5

虽然有大量关于如何建模的信息，但在 JPA (2) 中，一对一关系 或具有自然键的实体，我无法找到关于如何建模的清晰/简单的答案我们两者都有的情况，即父表具有自然键的一对一关系。很明显，我可能错过了这样的教程；如果是这样，将我指向一个也可能是答案。

而且，与 JPA 和像我这样的菜鸟一样，当一个人需要比最基本的模型多一点的时候，一个人很快就会碰壁。

因此，考虑以下数据库模型：

在此处输入图像描述

对应的 JPA 注释对象模型是什么？（因为我不想影响答案，所以我不考虑我尝试过的事情......）

也欢迎性能建议（例如“一对多可以执行得更快”等）！

谢谢，

4

1 回答 1

13

复合标识符由两个数字列构成，因此映射如下所示：

@Embeddable
public class EmployeeId implements Serializable {

    private Long companyId;

    private Long employeeId;

    public EmployeeId() {
    }

    public EmployeeId(Long companyId, Long employeeId) {
        this.companyId = companyId;
        this.employeeId = employeeId;
    }

    public Long getCompanyId() {
        return companyId;
    }

    public Long getEmployeeId() {
        return employeeId;
    }

    @Override
    public boolean equals(Object o) {
        if (this == o) return true;
        if (!(o instanceof EmployeeId)) return false;
        EmployeeId that = (EmployeeId) o;
        return Objects.equals(getCompanyId(), that.getCompanyId()) &&
                Objects.equals(getEmployeeId(), that.getEmployeeId());
    }

    @Override
    public int hashCode() {
        return Objects.hash(getCompanyId(), getEmployeeId());
    }
}

父类，如下所示：

@Entity(name = "Employee")
public static class Employee {

    @EmbeddedId
    private EmployeeId id;

    private String name;

    @OneToOne(mappedBy = "employee")
    private EmployeeDetails details;

    public EmployeeId getId() {
        return id;
    }

    public void setId(EmployeeId id) {
        this.id = id;
    }

    public String getName() {
        return name;
    }

    public void setName(String name) {
        this.name = name;
    }

    public EmployeeDetails getDetails() {
        return details;
    }

    public void setDetails(EmployeeDetails details) {
        this.details = details;
    }
}

而孩子是这样的：

@Entity(name = "EmployeeDetails")
public static class EmployeeDetails {

    @EmbeddedId
    private EmployeeId id;

    @MapsId
    @OneToOne
    private Employee employee;

    private String details;

    public EmployeeId getId() {
        return id;
    }

    public void setId(EmployeeId id) {
        this.id = id;
    }

    public Employee getEmployee() {
        return employee;
    }

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

    public String getDetails() {
        return details;
    }

    public void setDetails(String details) {
        this.details = details;
    }
}

一切正常：

doInJPA(entityManager -> {
    Employee employee = new Employee();
    employee.setId(new EmployeeId(1L, 100L));
    employee.setName("Vlad Mihalcea");
    entityManager.persist(employee);
});

doInJPA(entityManager -> {
    Employee employee = entityManager.find(Employee.class, new EmployeeId(1L, 100L));
    EmployeeDetails employeeDetails = new EmployeeDetails();
    employeeDetails.setEmployee(employee);
    employeeDetails.setDetails("High-Performance Java Persistence");
    entityManager.persist(employeeDetails);
});

doInJPA(entityManager -> {
    EmployeeDetails employeeDetails = entityManager.find(EmployeeDetails.class, new EmployeeId(1L, 100L));
    assertNotNull(employeeDetails);
});
doInJPA(entityManager -> {
    Phone phone = entityManager.find(Phone.class, "012-345-6789");
    assertNotNull(phone);
    assertEquals(new EmployeeId(1L, 100L), phone.getEmployee().getId());
});

代码在GitHub上可用。

于 2016-08-01T10:37:07.737 回答