~"IdClass not defined" in JpaRepository, for an inherited #OneToOne #Id - java

I'm trying to create a jpa repository but there is a problem with a foreign-key primary-key. Although it is specified in the abstract base class (MessageDestination), it seems to be invisible from the repository of specialized MessageDestination class (e.g. MessageDestinationRoom).
[...] nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'messageDestinationRoomDAO': Invocation of init method failed; nested exception is java.lang.IllegalArgumentException: This class [class com.chat.message.entity.MessageDestinationRoom] does not define an IdClass
#Entity #Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstract class Message implements Serializable {
#Id #GeneratedValue(strategy = GenerationType.TABLE)
private Long id;
#OneToOne(targetEntity = MessageDestination.class,
cascade=CascadeType.ALL, mappedBy="msg")
#NotNull
private MessageDestination dest;
//...
}
#Entity #Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstract class MessageDestination implements Serializable {
#Id #OneToOne(cascade=CascadeType.ALL)
private Message msg;
}
#Entity
public class MessageDestinationRoom extends MessageDestination {
#OneToOne #NotNull
private Room destRoom;
//...
}
public interface MessageDestinationRoomDAO
extends JpaRepository<MessageDestinationRoom, Message> {
public Set<MessageDestinationRoom> findMessageDestinationRoomByDestRoom
(Room dest);
}
To solve the issue I saw that I can annotate MessageDestination as a #MappedSuperclass, but this can't work because it needs to be an #Entity to be stored in Message. Sadly, it's not possible:
org.hibernate.AnnotationException: An entity cannot be annotated with both #Entity and #MappedSuperclass
Any ideas? Thanks...

Since you are using table per class inheritance strategy and you dont have any mapped superclass (so each entity must have its own id).
You can annonate your MessageDestination Entity as #MappedSuperClass and remove the #Entity from MessageDestination. As by default its each subclass will inherited all its field including the #Id field

Pending for a better answer because the only solution I found is quite ugly. That consists of splitting the primary and the foreign key, so there is redundancy...
This:
#Entity #Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstract class MessageDestination implements Serializable {
#Id #OneToOne(cascade=CascadeType.ALL)
private Message msg;
}
public interface MessageDestinationRoomDAO
extends JpaRepository<MessageDestinationRoom, Message> {
public Set<MessageDestinationRoom> findMessageDestinationRoomByDestRoom
(Room dest);
}
becomes this:
#Entity #Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
abstract class MessageDestination implements Serializable {
#Id #GeneratedValue(strategy = GenerationType.TABLE)
private Long id;
#OneToOne(cascade=CascadeType.ALL)
private Message msg;
}
interface MessageDestinationRoomDAO
extends JpaRepository<MessageDestinationRoom, Long> {
public Set<MessageDestinationRoom> findMessageDestinationRoomByDestRoom
(Room dest);
}

I was also getting same issue when I was using #oneToMany And #ManyToOne Annotation based Mapping.
Basically what I was doing mistake was in the class that was throwing the error "does not define an IdClass" was having composite Keys i.e More that one #Id annotation used over two member variables due to which it was getting considered as Composite Key and since hibernate expects a seperate Key class needs to be defined in case of composite key this failure was coming.

Related

JPA how to annotate generic entity field conditionally?

I have this generic entity:
#MappedSuperclass
abstract class Position<T> {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
#Enumerated(EnumType.STRING)
private T name;
}
But there's a case where the generic type is a String:
#Entity
class ChildPosition0 extends Position<String> {
}
And, JPA will complaint that String is not an enum in this case, but I need to annotate this name field if it's an enum, if not, the database will mark it as int type, and that's not ideal. How do I solve this? How to annotate the field conditionally?
My workaround:
Use Position as a parent class, and adding those field in child class individually, even though they share the same field:
#MappedSuperclass
abstract class Position {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
}
And extends it from child entity like this:
Child1:
#Entity
public class ChildPosition1 extends Position {
#Enumerated(EnumType.STRING)
private Priority name; // <- Priority is enum type
}
Child2:
#Entity
public class ChildPosition2 extends Position {
private String name;
}
This is too ugly IMO. And Java does not allow class field override from child class. So, back to the question: how to annotate generic field conditionally?

