Question

有两个实体：用户和员工。用户具有类型为Employee的字段。

@Entity
@Table(name="user")
public class User extends AuditableEntity {    
    Long idUser;
    String username;
    String password;
    Employee employee;

    @Id
    @GeneratedValue(strategy = GenerationType.IDENTITY)
    public Long getIdUser() { return idUser; }  
    public void setIdUser(Long idUser) { this.idUser = idUser; }

    @Column(name = "username")
    public String getUsername() { return username; }
    public void setUsername(String username) { this.username = username; }

    @Column(name = "password")
    public String getPassword() { return password; }
    public void setPassword(String password) { this.password = password; }

    @ManyToOne
    @JoinColumn(name = "idemployee")    
    public Employee getEmployee() { return employee; }
    public void setEmployee(Employee employee) { this.employee = employee; }
}

并且

@Entity
@Table(name = "employee")
public class Employee extends AuditableEntity { 
    Long        idEmployee;
    String      surname;
    String      name;
    String      patronymic;
    Date        birthdate;

    @Id
    @GeneratedValue (strategy=GenerationType.IDENTITY)
    @Column(name = "idemployee")
    public Long getIdEmployee() { return idEmployee; }
    public void setIdEmployee(Long idEmployee) { this.idEmployee = idEmployee; }

    @Column(name = "surname")
    public String getSurname() { return surname; }
    public void setSurname(String surname) { this.surname = surname; }

    @Column(name = "name")
    public String getName() { return name; }
    public void setName(String name) { this.name = name; }

    @Column(name = "patronymic")
    public String getPatronymic() { return patronymic; }
    public void setPatronymic(String patronymic) { this.patronymic = patronymic; }

    @JsonFormat(pattern = "dd.MM.yyyy")
    @Column(name = "birthday")
    public Date getBirthdate() { return birthdate; }
    public void setBirthdate(Date birthdate) { this.birthdate = birthdate; }
}

我需要将User序列化为XML / JSON。我使用的是JAXB，但它也在对员工进行序列化：

<User>
    <idUser>15</idUser>
    <username>user15</username>
    <password>password15</password>
    <employee>
        <idEmployee>23</idEmployee>
        <surname>Smith</surname>
        <name>John</name>
        <patronymic>H.</patronymic>
        <birthdate>01.01.1970</birthdate>
    </employee>
<User>

我需要这样的结果：

<User>
    <idUser>15</idUser>
    <username>user15</username>
    <password>password15</password>   
    <idEmployee>23</idEmployee>    
<User>

我尝试使用@XmlID，@ XmlIDREF - 但它只能用于String id列。还尝试使用@XmlTransient - 但它只排除了Employee。如何在没有员工的情况下序列化用户，仅使用idEmployee？

第二个问题是反序列化。有没有标准的方法可以做到这一点？

Answer 1

使用employee在课程User中注释属性@XmlTransient。此注释可应用于字段或其getter。

Answer 2

您可以将员工ID的其他getter添加到User实体

public class User
{   
    ...  
    @Transient // for JPA
    @XmlElement 
    Long getIdEmployee()
    {
       return employee.getIdEmployee();
    }
}

或者您可以使用@XmlValue

执行此操作

public class User
{  
    ...  
    @XmlElement(name = "idEmployee")
    public Employee getEmployee() { return employee; }
    public void setEmployee(Employee employee) { this.employee = employee; }
}

@XmlAccessorType(XmlAccessType.NONE) // to prevent marshalling of all properies
public class Employee 
{ 
   ...
   @XmlValue
   public Long getIdEmployee() { return idEmployee; }
   ...
}

序列化JPA实体 - 仅保存已获取实体的ID

2 个答案: