I'm trying, but have not been successful so far, using the following classes with Hibernate.
#MappedSuperclass
#Embeddable
public abstract class Foo {
// atributes...
}
#Embeddable
public class Poo extends Foo {
// atributes...
}
#Entity
#Table
public class None {
// atributes...
#Embedded
private Foo foo;
// constructor
public None(Foo foo) {
this.foo = foo;
}
}
// example of save
None none = new None(Poo poo);
save(none);
Hibernate returns: Cannot instantiate abstract class or interface
Is it possible to perform this operation with JPA?
I ran into the same problem.
It seems like #embedable does not work with #DiscriminatorColumn. The only way I could get this to work is to use #DiscriminatorColumn, but treat the #embedable like a separate entity on the same table.
What this means is that the query will likely join the table to itself.
#Entity
#Table(name="tbComputers")
public class Computer{
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
public long id;
public String motherboard;
#OneToOne
#JoinColumn(name="id")
public CPU cpu;
}
#Entity
#DiscriminatorColumn(name="cpu_type")
#Table(name="tbComputers")
public abstract class CPU {
#Id
private Long id;
#Column(name = "cpu")
public String name;
public abstract String someProcessorSpecificMethod();
}
#Entity
#DiscriminatorValue("Intel")
public class Intel extends CPU {
#Override
public String someProcessorSpecificMethod() {
return "Intel";
}
}
#Entity
#DiscriminatorValue("AMD")
public class AMD extends CPU {
#Override
public String someProcessorSpecificMethod() {
return "AMD";
}
}
EDIT: After further testing I found that while this works for reading data, it does not for persisting. It will create a separate INSERT. It seems like it is not supported https://hibernate.atlassian.net/browse/HHH-1910. The alternative is to to split the table.
Related
I am new with JPA, so maybe someone can explain me how to correctly annotate abstract classes using JPA?
I have an abstract class with generated id field:
public abstract class AbstractClass implements Serializable {
private static final long serialVersionUID = 1L;
private long id;
#Id
#GeneratedValue
public long getId() {
return id;
}
public void setId(long id) {
this.id = id;
}
}
and an abstract class with name which extends AbstractClass:
public abstract class AbstractNameClass extends AbstractClass {
private String shortName;
#Column(name = "shortName", nullable = false)
public String getShortName() {
return shortName;
}
public void setShortName(String shortName) {
this.shortName = shortName;
}
}
I have two types of classes, one extends AbstractClass and other classes extends AbstractNameClass:
#Entity
public class Model extends AbstractNameClass {
// this class should inherit id (from AbstractClass) and shortName (from AbstractNameClass)
}
and
#Entity
public class Vehicle extends AbstractClass {
// this class should inherit only id
}
If I add #MappedSuperclass annotation on AbstractClass, then I can create and save objects which are extending AbstractClass, but how to annotate AbstractNameClass? I tried to add #Entity annotation, but I got "No identifier specified for entity" error, also I tried to add #MappedSuperclass annotation and also got the same error.
So my question would be - how to correctly annotate abstract classes using JPA, without creating AbstractClass and AbstractNameClass tables (in my db I want to have only Model and Vehicle tables)?
I'm building RESTful service on java using JERSEY and need to implement relationships between entities storing just identifier on another entity not whole. Is there any way to implements it in the hibernate?
I'm using something like this but it is not working.
#Entity
#javax.persistence.Table(name = "manager_user")
public class ManagerUser extends User {
#ManyToOne(targetEntity = ShopAdminUser.class)
private Integer shopAdminUserId;
//...
}
#Entity
#javax.persistence.Table(name = "shop_admin_user")
public class ShopAdminUser extends User {
#Lob
private String contactData;
public String getContactData() {
return contactData;
}
public void setContactData(String contactData) {
this.contactData = contactData;
}
}
#Entity
#Inheritance(strategy= InheritanceType.TABLE_PER_CLASS)
public abstract class User {
#Id
#GeneratedValue(strategy = GenerationType.TABLE)
private Integer id;
//...
}
It will be very comfortable for me to implement this.
I have a problem, I have two entity Job and JobPK
Job class looks like this sample code :
#Entity
#IdClass(JobPK.class)
#Table(name="JOB")
#Inheritance
#DiscriminatorColumn(name="JOB_TYPE")
public abstract class Job implements Serializable {
#Id
#Column(name="FOLDER_ID")
private BigDecimal folderId;
#Id
#ColumnDefinition(position = 1)
private String name;
#Column(name="JOB_TYPE",insertable=false,updatable=false)
private String jobType;
...
