I have a Client and Affiliate class, inheriting from Person class. Joined inheritance strategy type is being used - each of them sharing primary key with the parent class. As there's no discriminator column we chose to use DescriptorCustomizer and ClassExtractor. But it doesn't really give any idea how it works, also, the code doesnt seem to compile. It would be nice if someone gives a nice example with code snippet for understanding.
According to the mentioned documentation:
If you are mapping to an existing database, and the tables do not have
a discriminator column you can still define inheritance using the
#ClassExtractor annotation or <class-extractor> element. The class
extractor takes a class that implements the ClassExtractor
interface. An instance of this class is used to determine the class
type to use for a database row. The class extractor must define a
extractClassFromRow() method that takes the database Record and
Session.
we need to annotate the root entity in a hierarchy with user defined using the class extractor:
#Entity
#Inheritance(strategy = InheritanceType.JOINED)
#ClassExtractor(PersonClassExtractor.class)
public abstract class Person {
#Id #GeneratedValue(strategy = GenerationType.AUTO)
private int id;
private String name;
private int age;
// ...
}
Notice that we don't use #Customizer annotations since as this is not required in case of JOINED inheritance strategy:
If a class extractor is used with SINGLE_TABLE inheritance, the rows
of the class type must be able to be filtered in queries. This can be
accomplished by setting an onlyInstancesExpression() or
withAllSubclassesExpression() for branch classes. These can be set
to Expression objects using a DescriptorCustomizer.
The class extractor must be able to determine and return the class type from the database row.
In general we need a replacement of a discriminator column, i.e.
column name unique for a given entity type among others
criteria based on values of a given column of the root entity
Suppose that each of inherited entity type in a hierarchy has a column with unique name:
#Entity
public class Client extends Person {
#Column(name = "CLIENT_SPECIFIC")
private String clientSpecific;
// ...
}
#Entity
public class Affiliate extends Person {
#Column(name = "AFFILIATE_SPECIFIC")
private float affiliateSpecific;
// ...
}
then class extractor may look as follows:
public class PersonClassExtractor extends ClassExtractor {
#Override
public Class<?> extractClassFromRow(Record databaseRow, Session session) {
if (databaseRow.containsKey("CLIENT_SPECIFIC")) {
return Client.class;
} else if (databaseRow.containsKey("AFFILIATE_SPECIFIC")) {
return Affiliate.class;
} else {
return Person.class; // this should never happen
}
}
}
retrieve a list of clients and affiliates
List<Person> polymorphicResults = em.createQuery("SELECT p FROM Person p")
.getResultList();
retrieve a list of affiliates or clients respectively
List<Affiliate> concreteResults = em.createQuery("SELECT a FROM Affiliate a")
.getResultList();
List<Client> concreteResults = em.createQuery("SELECT c FROM Client c")
.getResultList();
Related
I have an entity class as follows. What I need is, I need to use generics, and by using it I need to pass some classes that extend the SomeInfo class and persist a common field inside the SubClass table and when returning data I can populate T parameter and send it easily. But I can't extend SomeInfo class directly because it contains fields that I don't need to be created in the SubClass table and even I don't need to use the Inheritance Single Table solution. I have tried using #AttributeOverride but it threw an error and I used #Type annotation and the error was gone, but the field info_name isn't created.
#Entity
public class SubClass<T extends SomeInfo> extends ParentClass {
#Type(type = "com.package.SomeInfo")
#AttributeOverrides({
#AttributeOverride(name = "name", column = #Column(name = "info_name", nullable = false))
})
private T info;
}
#MappedSuperClass
public abstract class SomeInfo extends ParentClass {
// some fields
}
Is there a way to achieve what I need? Thanks in advance.
I want to use Java records as embeddable objects with JPA. For example I want to wrap the ID in a record to make it typesafe:
#Entity
public class DemoEntity {
#EmbeddedId
private Id id = new Id(UUID.randomUUID());
#Embeddable
public static record Id(#Basic UUID value) implements Serializable {}
}
But If I try to persist it with Hibernate 5.4.32 I get the following error:
org.hibernate.InstantiationException: No default constructor for entity: : com.example.demo.DemoEntity$Id
at org.hibernate.tuple.PojoInstantiator.instantiate(PojoInstantiator.java:85) ~[hibernate-core-5.4.32.Final.jar:5.4.32.Final]
at org.hibernate.tuple.component.AbstractComponentTuplizer.instantiate(AbstractComponentTuplizer.java:84) ~[hibernate-core-5.4.32.Final.jar:5.4.32.Final]
...
