我正在寻找一种在一对多关系上进行联接的方法,其中多方是通过继承定义的,而联接的右侧部分仅限于特定的子类(向下转型)。
假设我有以下实体(示例取自 here ):
@Entity
public class Project {
@Id
@GeneratedValue
private long id;
private String name;
@OneToMany(cascade = CascadeType.ALL)
private List<Employee> employees;
.............
}
@Inheritance(strategy = InheritanceType.SINGLE_TABLE)
@Entity
@DiscriminatorColumn(name = "EMP_TYPE")
public class Employee {
@Id
@GeneratedValue
private long id;
private String name;
.............
}
@Entity
@DiscriminatorValue("F")
public class FullTimeEmployee extends Employee {
private int annualSalary;
.............
}
@Entity
@DiscriminatorValue("P")
public class PartTimeEmployee extends Employee {
private int weeklySalary;
.............
}
@Entity
@DiscriminatorValue("C")
public class ContractEmployee extends Employee {
private int hourlyRate;
.............
}
我可以轻松构建涉及父类(super class) Employee
中定义的属性的联接查询,例如:
JPAQuery query = ...
QProject project = new QProject("p");
QEmployee employee = new QEmployee("e");
query.join(project.employees, employee);
query.where(employee.name.startsWith("A"));
但是,如果我想访问子类的属性,例如 FullTimeEmployee.annualSalary
,从而将连接限制为该子类型,我该怎么做?
如何构建与以下 JPQL 等效的内容:
SELECT DISTINCT p FROM Project p JOIN TREAT(p.employees AS FullTimeEmployee) e WHERE e.annualSalary > 100000
最佳答案
你可以这样做:
EntityManager em = ...;
QProject p = QProject.project;
QFullTimeEmployee e = QFullTimeEmployee.fullTimeEmployee;
List<FullTimeEmployee> emps = new JPAQuery<>(em)
.select(p)
.distinct()
.from(p)
.innerJoin(p.employees, e._super)
.where(e.annualSalary.gt(100000))
.fetch();
另请参阅 Querydsl 论坛上的这篇文章:https://groups.google.com/d/msg/querydsl/4G_ea_mQJgY/JKD5lRamAQAJ
关于java - Querydsl:如何向下转换连接实体?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/57724240/