Querydsl：如何向下转换连接实体？

Question

我正在寻找一种在一对多关系上进行连接的方法，其中多方是通过继承定义的，连接的右侧部分仅限于特定的子类（向下转换）。

假设我有以下实体（示例来自 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超类 Employee 中定义的属性的连接查询，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"));

但是如果我想访问属性的子类，比如 FullTimeEmployee.annualSalary，从而将连接限制为该子类型，我该怎么做？

如何构建以下 JPQL 的等价物：

SELECT DISTINCT p FROM Project p JOIN TREAT(p.employees AS FullTimeEmployee) e WHERE e.annualSalary > 100000

Answer 1

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 post：https://groups.google.com/d/msg/querydsl/4G_ea_mQJgY/JKD5lRamAQAJ

Querydsl：如何向下转换连接实体？

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java

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