我有三个表:
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;
是否可以像这样保存(使用 Hibernate Session 的 save() 方法)Catalog 对象
Catalog catalog = new Catalog();
catalog.setProductName("Product");
catalog.setGenre(new Genre());
catalog.setType(new Type());
save(catalog);
不用写SQL?我需要用流派和类型做什么?我应该设置两个实例的 ID 吗?
更新:
这段代码工作得很好
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();
最佳答案
当然,您可以使用 persist(Object obj)
将生成的对象保存在数据库中。
那么,您应该在 JUnit 测试中测试该函数。在业务代码中,它应该执行您的 DAO。
不需要,所有的Ids都是生成的,你不需要设置id。它由 Hibernate 管理。
对于您的示例,单元测试应如下所示:
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,您必须重命名 EntityManagerFactory 中的 PersistenceUnit 测试。它应该与您在 persistence.xml 中命名的 PersistenceUnit 相匹配)
其他有趣的讲座:
关于java - 如何使用 Hibernate 保存复杂的对象?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22263477/