我遇到了以下问题:我想简单地加入表markets(main)和telephones,但是在使用注解执行此操作时遇到了一些问题。
我只想使用comercio.id作为telefone.fkïid的对应外键,但它无法识别它。我还想问一下是否需要为这两个类创建一个存储库接口。
编辑。另外,对于这样一个例子,post-req看起来怎么样?
科梅西奥.java
@Entity
@Table(name = "comercio")
public class Comercio {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String nome;
private String cnpj;
private String endereco;
@OneToMany(targetEntity = Telefone.class , mappedBy = "comercio", fetch = FetchType.LAZY)
private List<Telefone> telefones;
private String email;
电话.java
@Entity
@Table(name = "telefone")
public class Telefone {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long telefone_id;
@ManyToOne
@JoinColumn(name="fk_id", referencedColumnName = "id")
private Comercio comercio;
private String telefone;
我得到的答案是:
{
"nome" : "Americanas",
"cnpj" : "000",
"endereco" : "SQN 112",
"telefones" : [ ],
"email" : "contato@americanas.com",
"comercio_id" : 1
}
慰问:
Hibernate: select comercio0_.id as id1_0_, comercio0_.cnpj as cnpj2_0_, comercio0_.email as email3_0_, comercio0_.endereco as endereco4_0_, comercio0_.nome as nome5_0_ from comercio comercio0_
Hibernate: select telefones0_.fk_id as fk_id3_1_0_, telefones0_.telefone_id as telefone1_1_0_, telefones0_.telefone_id as telefone1_1_1_, telefones0_.fk_id as fk_id3_1_1_, telefones0_.telefone as telefone2_1_1_ from telefone telefones0_ where telefones0_.fk_id=**?**
感谢你的帮助。
1条答案
按热度按时间ztyzrc3y1#
我认为电话本身并不是一个实体:你永远不会想要查询电话,它只作为comercio的一部分存在。
然后可能是
Embeddable
但是,与实体不同的是,它只有一个属性-数字-然后您可以简化为下面的内容,并将其Map为字符串的集合。那么就不需要一个Telefone
班级。