java - JPA持久化继承

Question

我有一个长期面临的问题，但没有得到答案。

我想使用 JPA 保留继承表，但收到错误消息。我为此使用 eclipselink 。

我的单元测试因该类而失败。

    -- Table: PERSON

CREATE TABLE PERSON (
    ID SERIAL  NOT NULL PRIMARY KEY
    DEFAULT nextval ('Person_Sequence'),
    FIRST_NAME VARCHAR  NOT NULL,
    LAST_NAME VARCHAR  NOT NULL,
    DTYPE VARCHAR
);

-- Table: BUYER

CREATE TABLE BUYER (
    ID INTEGER  NOT NULL PRIMARY KEY
);

我有一个类(class)和餐 table 人员以及该类(class)人员的子类(class)买家

人员类别:

   import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.Id;
import javax.persistence.Inheritance;
import javax.persistence.InheritanceType;

@Entity
@Inheritance(strategy=InheritanceType.JOINED)
public   class Person {

    @Id
    private int ID;
    @Column(name="FIRST_NAME")
    private String firstName;
    @Column(name="LAST_NAME")
    private String lastName;

    protected Person() {
    }

    public Person(int ID, String firstName, String lastName) {
        this.ID = ID;
        this.firstName = firstName;
        this.lastName = lastName;
    }

    public int getId() {
        return ID;
    }

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

    public String getFirstName() {
        return firstName;
    }

    public void setFirstName(String firstName) {
        this.firstName = firstName;
    }

    public String getLastName() {
        return lastName;
    }

    public void setLastName(String lastName) {
        this.lastName = lastName;
    }

    @Override
    public String toString() {
        return "Person [getId()=" + getId() + ", getFirstName()=" + getFirstName() + ", getLastName()=" + getLastName()
                + "]";
    }

}

买家类别:

import java.util.ArrayList;
import java.util.Collection;

import javax.persistence.Column;
import javax.persistence.Entity;
import javax.persistence.OneToMany;

@Entity
public class Buyer extends Person {

    protected Buyer() {
    };

    public Buyer(int id, String firstName, String lastName) {
        super(id, firstName, lastName);
    }

}

到目前为止还没有出现错误。但在这里您可以看到我的单元测试和我收到的错误消息:

import static org.junit.Assert.*;


import javax.persistence.EntityManager;
import javax.persistence.EntityManagerFactory;
import javax.persistence.EntityTransaction;
import javax.persistence.Persistence;

import org.junit.AfterClass;
import org.junit.BeforeClass;
import org.junit.Test;

public class BuyerTest {

    final private int id = 33;
    final private String firstName = "Bart";
    final private String lastName = "Simpson";

    static EntityManagerFactory factory;
    static EntityManager manager;
    static EntityTransaction transaction;

    @BeforeClass
    public static void setup() {
        factory = Persistence.createEntityManagerFactory("testEclipseLink");
        assertNotNull(factory);
        manager = factory.createEntityManager();
        assertNotNull(manager);
        transaction = manager.getTransaction();
    }

    @AfterClass
    public static void teardown() {
        if (manager == null)
            return;
        manager.close();
        factory.close();
    }

    @Test
    public void create() {
        transaction.begin();
        System.out.println("first " + firstName + "last " + lastName);
        Buyer buyer = new Buyer(id, firstName, lastName);
        assertNotNull(buyer);
        manager.persist(buyer);
        transaction.commit();
    }

    @Test
    public void modify() {
        Buyer buyer = manager.find(Buyer.class, id);
        assertNotNull(buyer);
        transaction.begin();
        buyer.setFirstName("Homer");
        transaction.commit();
        teardown();
        setup();
        buyer = manager.find(Buyer.class, id);
        assertEquals("Homer", buyer.getFirstName());
    }

    @Test
    public void remove() {
        Buyer buyer = manager.find(Buyer.class, id);
        assertNotNull(buyer);
        transaction.begin();
        manager.remove(buyer);
        transaction.commit();
        buyer = manager.find(Buyer.class, id);
        assertNull(buyer);
    }




}

所以，如果我运行测试，我会收到此错误:

[EL Warning]: 2016-10-16 19:53:54.691--UnitOfWork(2036292945)--Exception [EclipseLink-4002] (Eclipse Persistence Services - 2.6.4.v20160829-44060b6): org.eclipse.persistence.exceptions.DatabaseException
Internal Exception: org.postgresql.util.PSQLException: FEHLER: NULL-Wert in Spalte »first_name« verletzt Not-Null-Constraint
  Detail: Fehlgeschlagene Zeile enthält (33, null, null).
Error Code: 0
Call: INSERT INTO BUYER (ID) VALUES (?)
    bind => [1 parameter bound]

它告诉我，我在 Firstname 上遇到了空约束错误。但是如果我在控制台上打印名字，它就被设置了。

我已经尝试过，如果我删除继承并仅使用我的 Buyer 类，它就可以正常工作!所以我无法弄清楚买家和人员类别之间的问题出在哪里。另外如果是 Src- 或 Table 定义问题

Answer 1

我看到的第一件事是买家是一个人，因此您需要在架构中添加外键来反射(reflect)这一点。

CREATE TABLE BUYER (
ID INTEGER  NOT NULL PRIMARY KEY,
personId bigint unsigned NOT NULL,
constraint fk_personId foreign key (personId) references PERSON (id)

);

此外，在您的子类中，您还需要注释。

@PrimaryKeyJoinColumn(name="personId")

一个好主意是让表的所有 ID 自动递增