虽然fetch类型是懒惰的，但出于某种原因，hibernate会发送第二个请求并获取懒惰的请求。我已经处理这个问题几天了，出于某种原因，我不明白。我在互联网上尝试了一些方法，但不幸的是，我无法修复它，如果你能帮助我看看我的代码，我会非常感激。

用户实体

@Entity
@Table(name = "users")
@Data
public class User {
    @Id
    @GeneratedValue(strategy = GenerationType.SEQUENCE)
    private Long id;
    @Column(name = "username", length = 50)
    private String username;
    @Column(name = "email")
    private String email;
    @Column(name = "password")
    private String password;
    @ManyToMany(fetch = FetchType.LAZY)
    @JoinTable(name = "user_friends" , joinColumns = @JoinColumn(name = "user_id") , inverseJoinColumns = @JoinColumn(name = "friend_id"))
    @OnDelete(action = OnDeleteAction.CASCADE)
    private Set<User> friends = new HashSet<>();
}

字符串

用户服务

@Service
@Transactional
public class UserService implements ICrudService<User> {
    @Autowired
    private UserRepository repository;
    @Override
    @Transactional
    public List<User> findAll() {
        return repository.findAll();
    }
    @Override
    @Transactional
    public User findById(Long id) throws Exception {
        return repository.findById(id).orElseThrow(() -> new Exception("Kullanici Bulunamadi"));
    }
    @Override
    @Transactional
    public User create(User user) {
        return repository.save(user);
    }
    @Override
    public void delete(Long id) {
        repository.deleteById(id);
    }
}

型

用户控制器

@RestController
@RequestMapping("/users")
public class UserController {
    @Autowired
    private UserService userService;
    @GetMapping("/all")
    public List<User> getUsers(){
        return userService.findAll();
    }
    @GetMapping("/{userId}")
    public User getUserById(@PathVariable Long userId) throws Exception {
        return userService.findById(userId);
    }
    @PostMapping
    public User saveUser(@RequestBody User user){
        return userService.create(user);
    }
    @DeleteMapping("/{userId}")
    public void deleteUserById(@PathVariable Long userId){
        userService.delete(userId);
    }
}

型
我使用JpaReository作为服务中的存储库
“/users/all”的输出如下所示

输出

[
    {
        "id": 1,
        "username": "testUser",
        "email": "[email protected]",
        "password": "verystrongpasword",
        "friends": []
    }
]

型

这是休眠sql格式

Hibernate: 
    select
        u1_0.id,
        u1_0.email,
        u1_0.password,
        u1_0.username 
    from
        users u1_0
Hibernate: 
    select
        f1_0.user_id,
        f1_1.id,
        f1_1.email,
        f1_1.password,
        f1_1.username 
    from
        user_friends f1_0 
    join
        users f1_1 
            on f1_1.id=f1_0.friend_id 
    where
        f1_0.user_id=?

型