Spring Data无法延迟初始化角色的集合，无法初始化代理 – 没有会话。

huangapple
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英文:

Spring Data failed to lazily initialize a collection of role, could not initialize proxy - no Session

问题

这是您提供的代码部分的中文翻译：

  1. 我正在使用Spring Boot制作API，但似乎无法初始化懒加载集合。唯一对我有效的解决方案是将其更改为急加载，但这不是一个解决方法。
  3. @Entity
  4. @Getter
  5. @Setter
  6. @NoArgsConstructor
  7. @Table(name = "user")
  8. public class UserEntity {
  10.     @Id
  11.     @GeneratedValue
  12.     private Long id;
  14.     @Column(unique = true)
  15.     String email;
  17.     @ElementCollection(fetch = LAZY)
  18.     List<String> roles = new ArrayList<>();
  20.     public UserEntity(String email) {
  21.         this.email = email;
  22.     }
  23. }
  25. @Order(Ordered.LOWEST_PRECEDENCE)
  26. @Component
  27. public class AuthFilter extends OncePerRequestFilter {
  29.     Logger LOG = LoggerFactory.getLogger(AuthFilter.class);
  31.     @Autowired
  32.     UserRepository userRepository;
  34.     @Override
  35.     @Transactional
  36.     protected void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain)
  37.             throws ServletException, IOException {
  39.         Authentication authentication = SecurityContextHolder.getContext().getAuthentication();
  41.         if(authentication != null && authentication.isAuthenticated()){
  42.             String email = authentication.getName();
  43.             UserEntity user = userRepository.findByEmail(email).orElse(new UserEntity(email));
  44.             user.getRoles().clear();
  45.             user.getRoles().addAll(authentication.getAuthorities().stream().map(GrantedAuthority::getAuthority).collect(Collectors.toSet()));
  46.             userRepository.save(user);
  47.         }
  48.         filterChain.doFilter(request, response);
  49.     }
  50. }

这些是您提供的两个类。一个是用户实体，另一个是从认证对象创建用户的过滤器。当调用user.getRoles().clear();时，您会收到著名的错误消息org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: ch.ciip.ressources.api.user.UserEntity.roles, could not initialize proxy - no Session。

从我的理解来看，Spring Data会为每个懒加载关系创建代理，并且只有在访问时才会获取数据。但要访问它，它需要一个会话（session），显然，Spring在访问代理时无法创建会话，然而，当我在存储库上调用save时，它却没有问题地执行。

我尝试过以下方法：

  • 使用@Transactional，来自org.springframework.transaction.annotation.Transactional和javax.transaction.Transactional。但这完全没有改变任何事情。
  • 将获取类型更改为EAGER。这个方法有效，但不是我想要的方法。
  • 在application.properties文件中将enable_lazy_load_no_trans=true更改为true。不推荐此方法，因此我不会尝试它。
  • 在尝试修改之前在懒加载集合上调用一个方法。例如，调用user.getRoles().size();以使其初始化。但这不起作用。
  • 调用Hibernate.initialize(user.getRoles());来初始化集合。但它也不起作用。

我想知道为什么Spring会说No Session，因为它确实为userRepository.findByEmail(email)和userRepository.save(user)创建了会话。

我可以手动创建一个EntityManager并使用它来获取/持久化我的实体，但这不是目标，因为我正在使用Spring Data JPA。

英文:

I am making an API with Spring Boot and I never seem to manage to initialize lazy collections. The only solution that has ever worked for me is changing it to eager, but that is no solution.

  1. @Entity
  2. @Getter
  3. @Setter
  4. @NoArgsConstructor
  5. @Table(name = &quot;user&quot;)
  6. public class UserEntity {
  7. @Id
  8. @GeneratedValue
  9. private Long id;
  10. @Column(unique = true)
  11. String email;
  12. @ElementCollection(fetch = LAZY)
  13. List&lt;String&gt; roles = new ArrayList&lt;&gt;();
  14. public UserEntity(String email) {
  15. this.email = email;
  16. }
  17. }
  1. @Order(Ordered.LOWEST_PRECEDENCE)
  2. @Component
  3. public class AuthFilter extends OncePerRequestFilter {
  4. Logger LOG = LoggerFactory.getLogger(AuthFilter.class);
  5. @Autowired
  6. UserRepository userRepository;
  7. @Override
  8. @Transactional
  9. protected void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain)
  10. throws ServletException, IOException {
  11. Authentication authentication = SecurityContextHolder.getContext().getAuthentication();
  12. if(authentication != null &amp;&amp; authentication.isAuthenticated()){
  13. String email = authentication.getName();
  14. UserEntity user = userRepository.findByEmail(email).orElse(new UserEntity(email));
  15. user.getRoles().clear();
  16. user.getRoles().addAll(authentication.getAuthorities().stream().map(GrantedAuthority::getAuthority).collect(Collectors.toSet()));
  17. userRepository.save(user);
  18. }
  19. filterChain.doFilter(request, response);
  20. }
  21. }

These are my two classes. A user entity and a filter that will create user from an authentication object. When the line user.getRoles().clear(); is called, I get the famous error org.hibernate.LazyInitializationException: failed to lazily initialize a collection of role: ch.ciip.ressources.api.user.UserEntity.roles, could not initialize proxy - no Session.

