如何使用 querydsl 或 spring data jpa 规范对分层实体执行查询?

Mat*_*iMu 6 java hibernate querydsl spring-data-jpa

我有一个这样的实体层次结构。除了一些常见的属性外,一些属性仅由几个子类型共享:

@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> {

}
Run Code Online (Sandbox Code Playgroud)

使用 QueryDSL 我试图根据这样的动态过滤条件搜索人员:

@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);
  }
}
Run Code Online (Sandbox Code Playgroud)

这似乎不是正确的方法,但是,我收到了很多错误,例如“薪水”而不是 Person 的成员。

处理分层实体搜索的各种方法是什么?为了类型安全,我更喜欢 QueryDSL,但使用 Spring Data Specification 的解决方案也很好。

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。所以我需要一种程序化的方法来制定它们。

kas*_*tom 1

我无法重现“不是 Person 的成员”错误,但我已设法从您的查询中获取结果。

按照Baeldung 的教程并调整您问题中的代码,我成功地获得了没有错误的结果。是示例项目。希望这可以帮助

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() {
    }
}
Run Code Online (Sandbox Code Playgroud)
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() {
    }
}
Run Code Online (Sandbox Code Playgroud)
@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;
    }
}
Run Code Online (Sandbox Code Playgroud)