     this.email = email; 
     this.enabled = enabled; 
    } 
    public User(String login, String password, String email, boolean enabled, Set<UserRole> userRole){ 
     this.login = login; 
     this.password = password; 
     this.email = email; 
     this.enabled = enabled; 
     this.userRole = userRole; 
    } 

    @Id 
    @Column(name="login", unique=true, nullable=false, length = 50) 
    public String getLogin(){ 
     return this.login; 
    } 

    public void setLogin(String login){ 
     this.login = login; 
    } 

    @Column(name="password", nullable = false, length = 50) 
    public String getPassword(){ 
     return this.password; 
    } 

    public void setPassword(String password){ 
     this.password = password; 
    } 

    @Column(name = "email", nullable = false, length = 100) 
    public String getEmail(){ 
     return this.email; 
    } 

    public void setEmail(String email){ 
     this.email = email; 
    } 

    @Column(name = "enabled", nullable = false) 
    public boolean isEnabled(){ 
     return this.enabled; 
    } 

    public void setEnabled(boolean enabled){ 
     this.enabled = enabled; 
    } 

    @OneToMany(fetch = FetchType.LAZY, mappedBy = "user") 
    public Set<UserRole> getUserRole(){ 
     return this.userRole; 
    } 

    public void setUserRole(Set<UserRole> userRole){ 
     this.userRole = userRole; 
    } 

} 

    @Entity 
    @Table(name="user_roles", uniqueConstraints = @UniqueConstraint(columnNames = {"role","login"})) 
    public class UserRole implements Serializable{ 

     private static final long serialVersionUID = 1L; 

     private int roleId; 
     private User user; 
     private String role; 

     public UserRole(){} 
     public UserRole(User user, String role){ 
      this.user = user; 
      this.role = role; 
     } 

     @Id 
     @GeneratedValue(strategy = IDENTITY) 
     @Column(name = "role_id", unique = true, nullable = false) 
     public int getRoleId(){ 
      return this.roleId; 
     } 

     public void setRoleId(int roleId){ 
      this.roleId = roleId; 
     } 

     @ManyToOne(fetch = FetchType.LAZY) 
     @JoinColumn(name = "login", nullable = false) 
     public User getUser(){ 
      return this.user; 
     } 

     public void setUser(User user){ 
      this.user = user; 
     } 

     @Column(name = "role", nullable = false) 
     public String getRole(){ 
      return this.role; 
     } 

     public void setRole(String role){ 
      this.role = role; 
     } 

    } 

，並結束我的DAO：

@Override 
    @Transactional 
    @SuppressWarnings("unchecked") 
    public User findByLogin(String login){ 
     String query = "from User where login = ?"; 
     List<User> listUser = sessionFactory.getCurrentSession().createQuery(query).setParameter(0, login).list(); 

     if(listUser.size() > 0) 
      return listUser.get(0); 
     else 
      return null; 
    } 

必須在沒有user.getUserRole（）的情況下標記jsp，但是當我新插入此方法代碼段時，我看到錯誤試。同樣在這個項目中實現了SpringSecurity，但是沒有工作「登錄」。我想同樣的

Answer 1

問題您可以從LAZY to EAGER

fetch = FetchType.EAGER 

更改或使用Hibernate.intialize()初始化的userRole

if(listUser.size() > 0) 
      User user = listUser.get(0); 
Hibernate.initialize(user.getUserRole()); 
     else 
      return null;