So it looks like Hibernate would treat the record Id like an entity, although it is an #Embeddable.
The same happens with non-id fields and #Embedded:
#Embedded
private Thing thing = new Thing("example");
#Embeddable
public static record Thing(#Basic String value) implements Serializable {}
Is there a way to use #Embeddable records with JPA/Hibernate?
Java records with a single field can be used for custom ID types or any other value object with AttributeConverters.
In the entity class the ID type is used with #Id as usual:
#Entity
public class DemoEntity {
#Id
private Id id = new Id(UUID.randomUUID());
public static record Id(UUID value) implements Serializable {}
}
Note that the record Id doesn't have any annotation.
The converter makes it possible to use records:
#Converter(autoApply = true)
public class DemoEntityIdConverter implements AttributeConverter<DemoEntity.Id, String> {
#Override
public String convertToDatabaseColumn(DemoEntity.Id id) {
return id.value().toString();
}
#Override
public DemoEntity.Id convertToEntityAttribute(String s) {
return new DemoEntity.Id(UUID.fromString(s));
}
}
Don't forget to set autoApply = true to have this converter applied automatically (without referencing it explicitly on the respective field).
Records with more than one field could be mapped with a Hibernate UserType, but that is a bit cumbersome.
Entity or embeddable, in any case the record class wouldn't be suitable here because entities and their fields, including embeddable ones, are modifiable. The only exception would be for Id fields, but that doesn't seem like an important enough case to make this functionality for.
One of the Hibernate developers explains this here
I can create a repository via defining an interface on the appropriate JPA class A like the following:
public interface ARepository extends CrudRepository<A, Long>
{
}
and I can use that in my Controller (for example) via
#Autowired
private ARepository aRepository;
and just can do things like this:
aRepository.save(..);
aRepository.findAll();
..
No problem so far.
But my problem is that I have ca. 500 JPA classes and need to access each table which means to define 500 Repositories in the style of above.
So does exist an thing to create that either dynamically via some Spring Data "magic" which from my point of view should exist otherwise the above would not be possible. It looks like this is similar to my problem.
Apart from that one more issue related to the above. I can define findBy... methods in the interface and in the background there will be generated a query method for this particular attribute. The question is also if this can be done in a dynamic way related to the previous question, cause I have groups of tables which need supplemental query methods..
There is spring-data-generator which can automatically generate the interfaces for you.
Regarding your 2nd question I don't think you that can be done in a dynamic way. Java is statically compiled and there's no way to add members dynamically. There could be a tool that generates code for those methods but if that tool generates methods for all combinations of columns you will end up with a huge amount of methods.
You can make a base abstract entity for your 500 classes an then create one repo for this class. (I think it's a common practice to have a BaseEntity class with id, version etc. for every entity in the project).
For simple repo methods (like save, findAll etc.) it will work right from the box (note - entities must have the equal id type). For example:
#Entity
#Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstarct class BaseEntity {
#Id
#GeneratedValue(strategy = GenerationType.SEQUENCE)
private Long id;
}
#Entity
public class Entity1 extends BaseEntity {
private String name;
}
#Entity
public class Entity2 extends BaseEntity {
private String name;
}
public interface BaseEntityRepo extends JpaRepository<BaseEntity, Long> {
}
Note that BaseEntity must have #Inheritance(strategy = InheritanceType.TABLE_PER_CLASS) to prevent of using singe table base_entity for every entity. And their ids must not intersect (see #GeneratedValue(strategy = GenerationType.SEQUENCE)).
