I have a project that I've recently inherited that has 2 tables that share a lot of common fields. I'm new to hibernate and want to know if I can use composition to generate the table instead of inheritance? B and D are basically the same class with a different table name.
Current Hierarchy is
B extends A extends BaseClass
D extends C extends BaseClass
My problem at the moment is that a lot of other classes extend BaseClass which don't have the shared fields and the 2 child classes don't share a common parent so I cannot add another level into the Hierarchy and use #MappedSuperclass.
Because of this I'd like to know if I can group my common fields into a single class and compose my child classes with this new class somehow?
Apologies for the cryptic names but as always; confidentiality...
edit - found something simmilar with #Embeddable https://docs.jboss.org/hibernate/orm/5.2/userguide/html_single/chapters/domain/embeddables.html
If you are using the JPA interface to Hibernate, you can use #Embedded and #Embeddable, to get more or less what you want. Be aware that the change will not be transparent: where you had:
#Entity
public class B extends A {
#Basic
private int foo;
...
}
that you referenced in JPQL by using b.foo, you will have:
#Embeddable
public classs Common {
#Basic
private int foo;
...
}
#Entity
public class B extends A {
#Embedded
private Common common;
...
}
which you will have to reference in JPQL using b.common.foo.
Read more about embeddable entities here:
https://en.wikibooks.org/wiki/Java_Persistence/Embeddables
You could potentially use #Embedded and embed the same object for both B and D, perhaps something like:
#Embeddable
public class CommonFieldObject {
#Column(name="COMFIELD1")
private String commonField1;
#Column(name="COMFIELD2")
private String commonField2;
...
}
#Table
public class C extends A {
#Embedded
#AttributeOverrides({
#AttributeOverride(name="commonField1", column=#Column(name="CFO_COMFIELD1")),
#AttributeOverride(name="commonField2", column=#Column(name="CFO_COMFIELD2"))
})
private CommonFieldObject commonFieldObj; //CFO_ prefix for this reference - in case we have a second field referencing a CommonFieldObject - use a different prefix..
...
}
You should then get the columns CFO_COMFIELD1 and CFO_COMFIELD2 in your table and you can recycle the CommonFieldObject for class D.
Related
I have two entities
#Entity
#Table(name = "view_a")
public class A extends BaseStringIdTableClass
#Entity
#Table(name = "view_b")
public class B extends BaseStringIdTableClass
#Entity
#Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstract class BaseStringIdTableClass implements Serializable {
#Id
private String id;
And in the database i have two views
select * from view_a
|ID|ColumnA|....
|34222|some Value|....
select * from view_b
|ID|ColumnB|...
|34222|lla lla|...
I have therefore in the database different views. But the rows in this different views have the same ID.
Now i try to read the entities with the standard CRUD Repository.
A a = aRepository.findById("34222").get();
B b = bRepository.findById("34222").get();
In this case i can not find entity b. If i swop the two lines of code i can not find entity a.
I think the persistence context can at one time containt for a specific ID only one entity? Is this right. What can i do?
Repository definitions
public interface ARepository extends JpaRepository<A, String>, QuerydslPredicateExecutor<A> {
public interface BRepository extends JpaRepository<B, String>, QuerydslPredicateExecutor<B> {
Sleeping over one night always help....First sorry for my initial incomplete question.
The problem/error was the both entities extended the same abstract class. And in this abstract class the ID was definied.
Fix after this recogniton was easy. One of the entities does not extend my abstract class, but definies his own id. And now it works......
It's possible to use queryDSL generated classes to refer abstract methods from class?
Here is a example:
#Entity
#Inheritance(strategy = InheritanceType.JOINED)
class A {
#Id private Long id;
public getId/setId;
protected abstract Date finalDate();
}
#Entity
class B extends A {
private Date finalDate;
public getFinalDate/setFinalDate;
}
#Entity
class C extends A {
private B b;
public getFinalDate(){return b.getFinalDate());
}
I would like to use a query like this:
new JPAQuery<A>(em)
.select(a)
.where(a.finalDate.isNotNull())
.fetch();
But after build phase (Construct meta classes), the generated A class (QA.class), doesn't has nothing like.
JB Nizets comment is the answer:
No the query needs to be translated to SQL, and executed by the
database. The database doesn't know and care about your classes and
methods. All it knows about is its tables and columns. – JB Nizet 10
hours ago
I have the following
#MappedSuperclass
public abstract class A {
#Id #GeneratedValue
public Long id;
}
#Entity
public class B extends A {
}
#Entity
public class C extends A {
}
#Entity
public class D {
#ManyToOne
public A a;
}
The problem is class D and the field a (could be either of type B or C). What should be the mapping?
If you reference an A from another entity, then A should not be a MappedSuperclass, but an entity. You should annotate it with #Entity, and choose an inheritance strategy.
Other than that, the mapping will stay as is.
This is not valid in JPA, as relationships cannot be defined to #MappedSuperclass.
