【发布时间】:2023-11-23 08:55:01
【问题描述】:
你好,我有两个表
我正在尝试实现这样的查询
SELECT * FROM jobs.phonenumber where Person_ID='1';
但在 jpql 中。
我正在尝试制作类似的东西:
String query="SELECT p1 FROM PhoneNumber p1 WHERE p1.person=:id;
这是我的实体类的示例:
@Entity
public class PhoneNumber {
@Id
@GeneratedValue
@Column(name = "phoneId")
private long id;
@Column(name = "phoneNumber")
private int number;
@ManyToOne(fetch=FetchType.LAZY)
@JoinColumn(name = "Person_ID")
private Person person;
@Entity
public class Person {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "personId")
private long id;
@Column(name = "personName")
private String name;
@Column(name = "personAge")
private int age;
@OneToMany(mappedBy = "person")
private List<PhoneNumber> phones;
我有一个错误:参数值1 与预期类型不匹配 [model.Person (n/a)]
这是我的方法:
@Path("{id}/phones")
@GET
@Produces(MediaType.TEXT_PLAIN)
public List<PhoneNumber> getPhoneNumbers(@PathParam(value = "id") long id) {
List<PhoneNumber> resultList = new ArrayList<>();
EntityManager createEntityManager = emf.createEntityManager();
String query="select p1 from PhoneNumber p1 where p1.person=:id";
TypedQuery<PhoneNumber> createNamedQuery = createEntityManager.createQuery(query, PhoneNumber.class);
createNamedQuery.setParameter("id", id);
try {
resultList = createNamedQuery.getResultList();
} catch (Exception e) {
System.out.println("Could not retrive phone list");
}
return resultList;
}
有人可以帮我解决一下吗,我做错了什么?
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