JPA (Hibernate): Error when using #EmbeddedId in generic #MappedSuperclass - java

I am currently defining JPA entities for a legacy database (lots of composite keys, but also single-column keys). I have created the following entity superclass:
#MappedSuperclass
public abstract class AbstractEntity<ID extends Serializable> {
public abstract ID getId();
public abstract void setId(ID id);
}
And then a superclass for composite keys (as well as a superclass for long primary key, not listed here):
#MappedSuperclass
public abstract class AbstractEmbeddedIdEntity<ID extends Serializable> extends AbstractEntity<ID> {
#EmbeddedId
private ID id;
public AbstractEmbeddedIdEntity() {
id = newId();
}
#Override
public ID getId() {
return id;
}
#Override
public void setId(ID id) {
this.id = id;
}
protected abstract ID newId();
}
And finally concrete entities like this:
#Entity
#Table(name = "firstEntity")
public class FirstEntity extends AbstractEmbeddedIdEntity<FirstEntityId> {
public FirstEntity() {
}
#Embeddable
public static class FirstEntityId implements Serializable {
#Column(name = "firstId")
private String firstId;
public FirstEntityId() {
}
#Override
public boolean equals(Object obj) {
if (obj == this) {
return true;
}
if (!(obj instanceof FirstEntityId)) {
return false;
}
FirstEntityId other = (FirstEntityId) obj;
return
Objects.equals(firstId, other.firstId);
}
#Override
public int hashCode() {
return Objects.hash(firstId);
}
}
#Override
protected FirstEntityId newId() {
return new FirstEntityId();
}
}
Now the issue is that if I have multiple entities like this and try to access an ID property of an entity (currently with Spring Boot, e.g. findByIdFirstId(String firstId)), an exception is thrown:
java.lang.IllegalArgumentException: Unable to locate Attribute with the the given name [firstId] on this ManagedType [unknown]
I have debugged this and found out that in hibernate, the metamodel maps all of my entities to the same MappedSupperclass instance. During application startup, the #EmbeddedId returned by newId() is set to the MappedSupperclass, overwriting the ID of the previous entity. So in the end, all entities are mapped to the same MappedSupperclass, but the MappedSupperclass only has the #EmbeddedId of the last entity.
In the above example, accessing the ID property fails because the #EmbeddedId of the last entity doesn't have a property called "firstId" (it has been overwritten with the ID properties of the last entity).
Now I am wondering if my approach is wrong, if I am missing something or if this could be an issue with hibernate?
Complete example using spring boot available on github. Run with mvn spring-boot:run.

This looks to me like a bug in hibernate, therefore I have created a ticket in the hibernate bug tracker.
As a workaround I am now defining the ID attribute (#EmbeddedId) in the concreate entity classes instead of the abstract superclass.

Related

Unrecognized Hibernate Type when using enum in query

I have an Entity class with a set of enums:
#Entity
public class Something {
public enum Type{
FIRST_ENUM((short)1),
SECOND_ENUM((short)2);
private short id;
private Type(short id) {
this.id = id;
}
public short getId() {
return id;
}
}
#CollectionTable(name="table_name", joinColumns=#JoinColumn(name="something_id"))
#Column(name="type")
private Set<Type> types;
... //other props + getters and setters
}
For the enum I made a converter (whole converter package is loaded by #EntityScan annotation) :
#Converter(autoApply=true)
public class TypeConverter implements AttributeConverter<Type, Integer> {
#Override
public Integer convertToDatabaseColumn(Type st) {
//implementation
}
#Override
public Type convertToEntityAttribute(Integer i) {
//implementation
}
}
Now when I try to use the enum in a query
... AND {packagename}.Something$Type.FIRST_ENUM MEMBER OF {someobject}.something.types ...
I stumble upon following error:
org.hibernate.QueryException: Unrecognized Hibernate Type for handling query constant ({package}.Something$Type.FIRST_ENUM); expecting LiteralType implementation or AttributeConverter
Does anyone got a clue why I cannot use the enum in my query? It somehow seems the enum is not known to Hibernate. I don't understand why, because the class is loaded when I start my application.

