Hibernate Annotations - @MappedSuperclass How to override column identifier

Question

I'm working with Hibernate Annotations and the issue that I'm trying to solve goes as follows:

I need to have 2 different @Entity classes with the same columns mapping but with a different Identifier.

The first one should use id as identifier.

The second should use name as identifier.

So, I have an abstract class, annotated with @MappedSuperclass that have all of the columns including id and name, and in addition 2 @Entity classes that extends the super class and overriding the getters of the id and name.

@MappedSuperclass public class MappingBase { protected Integer id; protected String name; @Column (name = "ID") public void getId() { return this.id; } @Column (name = "NAME") public void getName() { return this.name; } } @Entity @Table (name = "TABLE") public class Entity1 extends MappingBase { @Id @Column (name = "ID") public void getId() { return this.id; } } @Entity @Table (name = "TABLE") public class Entity2 extends MappingBase { @Id @Column (name = "NAME") public void getName() { return this.name; } }

Note: I must have the members (id,name) in the super class. I know that i can add @Transient to the id and name getters but this means that i must add both of them in each class and it's not a good design :( In addition, the following insertable="false, updateable=false can help but i don't understand what is the meaning of this...

Please help me!

Did you got a way to do this, I too have same situation where I need to override the @Id field in the subclass — Ysak
– Ysak, Commented Nov 25, 2015 at 9:18
You can't override the id. You can if you put the @Id annotation on the getter, but then you can't have your column annotations on your fields, they all must be on the getters/setters as well, since jpa doesn't support mixed annotations on entities. Build 2 hierarchies with 2 mappedsupperclasses — metodski
– metodski, Commented Jan 24, 2020 at 8:51

misbah · Accepted Answer · 2016-01-12 06:58:26Z

Hibernate/JPA allows us to annotate either properties or accessors. If we have @Id annotation on a property, JPA will lookup all the properties of the class. Similarly, if we have @id annotation on a getter method, JPA will lookup all the getters.

We can solve the above problem by annotating properties instead. The superclass and the two subclasses will be as follows-

@MappedSuperclass public abstract class AbstractMappingBase { //properties other than id and name public abstract Integer getId(); public abstract String getName(); //other getters and setters } @Entity public class Entity1 extends AbstractMappingBase { @Id private Integer id; private String name; @Override public Integer getId() { return id; } public void setId(Integer id) { this.id = id; } @Override public String getName() { return name; } public void setName(String name) { this.name = name; } } @Entity public class Entity2 extends AbstractMappingBase { private Integer id; @Id private String name; @Override public Integer getId() { return id; } public void setId(Integer id) { this.id = id; } @Override public String getName() { return name; } public void setName(String name) { this.name = name; } }

Here JPA will look for properties instead of getters. There are no duplicate properties between superclass and its subclasses. So it will work fine.

dinukadev · Accepted Answer · 2013-01-09 07:26:44Z

1

You are much better off defining your base class as @Embeddable and using @Embedded in your implementation classes with the use of @AttributeOverride.

answered Jan 9, 2013 at 7:26

dinukadev

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2 Comments

kmualem Over a year ago

I thought that this is the way to create a composite identifier, in addition i can't wrap those fields, i need them as class members and not a part from another object. Can you please advise ?

WoooHaaaa Over a year ago

@Kobim , have you find a way to make the overide ?

kmualem · Accepted Answer · 2015-11-26 10:37:56Z

If i remember correctly, I simply defined 2 @Entity classes with the same table that inherits from one abstract @MappedSuperclass class. The super class contains the id member and each Entity class define it's own @Id @Column definition. It should work!

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