From my understanding, Spring Data will create a proxy for each lazy relation, and will only fetch the data once it is accessed. But to access it it needs a session and apparently, Spring is not able to create a session when accessing a proxy, however, it has no problem doing it when I call save on a repository.

What I've tried is :

  • @Transactional from both org.springframework.transaction.annotation.Transactional and javax.transaction.Transactional. But it changes absolutely nothing.
  • Change fetch type to EAGER. This method works, but it is not what I want to do.
  • Change enable_lazy_load_no_trans=trueto true in application.properties file. This method is not recommended, so I will not try it.
  • Calling a method on the lazy collection before trying to modify it. For example calling user.getRoles().size(); so it gets initialized. But it does not work.
  • Calling Hibernate.initialize(user.getRoles()); to initialize the collection. But it does not work.

I wonder why Spring says No Session because it does create a session for the userRepository.findByEmail(email) as well as for userRepository.save(user).

I could create an entitymanager manually and use it to fetch/persist my entities. But that is not the goal as I'm using Spring Data JPA.

答案1

得分: 5

将方法设为public，即：

  1. @Override
  2. @Transactional
  3. public void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain)
  4.         throws ServletException, IOException {
  6.     Authentication authentication = SecurityContextHolder.getContext().getAuthentication();
  8.     if (authentication != null && authentication.isAuthenticated()) {
  9.         String email = authentication.getName();
  10.         UserEntity user = userRepository.findByEmail(email).orElse(new UserEntity(email));
  11.         user.getRoles().clear();
  12.         user.getRoles().addAll(authentication.getAuthorities().stream().map(GrantedAuthority::getAuthority).collect(Collectors.toSet()));
  13.         userRepository.save(user);
  14.     }
  15.     filterChain.doFilter(request, response);
  16. }
英文:

Make the method public, i.e.:

  1. @Override
  2. @Transactional
  3. public void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain)
  4. throws ServletException, IOException {
  5. Authentication authentication = SecurityContextHolder.getContext().getAuthentication();
  6. if(authentication != null &amp;&amp; authentication.isAuthenticated()){
  7. String email = authentication.getName();
  8. UserEntity user = userRepository.findByEmail(email).orElse(new UserEntity(email));
  9. user.getRoles().clear();
  10. user.getRoles().addAll(authentication.getAuthorities().stream().map(GrantedAuthority::getAuthority).collect(Collectors.toSet()));
  11. userRepository.save(user);
  12. }
  13. filterChain.doFilter(request, response);
  14. }

答案2

得分: 2

Creating a separate service class and injecting it into this filter is a workaround I can propose.

  1. @Service
  2. class UserService {
  3.     @Autowired
  4.     private UserRepository userRepository;
  6.     @Transactional
  7.     public void findUserByEmailAndAddRoles(String email, Set<String> authorities) {
  8.         UserEntity user = userRepository.findByEmail(email)
  9.                    .orElse(new UserEntity(email));
  10.         user.getRoles().clear();
  11.         user.getRoles().addAll(authorities);
  12.         userRepository.save(user);
  13.     }
  14. }

Then use this in your filter

  1. @Override
  2. protected void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain)
  3.         throws ServletException, IOException {
  5.     Authentication authentication = SecurityContextHolder.getContext().getAuthentication();
  7.     if(authentication != null && authentication.isAuthenticated()){
  8.         String email = authentication.getName();
  9.         Set<String> authorities = authentication.getAuthorities().stream()
  10.            .map(GrantedAuthority::getAuthority).collect(Collectors.toSet());
  11.         userService.findUserByEmailAndAddRoles(email, authorities);
  12.     }
  13.     filterChain.doFilter(request, response);
  14. }
英文:

Creating a separate service class and injecting it into this filter is a workaround i can propose.

  1. @Service
  2. class UserService {
  3. @Autowired
  4. private UserRepository userRepository;
  5. @Transactional
  6. public void findUserByEmailAndAddRoles(String email, Set&lt;String&gt; authorities) {
  7. UserEntity user = userRepository.findByEmail(email)
  8. .orElse(new UserEntity(email));
  9. user.getRoles().clear();
  10. user.getRoles().addAll(authorities);
  11. userRepository.save(user);
  12. }
  13. }

Then use this in your filter

  1. @Override
  2. protected void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain)
  3. throws ServletException, IOException {
  4. Authentication authentication = SecurityContextHolder.getContext().getAuthentication();
  5. if(authentication != null &amp;&amp; authentication.isAuthenticated()){
  6. String email = authentication.getName();
  7. Set&lt;String&gt; authorities = authentication.getAuthorities().stream()
  8. .map(GrantedAuthority::getAuthority).collect(Collectors.toSet());
  9. userService.findUserByEmailAndAddRoles(email, authorities);
  10. }
  11. filterChain.doFilter(request, response);
  12. }

huangapple
  • 本文由 发表于 2020年8月5日 17:04:49
  • 转载请务必保留本文链接：https://go.coder-hub.com/63261801.html
  • hibernate
  • java
  • jpa
  • spring
  • spring-boot
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