如何使用Hibernate保存复杂对象？
如何使用Hibernate保存复杂对象？
提问于 2014-03-08 00:50:12
我有三张桌子：
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)
);另外，我还有Java类
@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;是否可以保存(使用Hibernate会话的save()方法)
Catalog catalog = new Catalog();
catalog.setProductName("Product");
catalog.setGenre(new Genre());
catalog.setType(new Type());
save(catalog);而不写SQL？我需要做什么与类型和类型？我应该设置这两个实例的id吗？
UPD：
这个代码运行得很好
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();回答 1
Stack Overflow用户
回答已采纳
发布于 2014-03-08 10:26:48
当然，您可以使用persist(Object obj)将生成的对象持久化到数据库中。那么，您应该在JUnit测试中测试这个函数。在业务代码中，它应该执行您的DAO。不，所有的id都是生成的，不需要设置id。它由Hibernate管理。对于您的示例，UnitTest应该如下所示：
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();
}
}(当然，您必须从PersistenceUnit重命名EntityManagerFactory测试。它应该与您的命名PersistenceUnit在persistence.xml中匹配)
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原文链接：
https://stackoverflow.com/questions/22263477
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