mysql - jparepository 接口(interface)上的自定义方法

Question

我有两个实体:人和汽车。我想使用 JpaRepository 按人员 ID 列出汽车。我创建了双向关系，如以下代码

public class Person {
.
.
.
@OneToMany(mapped = "person")
@JsonManagedReference
private List<Car> cars;
.
.
.
}

以及汽车的实体

public class Car {
.
.
.
@ManyToOne
@JoinColumn(name = "person_id")
@JsonBackReference
private Person person;
}

所以，我通过以下方式实现了 JpaRepository 到 CarRepository

public interface CarRepository extends JpaRepository<Car, Long> {
@Query("SELECT c FROM Car c WHERE c.person_id = :person_id")
 public List<Car> getCarsByidPerson(@Param("person_id")long idPerson);
}

字段 person_id 存在于数据库的汽车表中，但我在 spring-boot 启动时收到以下错误消息

org.hibernate.QueryException: could not resolve property: person_id of: com.jpa_study.Car [SELECT c FROM com.jpa_study.Car c WHERE c.person_id = :person_id]

我需要在选择时使用 join 吗？或者这是可能的？

Answer 1

我明白了!

确实需要 JOIN on QUERY。按照代码修复

public interface CarRepository extends JpaRepository<Car, Long> {
@Query("SELECT c FROM Car c INNER JOIN Person p ON (Car.person.id = Person.id) WHERE p.id = :person_id")
 public List<Car> getCarsByidPerson(@Param("person_id")long idPerson);
}