You could map A and #Entity (TABLE_PER_CLASS inheritance would give you the same data model, but also consider JOINED or SINGLE_TABLE, as they are normally more efficient).
In EclipseLink you can also use a #VariableOneToOne for this type of relationship.
See,
http://www.eclipse.org/eclipselink/documentation/2.4/jpa/extensions/a_variableonetoone.htm#CHDDFDGF
What you are doing is correct. It will be the mapping A. So, in Table D ; we will have an additional column which will tell us if A is B or C. In Toplink we have the column name as Type. Even in hibernate we have similar concept. And this column is automatically populated by the ORM.
I have a problem trying to map an inheritance tree. A simplified version of my model is like this:
#MappedSuperclass
#Embeddable
public class BaseEmbedded implements Serializable {
#Column(name="BE_FIELD")
private String beField;
// Getters and setters follow
}
#MappedSuperclass
#Embeddable
public class DerivedEmbedded extends BaseEmbedded {
#Column(name="DE_FIELD")
private String deField;
// Getters and setters follow
}
#MappedSuperclass
public abstract class BaseClass implements Serializable {
#Embedded
protected BaseEmbedded embedded;
public BaseClass() {
this.embedded = new BaseEmbedded();
}
// Getters and setters follow
}
#Entity
#Table(name="MYTABLE")
#Inheritance(strategy=InheritanceType.SINGLE_TABLE)
#DiscriminatorColumn(name="TYPE", discriminatorType=DiscriminatorType.STRING)
public class DerivedClass extends BaseClass {
#Id
#Column(name="ID", nullable=false)
private Long id;
#Column(name="TYPE", nullable=false, insertable=false, updatable=false)
private String type;
public DerivedClass() {
this.embedded = new DerivedClass();
}
// Getters and setters follow
}
#Entity
#DiscriminatorValue("A")
public class DerivedClassA extends DerivedClass {
#Embeddable
public static NestedClassA extends DerivedEmbedded {
#Column(name="FIELD_CLASS_A")
private String fieldClassA;
}
public DerivedClassA() {
this.embedded = new NestedClassA();
}
// Getters and setters follow
}
#Entity
#DiscriminatorValue("B")
public class DerivedClassB extends DerivedClass {
#Embeddable
public static NestedClassB extends DerivedEmbedded {
#Column(name="FIELD_CLASS_B")
private String fieldClassB;
}
public DerivedClassB() {
this.embedded = new NestedClassB();
}
// Getters and setters follow
}
At Java level, this model is working fine, and I believe is the appropriate one. My problem comes up when it's time to persist an object.
At runtime, I can create an object which could be an instance of DerivedClass, DerivedClassA or DerivedClassB. As you can see, each one of the derived classes introduces a new field which only makes sense for that specific derived class. All the classes share the same physical table in the database. If I persist an object of type DerivedClass, I expect fields BE_FIELD, DE_FIELD, ID and TYPE to be persisted with their values and the remaining fields to be null. If I persist an object of type DerivedClass A, I expect those same fields plus the FIELD_CLASS_A field to be persisted with their values and field FIELD_CLASS_B to be null. Something equivalent for an object of type DerivedClassB.
Since the #Embedded annotation is at the BaseClass only, Hibernate is only persisting the fields up to that level in the tree. I don't know how to tell Hibernate that I want to persist up to the appropriate level in the tree, depending on the actual type of the embedded property.
I cannot have another #Embedded property in the subclasses since this would duplicate data that is already present in the superclass and would also break the Java model.
I cannot declare the embedded property to be of a more specific type either, since it's only at runtime when the actual object is created and I don't have a single branch in the hierarchy.
Is it possible to solve my problem? Or should I resignate myself to accept that there is no way to persist the Java model as it is?
Any help will be greatly appreciated.
Wow. This is the simplified version? I assume that the behavior that you are seeing is that BaseEmbedded field is persisted but not the FIELD_CLASS_A or B?
The problem is that when Hibernate maps the DerivedClassA and B classes, it reflects and sees the embedded field as a BaseEmbedded class. Just because you then persist an object with the embedded field being a NestedClass, the mapping has already been done and the FIELD_CLASS_A and B are never referenced.
What you need to do is to get rid of the NestedClass* and embedded field and instead have the fieldClassA and B be normal members of DerivedClassA and B. Then add add a name field to the #Entity which will put them both in the same table I believe. This will allow you to collapse/simplify your class hierarchy a lot further.
See: http://docs.jboss.org/hibernate/stable/annotations/reference/en/html_single/#d0e1168
#Entity(name = "DerivedClass")
#DiscriminatorValue("A")
public class DerivedClassA extends DerivedClass {
#Column(name="FIELD_CLASS_A")
private String fieldClassA;
...
#Entity(name = "DerivedClass")
#DiscriminatorValue("B")
public class DerivedClassB extends DerivedClass {
#Column(name="FIELD_CLASS_B")
private String fieldClassB;
...
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/