[英]How to save complex objects using Hibernate?
我有三張桌子:
CREATE TABLE catalog (
id INT PRIMARY KEY NOT NULL AUTO_INCREMENT,
type_id INT,
genre_id INT,
product_name VARCHAR(100),
FOREIGN KEY ( genre_id ) REFERENCES genres ( genre_id ),
FOREIGN KEY ( type_id ) REFERENCES types ( type_id )
);
CREATE TABLE genres (
genre_id INT PRIMARY KEY NOT NULL AUTO_INCREMENT,
genre_name VARCHAR(50)
);
CREATE TABLE types (
type_id INT PRIMARY KEY NOT NULL AUTO_INCREMENT,
type_name VARCHAR(50)
);
我也有Java課程
@Entity
@Table(name = "catalog", catalog = "media_store_db")
public class Catalog implements Serializable {
@Id
@Column(name = "id")
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
@Column(name = "product_name", length = 100)
private String productName;
@ManyToOne(cascade = {CascadeType.PERSIST, CascadeType.MERGE})
@JoinColumn(name = "genre_id", referencedColumnName = "genre_id")
private Genre genre;
@ManyToOne(cascade = {CascadeType.PERSIST, CascadeType.MERGE})
@JoinColumn(name = "type_id", referencedColumnName = "type_id")
private Type type;
@Entity
@Table(name = "genres", catalog = "media_store_db")
public class Genre implements Serializable {
@Id
@Column(name = "genre_id")
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
@Column(name = "genre_name")
private String name;
@Entity
@Table(name = "types", catalog = "media_store_db")
public class Type implements Serializable {
@Id
@Column(name = "type_id")
@GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
@Column(name = "type_name")
private String name;
是否可以像這樣保存(使用Hibernate Session的save()方法)Catalog對象
Catalog catalog = new Catalog();
catalog.setProductName("Product");
catalog.setGenre(new Genre());
catalog.setType(new Type());
save(catalog);
沒有寫SQL? 我需要做什么類型和類型? 我應該設置兩個實例的ID嗎?
UPD:
這段代碼工作得很好
Catalog catalog = new Catalog();
catalog.setProductName("12 Years a Slave");
catalog.setGenre(genreRepository.get(Long.valueOf(1)));
catalog.setType(typeRepository.get(Long.valueOf(1)));
Session session = cfg.getSession();
Transaction tx = session.beginTransaction();
session.save(catalog);
tx.commit();
session.close();
當然,您可以使用persist(Object obj)
將生成的Object保存在數據庫中。 那么,您應該在JUnit測試中測試該函數。 在商業代碼中,它應該做你的DAO。 不,生成了所有ID,您不需要設置ID。 它由Hibernate管理。 對於您的示例,UnitTest應如下所示:
public class DataGenerationTest {
private EntityManager em;
@Before
public void init(){
EntityManagerFactory emf = Persistence.createEntityManagerFactory("test");
em = emf.createEntityManager();
}
@Test
public void shouldAddSomeCatalogs(){
em.getTransaction().begin();
Catalog catalog = new Catalog();
catalog.setProductName("Proguct");
catalog.setGenre(new Genre());
catalog.setType(new Type());
em.persist(catalog);
em.getTransaction().commit();
em.close();
}
}
(當然你必須從EntityManagerFactory重命名PersistenceUnit測試。它應該與persistence.xml中命名的PersistenceUnit匹配)
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