我有这样的entity hierarchy。除了一些通用属性外,某些属性仅由一些子类型共享:
@Entity
@Inheritance(strategy = InheritanceType.JOINED)
public class Person {
private String firstName;
private String lastName
... further properties, getters and setters...
}
@Entity
public class Employee extends Person {
private String salary;
... further properties, getters and setters...
}
@Entity
public class BoardMember extends Person {
private String salary;
... further properties, getters and setters...
}
@Entity
public class ExternalMember extends Person {
private String clearanceLevel;
... further properties, getters and setters...
}
@Repository
public interface PersonRepository extends JpaRepository<Person, Long>, QuerydslPredicateExecutor<Person> {
}
使用QueryDSL,我试图根据像这样的动态过滤器标准来搜索人员:
@Service
@Transactional
public class PersonService {
@Autowired
PersonRepository personRepository;
public Page<Person> search(String firstName, String salary) {
var searchCriterias = new BooleanBuilder();
if (firstName != null) {
searchCriterias.and(QPerson.firstName.eq(firstName));
}
if (salary != null) {
searchCriterias.andAnyOf(
QPerson.person.as(QEmployee.class).salary.eq(salary),
QPerson.person.as(QBoardMember.class).salary.eq(salary),
);
}
personRepository.findAll(searchCriterias);
}
}
这似乎不是正确的方法,但是,我遇到很多错误,例如“工资”不是Person的成员。
处理分层实体的搜索有多种方法?我希望QueryDSL具有类型安全性,但是使用Spring Data Specification的解决方案也可以。
编辑:使用15种以上不同的搜索条件,搜索条件可能会变得非常复杂。因此,我需要一种编程方法来制定它们。
我不知道您的数据库架构,但是由于您使用的是InheritanceType.JOINED
,因此有一个用于超类的表和一个用于每个子实体的表,它们都被联接了。假设子实体具有一列person_id
,该列将其与父实体连接在一起。然后,您可以编写如下查询:
entityManager.createNativeQuery(
"select p.* " +
"from person p " +
"left join employee e on e.person_id = p.id " +
"left join board_member bm on bm.person_id = p.id " +
"where (:firstName is null or p.first_name = :firstName) " +
"and (:salary is null or e.salary = :salary or bm.salary = :salary);"
)
.unwrap(org.hibernate.query.Query.class)
.setParameter("firstName", firstName, StringType.INSTANCE)
.setParameter("salary", salary, IntegerType.INSTANCE)
.getResultList();
注意unwrap
,让我们将setParameter
与类型为Type
的第三个参数一起使用。当传递的参数可能为null
时,它是必需的,以便休眠模式始终可以确定其类型。
我无法重现“不是Person的成员”错误,但我设法从您的查询中得到结果。
跟随Baeldung's tutorial并根据您的问题改编代码,我已经设法获得没有错误的结果。 Here是示例项目。希望对您有所帮助
import javax.persistence.*;
@Entity
@Inheritance(strategy = InheritanceType.JOINED)
public class Person {
@Id
@GeneratedValue(strategy = GenerationType.AUTO)
private int id;
@Column
private String firstName;
@Column
private String lastName;
public Person(String firstName, String lastName) {
this.firstName = firstName;
this.lastName = lastName;
}
public Person() {
}
}
import javax.persistence.Column;
import javax.persistence.Entity;
@Entity
public class BoardMember extends Person {
@Column
private String salary;
public BoardMember(String firstName, String lastName, String salary) {
super(firstName, lastName);
this.salary = salary;
}
public BoardMember() {
}
}
@Service
@Transactional
public class PersonService {
@Autowired
PersonRepository personRepository;
public List<Person> search(String firstName, String salary) {
var searchCriterias = new BooleanBuilder();
if (firstName != null) {
searchCriterias.and(QPerson.person.firstName.eq(firstName));
}
if (salary != null) {
searchCriterias.andAnyOf(
QPerson.person.as(QEmployee.class).salary.eq(salary),
QPerson.person.as(QBoardMember.class).salary.eq(salary)
);
}
var result = new ArrayList<Person>();
for (Person person : personRepository.findAll(searchCriterias)) {
result.add(person);
}
return result;
}
}