Java Springboot 2.1.9.RELEASE-使用JPA持久化具有父子关系的实体

我在mysql中的

Employee

和

Address

表之间有一对多关系。我希望在使用子数据持久化父对象时，首先创建父对象，并在子表中自动使用其

parent\u id

。我得到以下错误

任何帮助都将不胜感激：

父代码：

@Entity
@Table(name = "employee")
public class Employee implements Serializable {
    private static final long serialVersionUID = 1L;
    @Id
    @Column(length = 6)
    @GeneratedValue(strategy = GenerationType.IDENTITY) 
    private long id;

    @Temporal(TemporalType.DATE)
    private Date dob;
    
    @Column(name="email_address")
    private String emailAddress;
    
    @Column(name="first_name")
    private String firstName;

    @Column(name="gender")
    private String gender;


    @Column(name="last_name")
    private String lastName;
    
    @Column(name="user_id")
    private String userName;

    @Column(name="password")
    private String password;

    @Column(name="email_verified")
    private String emailVerification;
   
    @OneToMany(targetEntity = Address.class, mappedBy = "employee", orphanRemoval = true, 
   cascade = 
    CascadeType.ALL, fetch = FetchType.EAGER)
    private Set<Address> addresses = null;

    public Employee() {
    }

    public Employee(long id) {
        this.id = id;
    }

控制器：

@CrossOrigin (origins = "*" )
@PostMapping("/employee")
public Employee createEmployee(@Valid @RequestBody Employee employee) {
    return employeeRepository.save(employee);
} 

JSON输入：

{
  "dob": "2001-11-12",
  "emailAddress": "aaa@gmail.com",
  "firstName": "John",
  "gender": "M",
  "lastName": "Wilson",
  "userName": "jon",
  "password": "pass",
  "emailVerification": null,
  "addresses": [
    {
      "address1": "123 My Street",
      "address2": null,
      "city": "MyCity1",
      "zip": "11111",
      "stateId": 1
    },
    {
      "address1": "567 My Street",
      "address2": null,
      "city": "MyCity2",
      "zip": "22222",
      "stateId": 1
    }
  ]
}

错误

---

一个选项是遍历所有地址并手动设置

employee

属性。您应该手动执行，因为JSON反序列化将不会触及该字段，即使没有

@JsonIgnore

注释。但这只是问题的表面

您正在

employee

属性上设置

updateable=false，insertable=false

。另一个属性是

字符串empId

，它映射到同一列，表示您希望手动管理关系。列/属性的类型应与相关实体的PK/id相同，因此

long

！而且，考虑到JSON也不包含

empId

，您仍然需要手动设置它；但是你在创造的时候还不知道！如果删除

Address.empId

属性和

updateable=false，insertable=false

并手动设置

employee

，它应该可以工作

如果您希望Jackson为您这样做，除了上一段中讨论的任务外，还可以在

Employee.addresses

属性上使用

@JsonManagedReference

，在

Address.Employee

属性上使用

@JsonBackReference

（删除

updateable=false，insertable=false

和

字符串empId

属性）并当然删除

@JsonIgnore

。请参阅详细说明：

@JsonManagedReference
@OneToMany（targetEntity=Address.class，mappedBy=“employee”，orphanRemoving=true，cascade=CascadeType.ALL，fetch=FetchType.EAGER）
专用地址集=空；
@JsonBackReference
@manytone（fetch=FetchType.LAZY，可选=false）
@JoinColumn（name=“emp_id”，nullable=false）
私人雇员；

---

免责声明：我反对混合使用JPA和Jackson注释。这确实很方便，因为您不需要单独的DTO类，不需要映射逻辑，也不需要担心执行转换实体所需的CPU/内存↔ DTO。另一方面，数据传输和数据持久性是两个完全不同的关注点，因此SOLID将付诸东流。并且有可能发送客户端不应该看到的字段（例如用户的密码或电子邮件）。

有效。非常感谢！！！

{
  "dob": "2001-11-12",
  "emailAddress": "aaa@gmail.com",
  "firstName": "John",
  "gender": "M",
  "lastName": "Wilson",
  "userName": "jon",
  "password": "pass",
  "emailVerification": null,
  "addresses": [
    {
      "address1": "123 My Street",
      "address2": null,
      "city": "MyCity1",
      "zip": "11111",
      "stateId": 1
    },
    {
      "address1": "567 My Street",
      "address2": null,
      "city": "MyCity2",
      "zip": "22222",
      "stateId": 1
    }
  ]
}

Hibernate: 
    insert 
    into
        address
        (address1, address2, city, emp_id, state_id, zip) 
    values
        (?, ?, ?, ?, ?, ?)
2021-04-28 22:17:19.067 RACE 25508 --- [nio-9090-exec-2] o.h.type.descriptor.sql.BasicBinder      : binding parameter [1] as [VARCHAR] - [567 My Street]
2021-04-28 22:17:19.067 TRACE 25508 --- [nio-9090-exec-2] o.h.type.descriptor.sql.BasicBinder      : binding parameter [2] as [VARCHAR] - [null]
2021-04-28 22:17:19.067 TRACE 25508 --- [nio-9090-exec-2] o.h.type.descriptor.sql.BasicBinder      : binding parameter [3] as [VARCHAR] - [MyCity2]
2021-04-28 22:17:19.067 TRACE 25508 --- [nio-9090-exec-2] o.h.type.descriptor.sql.BasicBinder      : binding parameter [4] as [VARCHAR] - [null]
2021-04-28 22:17:19.067 TRACE 25508 --- [nio-9090-exec-2] o.h.type.descriptor.sql.BasicBinder      : binding parameter [5] as [INTEGER] - [1]
2021-04-28 22:17:19.067 TRACE 25508 --- [nio-9090-exec-2] o.h.type.descriptor.sql.BasicBinder      : binding parameter [6] as [VARCHAR] - [22222]
2021-04-28 22:17:19.071  WARN 25508 --- [nio-9090-exec-2] o.h.engine.jdbc.spi.SqlExceptionHelper   : SQL Error: 1048, SQLState: 23000
2021-04-28 22:17:19.071 ERROR 25508 --- [nio-9090-exec-2] o.h.engine.jdbc.spi.SqlExceptionHelper   : Column 'emp_id' cannot be null
2021-04-28 22:17:19.094  WARN 25508 --- [nio-9090-exec-2] .m.m.a.ExceptionHandlerExceptionResolver : Resolved [org.springframework.dao.DataIntegrityViolationException: could not execute statement; SQL [n/a]; constraint [null];