Usage:
#RunWith(SpringRunner.class)
#SpringBootTest
public class BaseEntityRepoTest {
#Autowired private BaseEntityRepo repo;
#Before
public void setUp() throws Exception {
repo.save(asList(
new Entity1("entity1"),
new Entity2("entity2")
));
}
#Test
public void readingTest() throws Exception {
List<BaseEntity> entities = repo.findAll();
assertThat(entities).hasSize(2);
}
}
Related to your second question you can use this approach:
public interface BaseEntityRepo extends JpaRepository<BaseEntity, Long> {
<T> T findById(Long id, Class<T> type);
}
Usage:
#Test
public void findById() {
final Entity1 entity1 = repo.findById(1L, Entity1.class);
final Entity2 entity2 = repo.findById(2L, Entity2.class);
assertThat(entity1).isNotNull();
assertThat(entity2).isNotNull();
}
But you can build repo query methods only for 'common' properties of inherited entities which are present in the base class. To make this method work you must move the name parameter to the BaseEntity:
<T> List<T> findAllByNameLike(String name, Class<T> type);
I am using Hibernate to connect to my database.
I have an inheritance structure in my application.The problem is that when i do a query like "from Animal", it does a left outer join for the class Animal,its sub classes and all the associations for Animal and its subclasses.
How do i avoid this situation.I want to load the data only when i specify it through a fetchmode in my criteria query?
Yes, Hibernate supports polymorphic queries. From the documentation:
14.8. Polymorphic queries
A query like:
from Cat as cat
returns instances not only of Cat, but
also of subclasses like DomesticCat.
Hibernate queries can name any Java
class or interface in the from clause.
The query will return instances of all
persistent classes that extend that
class or implement the interface. The
following query would return all
persistent objects:
from java.lang.Object o
The interface Named might be
implemented by various persistent
classes:
from Named n, Named m where n.name = m.name
These last two queries will require
more than one SQL SELECT. This means
that the order by clause does not
correctly order the whole result set.
It also means you cannot call these
queries using Query.scroll().
This is the default behavior (called implicit polymorphism) and Hibernate supports both implicit and explicit polymorphism:
Implicit polymorphism means that instances of the class will be
returned by a query that names any
superclass or implemented interface or
class, and that instances of any
subclass of the class will be returned
by a query that names the class
itself. Explicit polymorphism means
that class instances will be returned
only by queries that explicitly name
that class. Queries that name the
class will return only instances of
subclasses mapped inside this
<class> declaration as a
<subclass> or <joined-subclass>.
For most purposes, the default
polymorphism="implicit" is
appropriate. Explicit polymorphism is
useful when two different classes are
mapped to the same table This allows a
"lightweight" class that contains a
subset of the table columns.
This can be configured at the class level. Use polymorphism="explicit" if you are if you are using xml mappings, see 5.1.3 Class. Use Hibernate's #Entity annotation if you're using annotations, see 2.4.1. Entity. Below an example:
#javax.persistence.Entity
#org.hibernate.annotations.Entity(polymorphism = PolymorphismType.EXPLICIT)
#Inheritance(strategy = InheritanceType.JOINED)
public class Foo {
...
}
Assume you have a class structure as follows:
class Animal { }
class Dog : Animal { }
class Cat : Animal { }
then when you select all Animals, you'd expect to also load all Dogs and Cats. After all they are Animals.
A different story are the associations. You can created you mappings such that the associations are lazy load instead of eager load.
Basically it's the default ORM inheritance design pattern used by Hibernate called class inheritance (all the classes are mapped to a single table), if you want to change that you can google:
- single class hierarhy or table per class (this will map every class to a separate table in the DB)
- concrete class hierarhy (this will map only the concrete implementations to a table).
