Spring 数据 JPA 延迟加载

Question

我是 Java Spring 的新手，我尝试了几天让我的关系正常运转，但没有成功。我正在使用 Java Spring 和 Mysql.

管理员应将员工添加到数据库中。员工将只能看到特定数据。我的问题是，我不明白如何从 EmployeesController 发出正确的 POST 请求并从用户模型中获取 user_id。我尝试了一些不同的实现，但仍然无法正常工作。

我有一个用户、角色和员工实体。员工始终是用户，但用户可能不是员工。所以我的数据库结构如下：

用户：

+----------+--------------+------+-----+---------+----------------+
| Field    | Type         | Null | Key | Default | Extra          |
+----------+--------------+------+-----+---------+----------------+
| id       | int          | NO   | PRI | NULL    | auto_increment |
| username | varchar(30)  | NO   |     | NULL    |                |
| email    | varchar(100) | NO   |     | NULL    |                |
| password | varchar(100) | NO   |     | NULL    |              
+----------+--------------+------+-----+---------+----------------+

员工：

+-------------------+--------------+------+-----+---------+----------------+
| Field             | Type         | Null | Key | Default | Extra          |
+-------------------+--------------+------+-----+---------+----------------+
| id                | int          | NO   | PRI | NULL    | auto_increment |
| academic_title    | varchar(100) | NO   |     | NULL    |                |
| department        | varchar(100) | NO   |     | NULL    |                |
| user_id           | int          | NO   |     | NULL    |                |
+-------------------+--------------+------+-----+---------+----------------+

只有管理员才能将员工添加到系统中，员工只能登录并查看部分数据。因此，正如我在 Java Spring 中理解的那样，员工 table 中的 user_id 等参数并未额外写入模型中。这就是我现在拥有的：

Employee.java

@Entity
@Table(name = "employees")
public class Employee{
  
    @OneToOne(fetch = FetchType.LAZY, optional = false)
    @JoinColumn(name = "user_id", nullable = false)
    private User user;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;

    @Size(min=3, max = 100)
    private String academic_title;

    @Size(min=3, max = 100)
    private String department;

    public Employee() {}
 
    public Employee(String academic_title, String department) {
        super();
        this.academic_title = academic_title;
        this.department = department;
    }
    
    public Long getId() {
        return id;
    }
 
    public void setId(Long id) {
        this.id = id;
    }

    public String getTitle() {
        return title;
    }
 
    public void setTitle(String title) {
        this.title = title;
    }

    public String getDepartment() {
        return department;
    }
 
    public void setDepartment(String department) {
        this.department = department;
    }

    public User getUser() {
        return user;
    }

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

User.java

@Entity
@Table(name = "users")
public class User{
  @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    private Long id;
  
    @NotBlank
    @Size(min=3, max = 50)
    private String username;
 
    @NaturalId
    @NotBlank
    @Size(max = 50)
    @Email
    private String email;
 
    @NotBlank
    @Size(min=6, max = 100)
    private String password;
 
    @ManyToMany(fetch = FetchType.LAZY)
    @JoinTable(name = "user_roles", 
      joinColumns = @JoinColumn(name = "user_id"), 
      inverseJoinColumns = @JoinColumn(name = "role_id"))
    private Set<Role> roles = new HashSet<>();

    @OneToOne(fetch = FetchType.LAZY,
            cascade =  CascadeType.ALL,
            mappedBy = "user")
    private Employee employee;
 
    public User() {}
 
    public User(String username, String email, String password) {
        super();
        this.username = username;
        this.email = email;
        this.password = password;
    }
 
    public Long getId() {
        return id;
    }
 
    public void setId(Long id) {
        this.id = id;
    }
 
    public String getUsername() {
        return username;
    }
 
    public void setUsername(String username) {
        this.username = username;
    }
 
    public String getEmail() {
        return email;
    }
 
    public void setEmail(String email) {
        this.email = email;
    }
 
    public String getPassword() {
        return password;
    }
 
    public void setPassword(String password) {
        this.password = password;
    }
 
    public Set<Role> getRoles() {
        return roles;
    }
 
    public void setRoles(Set<Role> roles) {
        this.roles = roles;
    }

    public Employee getEmployee() {
        return employee;
    }

    public void setEmployee(Employee employee) {
        this.employee = employee;
    }
}

EmployeeController.java（创建员工函数）

    public Employee createEmployee(@PathVariable (value = "user_id") Long user_id,
                                 @Valid @RequestBody Employee employee) {
                                     
        userRepository.findById(user_id);
        employeeRepository.save(employee);
        return employee;

    }

userRepository.java

@Repository
public interface UserRepository extends JpaRepository<User, Long> {
    Optional<User> findByUsername(String username);
    Optional<User> findById(Long id);
    Boolean existsByUsername(String username);
    Boolean existsByEmail(String email);
}

我的邮递员POST请求添加员工：

{
    "academic_title": "xy",
    "department": "xx",
    "user_id": 5
}

当我尝试将 user_id 保存为普通 Long 参数时，我可以将其保存到数据库中。但是当我获取现有员工时，连接功能不起作用，所以我看不到 user_id 5 是谁。当我以另一种方式尝试时，就像上面的代码一样，我收到如下错误： not-null 属性 references a null or transient value 或类似的东西 员工列：user_id（应映射为 insert="false" update="false"）。所以我不知道该怎么办，我在这个问题上卡了5天。

在此先感谢您的帮助！

Answer 1

如果您决定使用 Hibernate 和注解 @OneToOne 来定义您的关系，您现在应该牢记 Employee 有一个 class 属性（属性）用户。为了在用户 table 上存储来自员工 table 的外键，您需要将 User 实例设置为 Employee.

你应该在你的情况下尝试这样的事情：

public Employee createEmployee(@PathVariable (value = "user_id") Long user_id,
                             @Valid @RequestBody Employee employee) {
                                 
    // find user by id or else throw exception
    User userById = userRepository.findById(user_id).orElseThrow(() -> new 
    RuntimeException("Not found."));
    
    // set user into employee. it's like binding two entities
    employee.setUser(userById);

    // after that you can save your employee entity.
    return employeeRepository.save(employee);
    
}

Answer 2

the join function doesn't work

User 和 Employee 之间的获取类型是 Lazy。这意味着它在加载员工时实际上并没有加载用户。相反，它会在请求时加载它们。

正如@Zooger 提到的，您应该先从数据库加载用户。

Spring 数据 JPA 延迟加载

Spring data JPA lazy loading

java

mysql

spring

hibernate

one-to-one