[FIXED] JPA Criteria: How to filter on multiple entity subclasses @Inheritance(strategy = InheritanceType.SINGLE_TABLE)

Issue

I wonder if there is any elegant way, to filter on Employee entity following data model below.

So if i need to filter on "name", "hourlyRate" and "salary" i can't use jpa specification, i need write a raw sql

model

 @Inheritance(strategy = InheritanceType.SINGLE_TABLE)
   @Entity
   @DiscriminatorColumn(name = "EMP_TYPE")
        JsonTypeInfo(use = JsonTypeInfo.Id.NAME, include = JsonTypeInfo.As.EXISTING_PROPERTY, property = "EMP_TYPE", visible = true)
        @JsonSubTypes({
          @JsonSubTypes.Type(value = FullTimeEmployee .class, name = "F"),
          @JsonSubTypes.Type(value = PartTimeEmployee .class, name = "P"),
    
        })
    
    public class Employee {
     @Id
     @GeneratedValue
     private long id;
     private String name;

    }

Type classes

    @Entity
@DiscriminatorValue("F")
public class FullTimeEmployee extends Employee {
  private int salary;
    .............
}

@Entity
@DiscriminatorValue("P")
public class PartTimeEmployee extends Employee {
  private int hourlyRate;
    .............
}

'SHOW TABLES'
[EMPLOYEE, PUBLIC]
'SHOW COLUMNS from EMPLOYEE'
[EMP_TYPE, VARCHAR(31), NO, , NULL]
[ID, BIGINT(19), NO, PRI, NULL]
[NAME, VARCHAR(255), YES, , NULL]
[HOURLYRATE, INTEGER(10), YES, , NULL]
[SALARY, INTEGER(10), YES, , NULL]

Solution

You can write a HQL query like the following:

from Employee e 
where (type(e) = FullTimeEmployee and e.salary > 1000) 
   or (type(e) = PartTimeEmployee and e.hourlyRate > 100)

Answered By - Christian Beikov
Answer Checked By - Terry (JavaFixing Volunteer)