To avoid multiple joins during class hierarchy fetching you can apply SINGLE_TABLE hierarchy mapping strategy, and then define secondary tables on subclasses with SELECT fetching strategy. However, this turns you "heavy join" model into "N+1 select" model. The example:
#Entity
#Inheritance(strategy = InheritanceType.SINGLE_TABLE)
#DiscriminatorColumn(name = Super.DISCRIMINATOR_COLUMN, discriminatorType = DiscriminatorType .STRING, length = 255)
public class Super {
public static final String DISCRIMINATOR_COLUMN = "classname";
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
protected long id;
#Column(updatable = false, insertable = false)
protected String classname;
protected String superProp = "superProp";
public long getId() {
return id;
}
public String getClassname() {
return classname;
}
public String getSuperProp() {
return superProp;
}
public void setSuperProp(String superProp) {
this.superProp = superProp;
}
}
#Entity
#SecondaryTable(name = SubA.TABLE)
#Table(appliesTo = SubA.TABLE, fetch = FetchMode.SELECT)
public class SubA extends Super {
public static final String TABLE = "SUBA";
#Column(table = TABLE)
protected String subAProp = "subAProp";
public String getSubAProp() {
return subAProp;
}
public void setSubAProp(String subAProp) {
this.subAProp = subAProp;
}
}
#Entity
#SecondaryTable(name = SubB.TABLE)
#Table(appliesTo = SubB.TABLE, fetch = FetchMode.SELECT)
public class SubB extends Super {
public static final String TABLE = "SUBB";
#Column(table = TABLE)
protected String subBProp = "subBProp";
public String getSubBProp() {
return subBProp;
}
public void setSubBProp(String subBProp) {
this.subBProp = subBProp;
}
}
And what SQL is done on from Super HQL query:
select [...] from SUPER super0_
select super_1_.subaprop as subaprop1_83_ from SUBA super_1_ where super_1_.id=1
select super_2_.subbprop as subbprop1_84_ from SUBB super_2_ where super_2_.id=2
More about this approach and general hibernate performance hints you can read in my article.
Suppose a Table per subclass inheritance relationship which can be described bellow (From wikibooks.org - see here)
Notice Parent class is not abstract
#Entity
#Inheritance(strategy=InheritanceType.JOINED)
public class Project {
#Id
private long id;
// Other properties
}
#Entity
#Table(name="LARGEPROJECT")
public class LargeProject extends Project {
private BigDecimal budget;
}
#Entity
#Table(name="SMALLPROJECT")
public class SmallProject extends Project {
}
I have a scenario where i just need to retrieve the Parent class. Because of performance issues, What should i do to run a HQL query in order to retrieve the Parent class and just the Parent class without loading any subclass ???
A workaround is described below:
Define your Parent class as MappedSuperClass. Let's suppose the parent class is mapped To PARENT_TABLE
#MappedSuperClass
public abstract class AbstractParent implements Serializable {
#Id
#GeneratedValue
private long id;
#Column(table="PARENT_TABLE")
private String someProperty;
// getter's and setter's
}
For each subclass, extend the AbstractParent class and define its SecondaryTable. Any persistent field defined in the parent class will be mapped to the table defined by SecondaryTable. And finally, use AttributeOverrides if needed
#Entity
#SecondaryTable("PARENT_TABLE")
public class Child extends AbstractParent {
private String childField;
public String getChildProperty() {
return childField;
}
}
And define another Entity with the purpose of retrieving just the parent class
#Entity
#Table(name="PARENT_TABLE")
#AttributeOverrides({
#AttributeOverride(name="someProperty", column=#Column(name="someProperty"))
})
public class Parent extends AbstractParent {}
Nothing else. See as shown above i have used just JPA specific annotations
Update: It appears the first option doesn't work as I thought.
First option:
Specify the class in the where clause:
select p from Project p where p.class = Project
Second option:
Use explicit polymorphism that you can set using Hibernate's #Entity annotation:
#javax.persistence.Entity
#org.hibernate.annotations.Entity(polymorphism = PolymorphismType.EXPLICIT)
#Inheritance(strategy = InheritanceType.JOINED)
public class Project {
#Id
private long id;
...
}
This is what Hibernate Core documentation writes about explicit polymorphism:
Implicit polymorphism means that
instances of the class will be
returned by a query that names any
superclass or implemented interface or
class, and that instances of any
subclass of the class will be returned
by a query that names the class
itself. Explicit polymorphism means
that class instances will be returned
only by queries that explicitly name
that class.
See also
How to get only super class in table-per-subclass strategy?
Actually, there is a way to get just the superclass, you just need to use the native query from JPA, in my case I'm using JPA Repositories it would be something like that:
#Query(value = "SELECT * FROM project", nativeQuery = true)
List<Resource> findAllProject();
The flag nativeQuery as true allow running the native SQL on database.
If you are using Entity Manager check this out: https://www.thoughts-on-java.org/jpa-native-queries/