java - Spring数据JPA延迟加载
问题描述
我对 Java Spring 很陌生,我尝试了几天来让我的关系正常工作,但它没有。我正在使用 Java Spring 和 Mysql。
管理员应将员工添加到数据库中。员工只会看到特定数据。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,但我仍然无法让它工作。
我有一个用户、角色和员工的实体。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 |
+----------+--------------+------+-----+---------+----------------+
雇员:
+-------------------+--------------+------+-----+---------+----------------+
| 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 | |
+-------------------+--------------+------+-----+---------+----------------+
只有管理员可以将员工添加到系统中,员工只能登录并查看一些数据。因此,就像我在 Java 中所理解的那样,员工表中的 user_id 之类的 Spring 参数并没有额外写入模型中。所以这就是我现在所拥有的:
雇员.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;
}
}
用户.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Employee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 保存为普通的 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,join 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,我收到一个错误,例如:not-null property references a null or transient value or something like Employee column: user_id (should be mapped with insert="false" update= “假”)。所以我不知道该怎么办,我被这个问题困了5天。
在此先感谢您的帮助!
解决方案
如果你决定使用 Hibernate 和 annotation 来定义你的关系@OneToOne
,你应该记住现在 Employee 有一个类属性(属性)User。对于在用户表上存储员工表中的外键,您需要将用户实例设置为员工。
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);
}
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