我遇到以下问题:我想简单地将表Markets(主)和Telephones联接在一起,但是在使用批注进行操作时却遇到了一些问题。
我只是想将Comercio.id用作Telefone.fk_id的相应外键,但无法识别。我还想问一下是否需要为两个类创建一个存储库接口。
编辑。另外,对于这样的示例,POST请求看起来如何?
Comercio.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;
Telefone.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:
{
"nome" : "Americanas",
"cnpj" : "000",
"endereco" : "SQN 112",
"telefones" : [ ],
"email" : "[email protected]",
"comercio_id" : 1
}
安慰:
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=**?**
感谢任何帮助。
最佳答案
我建议电话本身不是一个实体:您永远不会查询电话,它仅作为Comercio的一部分存在。
然后,它可能是Embeddable
而不是实体,因为它只有一个属性-数字-然后您可以简化为以下内容并将其映射为字符串的集合。这样就不需要Telefone
类。
@Entity
@Table(name = "comercio")
public class Comercio {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String nome;
private String cnpj;
private String endereco;
@ELementCollection
@CollectionTable(name = "comercio_telefone",
joinColumns=@JoinColumn(name="comercio_id"))
private List<String> telefones;
private String email;
}
关于java - JoinColumn Hibernate/API,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/60217911/