JPA eclipselink Inheritance between entities : oracle database

I'm facing a little problem with my web application which is in vaadin and i'm using jpa and eclipselink for the mapping. I have three entities :
encaiss (#MappedSuperclass contains just Id)
|
|
Encaissement (it contains the main and common properties)
/ \
/ \
Encaissement_Technique Encaissement_espece
When i create an entity "Encaissement" with "Espece" as type, it is well created in the table Encaissement but it doesn't exist in the table Encaissement_espece.
I guess that I should join the two tables according to the identifier (ID) which is in a #MappedSuperclass class. I would appreciate any help for managing my subordinate class (that is Encaissement_Technique and Encaissement_espece) because my next step would be to add records to those two tables from a simple form (so if i have a field "libelle" which is present in Encaissement but not in Encaissement_Espece how can make such instruction :
Encaissement_Espece espece= new Encaissement_Espece();
espece.setLibelle(field.getValue().toString());
Those are my entities :
encaiss, this class contain just the Id for all the classes
#MappedSuperclass
public abstract class encaiss {
#Id
#GeneratedValue(strategy=GenerationType.AUTO, generator="encaiss_seq_gen")
#SequenceGenerator(name="encaiss_seq_gen", sequenceName="ENCAISSEMENT_SEQ", allocationSize = 1, initialValue = 1)
protected Integer id_encaissement;
public Integer getId_encaissement() {
return id_encaissement;
}
public void setId_encaissement(Integer id_encaissement) {
this.id_encaissement = id_encaissement;
}
}
Encaissement (wich extend encaiss just to have an Id)
#Entity
#Inheritance(strategy=InheritanceType.JOINED)
#DiscriminatorColumn(name="ENCAISS_TYPE")
#Table(name="ENCAISSEMENT")
public class Encaissement extends encaiss implements Serializable{
#ManyToOne(cascade = CascadeType.PERSIST, fetch = FetchType.LAZY)
#JoinColumn(name = "ID_CLIENT")
private Client Client;
#Column(name="ENCAISS_TYPE")
protected String encaiss_type;
#Column(name="LIBELLE")
protected String libelle;
#Column(name="PIECE_JOINTE")
protected String piece_jointe;
#Embedded
protected Avis_Recette avis_recette;
public Encaissement(String encaiss_type, String libelle, String piece_jointe){
this.encaiss_type=encaiss_type;
this.libelle=libelle;
this.piece_jointe=piece_jointe;
}
public Encaissement(){
}
}
Encaissement_Espece, inherits from Encaissement
#Entity
#DiscriminatorValue("Espece")
#Table(name="ENCAISSEMENT_ESPECE")
public class Encaissement_Espece extends Encaissement{
public Caisse getCaisse() {
return caisse;
}
public void setCaisse(Caisse caisse) {
this.caisse = caisse;
}
public float getMontant() {
return montant;
}
public void setMontant(float montant) {
this.montant = montant;
}
#ManyToOne(cascade = CascadeType.PERSIST, fetch = FetchType.LAZY)
#JoinColumn(name = "ID_CAISSE")
private Caisse caisse;
#Column(name = "MONTANT")
private float montant;
public Encaissement_Espece(float montant){
this.montant=montant;
}
public Encaissement_Espece(){
}
}
Encaissement_Technique, inherits from Encaissement
#Entity
#DiscriminatorValue("Technique")
#Inheritance(strategy=InheritanceType.JOINED)
#DiscriminatorColumn(name="ENCAISS_TECHNIQUE_TYPE")
#Table(name="ENCAISSEMENT_TECHNIQUE")
public class Encaissement_Technique extends Encaissement implements Serializable{
public Banque getBanque() {
return banque;
}
public void setBanque(Banque banque) {
this.banque = banque;
}
public float getPrimeCoass() {
return primeCoass;
}
public void setPrimeCoass(float primeCoass) {
this.primeCoass = primeCoass;
}
public Set<Periode> getPeriode() {
return periode;
}
public void setPeriode(Set<Periode> periode) {
this.periode = periode;
}
public String getEncaiss_technique_type() {
return encaiss_technique_type;
}
public void setEncaiss_technique_type(String encaiss_technique_type) {
this.encaiss_technique_type = encaiss_technique_type;
}
#Column(name="PRIMECOASS")
protected float primeCoass;
#Column(name="ENCAISS_TECHNIQUE_TYPE")
protected String encaiss_technique_type;
public Encaissement_Technique(float primeCoass, String encaiss_technique_type){
this.primeCoass=primeCoass;
this.encaiss_technique_type=encaiss_technique_type;
}
public Encaissement_Technique(){
}
}
I hope i will find a pertinent answer as i searched for this in vain. It'll help me a lot.
Thank you.
"When i create an entity "Encaissement" with "Espece" as type, it is well created in the table Encaissement but it doesn't exist in the table Encaissement_espece." This statement suggests you have an instance of Encaissement and expect JPA to turn it into an instance of Encaissement_Espece just by changing the encaiss_type value. Java object inheritance doesn't work that way, which is what JPA inheritance tries to map to a relational database. An object in java cannot change what it is simply by setting a flag - you need to create a new instance if you want the data represented differently.
In this case, you need to create an instance of the Encaissement_Espece class. Because this class maps to the Encaissement and Encaissement_espece tables, JPA will automatically insert a row into both to represent this object. When you create Encaissement instance, a row goes into the Encaissement table, while when you create Encaissement_Technique instances, a row goes into both Encaissement_Technique and Encaissement tables. If you wish to change the object's type once it is persisted, you need to remove the old instance, flush, then persist the new one.
As mentioned in another answer, the encaiss_type is controlled through the class type itself and so does not need a mapping. Having one might be handy for queries or access (though you can just use instance of etc); it should be marked as insertable=false, updatable=false so that you do not attempt to modify the value directly.
remove the
#Column(name="ENCAISS_TYPE")
protected String encaiss_type;
from the Encaissment.
It will be handle automatically by JPA. It should solve the problem.

