获取最小值的 HQL

Question

我有以下实体：

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 不包含任何关系。

另外，我有一些实体分数对于公司-员工的每个组合都是唯一的。它看起来像这样： 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List，它将是Company和Employee实例的组合列表，有时Company或Employee可以重复。目标是获取 List，按 Employee 中的 Factor 过滤，并仅显示按升序排列的每个员工的最小分数。说，如果存在组合

employee1-company1, score=1
employee1-company2, score=2
employee2-company1, score=3
employee2-company4, score=5
employee3-company4, score6

ResultList 应如下所示：

employee1-company1, score=1
employee2-company1, score=3
employee3-company4, score=6

所以员工不应该重复，但公司可以在列表中重复。我不太确定，该怎么做。我取得的成就是按升序显示独特的结果，但没有显示最低分数。我使用了 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

任何人都可以帮助实现我所需要的吗？谢谢。

更新1：

我决定走另一条路。现在我通过 Employee 使用此查询获取它：

select distinct e from Employee e join e.scores es order by es.score asc

看来正是我需要的，但是如何在查询中放入最小es.score到Employee对象的字段分数呢？也许有一些方法可以将 e.score 替换为 es.score?

Answer 1

您可以使用以下内容：

Select a from (Select b from Score b order by score) as a group by a.employee

解释：

Select b from Score b order by score ：以 score
根据以上结果分组将给出唯一的员工

Answer 2

作为解决方案，我切换到 entityManager.createNativeQuery("some native sql string")。我对结果很满意。就此而言，关于 SQL 查询的问题是 here 唯一的缺点是无法使用join fetch，所以N+1select的问题就在这里，不过我打算取的数据量比较小，所以还可以接受。

获取最小值的 HQL

HQL for getting min

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