}
and JobPk :
public class JobPK implements Serializable {
private static final long serialVersionUID = -3266336718203527905L;
#Column(name="JOB_TYPE",insertable=false,updatable=false)
private String jobType;
#Id
private String name;
#Id
#Column(name="FOLDER_ID")
private BigDecimal folderId;
......
}
I have two class which extends Job : CalculatingJob and ImportingJob
Now I wont to use :
getEntityManager().find(CalculatingJob.class, new JobPK (BigDecimal.valueOf(folderId),name))
and I have problem because I must fill i JobPK descriminator value field. If I don't do that I've got Null Pointer Exception. Descriminator value is in key by default I think but I don't want put information about descriminator value explicite during JobPk creating. I thought that Entity which extends from Job will fill this field automaticaly. Any Idea to bypass this problem, maybe I can get Annotation #DescriminatorVale from CalculatingJob and then put into constructor JobPk
Thanks for Help
Try this configuration for Hierarchy structure
Job.java
#Table(name = "JOB")
#Inheritance
#IdClass(JobPK.class)
#DiscriminatorColumn(name = "JOB_TYPE", discriminatorType = DiscriminatorType.STRING)
public abstract class Job implements java.io.Serializable {
}
CalculatingJob.java
#Entity
#DiscriminatorValue("CalculatingJob")
public class CalculatingJob extends Job {
}
ImportingJob.java
#Entity
#DiscriminatorValue("ImportingJob")
public class ImportingJob extends Job {
}
JobPK.java
public class JobPK implements Serializable {
}
The discriminator value is entered by hibernate.
I am using JPA 2 and my problem is about inheritence and overidde Mapping in JPA 2.
I have one Abstract Class [AbstractCompte] and two leafs classes [Compte , CompteTmp].
I want to redefine the Mapping for one Field nrCompte.
nrCompte must be unique in Compte Class .
nrCompte is non unique in CompteTmp class .
I already test putting the #Column in the getter methods of COmpte and CompteTmp and it doesn't work and the result is that nrCompte is always not unique .
#MappedSuperclass
public abstract class AbstractCompte{
#Id
#GeneratedValue
private Long id;
private String nrCompte;
....
....
}
#Entity
public class CompteTmp extends AbstractCompte {
#Column(length=16, unique = false)
public String getNrCompte() {
return super.getNrCompte();
}
}
#Entity
public class Compte extends AbstractCompte {
#Column(length=16, unique = true)
public String getNrCompte() {
return super.getNrCompte();
}
}
Thanks in advance for your help .
JPA offers AttributeOverride, so you can map it like this:
#Entity
#AttributeOverride(name="nrCompte", column=#Column(unique=false))
public class CompteTmp extends AbstractCompte { ... }
#Entity
#AttributeOverride(name="nrCompte", column=#Column(unique=true))
public class Compte extends AbstractCompte { ... }
I've got two tables A and B with simple PK's.
#Entity
public class A {
#Id
public int idA;
}
#Entity
public class B {
#Id
public int idB;
}
I want to map a new association class AB that simply stores the relations between A and B, with composite PK idA+idB. AB doesn't have any extra columns, just the relation between idA and idB.
Is it possible to map AB using a single class? I want to avoid having to create a new ABId class just to use it as #IdClass or #EmbeddedId in AB, and I don't want to map this with a #ManyToMany association on A or B.
Why do you want to map such a join table? Just use a ManyToMany association between A and B. This join table will then be handled automatically when you'll add/remove a B to/from the list of Bs contained in A.
See http://docs.jboss.org/hibernate/core/3.6/reference/en-US/html_single/#d0e11402
If you really want to do that, then just map the two IDs with the #Id notation. The primary class will be the entity class itself (which must be serializable), as explained in the hibernate reference documentation. Note that this is Hibernate-specific.
It would be nice if #JBNizet 's suggestion worked. Unfortunately, there's an old bug which makes it impossible to adopt it in the version I'm using (3.3.1-GA)
I've finally sorted this out by defining an inner static ID class and using it as #IdClass:
#Entity
#Table(name="TABLE_AB")
#IdClass(value=ClassAB.ClassABId.class)
public class ClassAB implements Serializable {
private String idA;
private String idB;
#Id
public String getIdA(){ return idA; }
public void setIdA(String idA){ this.idA = idA; }
#Id
public String getIdB(){ return idB; }
public void setIdB(String idB){ this.idB = idB; }
static class ClassABId implements Serializable {
private String idA;
private String idB;
#Column(name="ID_A")
public String getIdA(){ return idA; }
public void setIdA(String idA){ this.idA = idA; }
#Column(name="ID_B")
public String getIdB(){ return idB; }
public void setIdB(String idB){ this.idB = idB; }
// HashCode(), equals()
}
}
This way I don't have to define a new public class, and I don't have to modify the mappings file to include the ID class.