#Id field not seen if it's in superclass?

i've a base class that provide only identification:
public abstract class Identifable<T> {
#Id
private T id = null;
public T getId() {
return id;
}
public void setId(T id) {
this.id = id;
}
public boolean hasId() {
return id != null;
}
}
and several subclasses that extends it like:
#Entity
#Cache
public class MyEntity extends Identifable<String> {
/* some specific attributes and methods */
}
I get an java.lang.IllegalStateException: #Id field 'id' in com.mypkg.MyEntity must be of type Long, long, or String.
Why? Can't Objectify see the inherited #Id field?
Thanks
The cause:
Objectify only inspects types at runtime using reflection. Because of type erasure all unbounded type parameters are during compilation converted to Object type, which is what objectify sees and complains.
The solution:
Use concrete type for id field. Possibly move it to a child class, as proposed by #Anthony.
In JPA, you must use for a field marked with #Id one of the following types:
any Java primitive type; any primitive wrapper type; java.lang.String; java.util.Date; java.sql.Date; java.math.BigDecimal; java.math.BigInteger
Just remove the generics from the base class and use one of the mentioned types for your id field.
Let's reason about if for a while... You are trying to build a super type in which the type of the ID varies. Are you sure that this is what you want objectify to build (a hierarchy of objects in which the root entity has a unknown ID type)? While I've seen this kind of code in several ORM frameworks, this is how I would build what you want.
Interface (not part of the object hierarchy):
public interface Identifable<T> {
public T getId();
public void setId(T id);
public boolean hasId();
}
Root of your hiearchy implements Identifable with a concrete type for the id:
#Entity
public class MyBaseClass implements Identifable<String> {
#Id
private String id = null;
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
}
public boolean hasId() {
return id != null;
}
}
And subclass comes naturally out of it:
#EntitySubclass(index=true)
public class MyEntity extends MyBaseClass {
// fields, accessors and mutators
}

JPA #Entity Inheritance

I have been looking into JPA/Hibernate #Entity inheritance for a while now and can't seem to find anything that addresses what I am trying to achieve.
Basically I want to be able to define an #Entity with all of the column and table mappings as required. Then I want to be able to extend the #Entity in a number of different locations with different sets of #Transient methods defined in the body of each "sub-Entity". This is a basic example of what I am trying to achieve but with no success thus far:
#Entity
#Table(name = "mountain")
public class MountainEntityBase implements Serializable {
public Integer mountainId = 0;
public Integer height = 0;
public List<ExplorerEntityBase> explorers = new ArrayList<ExplorerEntityBase>();
#Id
#GeneratedValue
#Column(name = "mountain_id")
public Integer getMountainId() { return mountainId; }
public void setMountainId(Integer mountainId) { this.mountainId = mountainId; }
#Column(name="height")
public String getHeight() { return height; }
public void setHeight(String height) { this.height = height; }
#OneToMany(mappedBy="mountainId")
public List<ExplorerEntityBase> getExplorers() { return this.explorers; }
public void setExplorers(List<ExplorerEntityBase> explorers) { this.explorers = explorers; }
}
.
#Entity
public class MountainEntity extends MountainEntityBase implements Serializable {
public List<MountainEntity> allMountainsExploredBy = new ArrayList<MountainEntity>();
#Transient
public List<MountianEntity> getAllMountainsExploredBy(String explorerName){
// Implementation
}
}
So any extended class will define only #Transients in its body. But also I want to allow for situations where the child class is empty:
#Entity
public class MountainEntity extends MountainEntityBase implements Serializable {
}
Thanks in advance for any help with this.
Inheritance in JPA is specified on the root entity using the #Inheritance annotation. There you can specify the database representation of the hierarchy. Check the documentation for more details.
If your child classes define only transient fields (not methods) (i.e. not saved in the db), then perhaps a discriminator column is the best option. But it may be the case that you don't actually need inheritance - the main entity can have all the methods (because it has all the fields the methods operate on)

