我有以下實體:HQL用於獲取分鐘
Company.class:
public class Company {
@JoinTable(name = "company_employee", joinColumns = @JoinColumn(name = "company_id") , inverseJoinColumns = @JoinColumn(name = "employee_id"))
@ManyToMany(fetch = FetchType.LAZY)
private Set<Employee> employees;
@OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, orphanRemoval = true, mappedBy = "company")
private List<Score> scores;
@JoinTable(name = "company_factor", joinColumns = @JoinColumn(name = "company_id") , inverseJoinColumns = @JoinColumn(name = "factor_id"))
@ManyToOne(fetch = FetchType.LAZY)
private Factor factors;
}
和Employee.class
public class Employee {
@ManyToMany(fetch = FetchType.EAGER, mappedBy="employees")
private Set<Company> companies;
@OneToMany(fetch = FetchType.LAZY, cascade = CascadeType.ALL, orphanRemoval = true, mappedBy = "company")
private List<Score> scores;
@JoinTable(name = "employee_factor", joinColumns = @JoinColumn(name = "employee_id") , inverseJoinColumns = @JoinColumn(name = "factor_id"))
@ManyToMany(fetch = FetchType.LAZY)
private Set<Factor> factors;
@Transient
private int score;
}
Factor.class不包含任何關係。
此外,我有一些實體評分是每個組合公司員工獨特的實體評分。 它看起來像這樣: Score.class:
@ManyToOne(fetch=FetchType.LAZY)
@JoinColumn(name = "company_id", insertable = false, updatable = false)
private Company company;
@ManyToOne(fetch=FetchType.EAGER)
@JoinColumn(name = "employee_id", insertable = false, updatable = false)
private Employee employee;
@Column(name = "score")
private BigDecimal score;
如果我得到的名單,這將是組合公司和員工實例的列表,有時公司或員工可以重複。 目標是獲得列表,按員工中的因子進行過濾,並僅顯示按升序排列的每位員工的最低分數。 說,如果存在組合
employee1-company1, score=1
employee1-company2, score=2
employee2-company1, score=3
employee2-company4, score=5
employee3-company4, score6
ResultList應該看起來像:
employee1-company1, score=1
employee2-company1, score=3
employee3-company4, score=6
所以員工應該不會重複,但公司可以在列表中重複。 我不太確定,該怎麼做。我所取得的成績是以升序顯示獨特的結果,但他們沒有顯示最低分數。我使用HQL:
select distinct e from Score e
left outer join fetch e.company
left outer join fetch e.company.factors
left outer join fetch e.employee
left outer join fetch e.employee.factors ef
where ef.factor_id=:factor_id
group by e.employee.employee_id
order by e.score asc
任何人都可以幫助如何實現我所需要的?謝謝。
UPDATE1:
我決定走另一條路。
select distinct e from Employee e join e.scores es order by es.score asc
現在看來,這正是我需要的,但如何在查詢中把最低es.score到現場比分Employee
對象: 現在我通過員工使用此查詢得到它?也許有一種方法可以用es.score
替代e.score
?通過對上述獲取ASC的結果(默認)得分次序
組:
不在JPA中,因爲您不能將子查詢作爲FROM。 –
謝謝,但它不起作用,說'('是意外的標記:'org.springframework.dao.InvalidDataAccessApiUsageException:org.hibernate.hql.internal.ast.QuerySyntaxException:意外的標記:'只爲案件:我在'entityManager.createQuery()'中使用它 –