JPA - Using insertable/updatable

I am writing a webservice to maintain a database. I am trying to use JPA (EclipseLink) for the entity classes. However, the database uses natural primary keys and therefore there's potential that an update on the ID fields will fail due to foreign key constraints. Our DBA has provided a function to update the ID fields which will create a new parent record with the updated ID, update the child records to point to the new parent and delete the old parent.
If the ID fields could be updated "normally", I would have a situation like this:
#Entity
#Table(name = "PARENT")
public class Parent implements Serializable
{
private static final long serialVersionUID = 1L;
private String parent;
private String attribute;
private Set<Child> childs;
public Parent()
{
}
#Id
#Column(name = "PARENT")
public String getParent()
{
return this.parent;
}
public void setParent(String parent)
{
this.parent = parent;
}
#Column(name = "ATTRIBUTE")
public String getAttribute()
{
return this.attribute;
}
public void setAttribute(String attribute)
{
this.attribute = attribute;
}
#OneToMany(mappedBy = "parentBean")
public Set<Child> getChilds()
{
return this.childs;
}
public void setChilds(Set<Child> childs)
{
this.childs = childs;
}
}
#Entity
#Table(name = "CHILD")
public class Child implements Serializable
{
private static final long serialVersionUID = 1L;
private String child;
private String attribute;
private Parent parentBean;
public Child()
{
}
#Id
#Column(name = "CHILD")
public String getChild()
{
return this.child;
}
public void setChild(String child)
{
this.child = child;
}
#Column(name = "ATTRIBUTE")
public String getAttribute()
{
return this.attribute;
}
public void setAttribute(String attribute)
{
this.attribute = attribute;
}
#ManyToOne
#JoinColumn(name = "PARENT")
public Parent getParent()
{
return this.parent;
}
public void setParent(Parent parent)
{
this.parent = parent;
}
}
I also have a GenericServiceBean class with a method to call functions:
#Stateless
public class GenericServiceBean implements GenericService
{
#PersistenceContext(unitName = "PersistenceUnit")
EntityManager em;
public GenericServiceBean()
{
// empty
}
#Override
public <T> T create(T t)
{
em.persist(t);
return t;
}
#Override
public <T> void delete(T t)
{
t = em.merge(t);
em.remove(t);
}
#Override
public <T> T update(T t)
{
return em.merge(t);
}
#Override
public <T> T find(Class<T> type, Object id)
{
return em.find(type, id);
}
. . .
#Override
public String executeStoredFunctionWithNamedArguments(String functionName,
LinkedHashMap<String, String> namedArguments)
{
Session session = JpaHelper.getEntityManager(em).getServerSession();
StoredFunctionCall functionCall = new StoredFunctionCall();
functionCall.setProcedureName(functionName);
functionCall.setResult("RESULT", String.class);
for (String key : namedArguments.keySet())
{
functionCall.addNamedArgumentValue(key, namedArguments.get(key));
}
ValueReadQuery query = new ValueReadQuery();
query.setCall(functionCall);
String status = (String)session.executeQuery(query);
return status;
}
}
If I set the ID fields to be not editable:
#Id
#Column(name = "PARENT", udpatable=false)
public String getParent()
{
return this.parent;
}
and call parent.setParent(newParent) will this still update the ID in the entity object? How does this affect any child entities? Will they also be updated (or not)?
Another scenario I don't know how to deal with is where I need to update both the ID and another attribute. Should I call the function which updates (and commits) the ID in the database then make calls to set both the ID and attribute via the normal set* methods and then the persistence context will only commit the attribute change?
Perhaps this is a situation where JPA is not appropriate?
Any advice on this is greatly appreciated.
If I set the ID fields to be not editable (...) and call parent.setParent(newParent) will this still update the ID in the entity object? How does this affect any child entities? Will they also be updated (or not)?
updatable=false means that the column won't be part of the SQL UPDATE statement regardless of what you do at the object level so the Id shouldn't be updated. And I'm also tempted to say that child entities shouldn't be affected, especially since you're not cascading anything.
Another scenario I don't know how to deal with is where I need to update both the ID and another attribute (...)
Well, my understanding is that you'd have to call the function anyway so I would call it first.
Perhaps this is a situation where JPA is not appropriate?
I'm not sure raw SQL would deal better with your situation. Actually, the whole idea of changing primary keys sounds strange if I may.

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