JPA #Id on #OneToOne Entity with a Composite Key

I have an #Entity class, with an #Id annotation and a #OneToOne annotation on the same field. Usually this would not be a problem, but the entity class of the field with these annotations uses a composite key. This is causing more complications than I anticipated.
Here is the entity class that is posing the problem:
#Entity
public class ReportDetails implements Serializable {
#Id
#OneToOne
private MachineLine machineLine;
}
And here is the MachineLine entity class that is being used as an ID in ReportDetails:
#Entity
#IdClass(MachineLine.MachineLineKey.class)
public class MachineLine {
#Id
#ManyToOne
private Machine machine;
#Id
private long lineIndex;
public static class MachineLineKey implements Serializable {
private Machine machine;
private long lineIndex;
}
}
I have left out any extra fields and the getters and setters from these class definitions, to save space.
When I try to run my application it gives the following exception:
java.lang.IllegalArgumentException: This class [class ReportDetails] does not define an IdClass
When I put an #IdClass annotation on ReportDetails it then requires defining the individual fields of whatever class I define in #IdClass, like in MachineLine. However, I am trying to avoid doing this, in favour of having the whole MachineLine entity returned whenever a ReportDetails entity is retrieved from the database.
Is there a way of having MachineLine as the ID field of ReportDetails, without having to define extra fields within ReportDetails?
This is what JPA calls a "derived identity". You might try something like this:
ReportDetails:
#Entity
public class ReportDetails implements Serializable {
// all attributes map by the relationship: AttributeOverride is not allowed
#EmbeddedId
private MachineLine.Id id;
#MapsId
#JoinColumns({
#JoinColumn(name="machineId", referencedColumnName="machineId"),
#JoinColumn(name="machineLineIndex", referencedColumnName="index")
})
#OneToOne
private MachineLine machineLine;
// ...
}
MachineLine:
#Entity
public class MachineLine {
#EmbeddedId
private Id id;
#MapsId("machineId") // maps machineId attribute of embedded id
#ManyToOne
private Machine machine;
// ...
#Embeddable
public static class Id implements Serializable {
private long machineId; // corresponds to PK type of Machine
private long index;
// ...
}
}
Machine:
#Entity
public class Machine {
#Id
private long id;
#OneToMany(mappedBy = "machine")
private List<MachineLine> lines;
// ...
}
Derived identities are discussed (with examples) in the JPA 2.2 spec in section 2.4.1.

Hibernate + JPA can't map different id for sub-class

For reasons that were around before I got to this project, there are tables that are similar types but have different ID columns.
So, when I try this
#Entity
#Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstract class Element implements Serializable {
public String title;
}
#Entity
public class PrimaryElement extends Element {
#Id
long pid;
}
#Entity
public class OtherElement extends Element {
#Id
long oid;
}
But then I get an obvious error
No identifier specified for entity: Element
Now, I can't very well put the ID in the Element class because they are obviously mapped to different columns.
I have tried various flavors of #Id and 'abstract' and #MappedSuperClass and so on..
I am at a complete loss. Is there a way around this?
Any insights would be appreciated.
Thanks!
You can either replace the #Entity and #Inheritance annotations of your Element class by #MappedSuperclass (this annotation is responsible for technical mappings and only PrimaryElement and OtherElement will be fully featured entities) or move the oid field with it's #Id annotation to the class Element and use the #AttributeOverride annotation in it's subclasses to modify the column names (in this case also the abstract class Element will be a fully featured entity).
#AttributeOverride(name="oid", column=#Column(name="primary_element_id"))
UPDATE:
#MappedSuperclass
public abstract class Element implements Serializable {
#Id
private long id;
public String title;
}
#Entity
#AttributeOverride(name="id", column=#Column(name="pid"))
public class PrimaryElement extends Element {
}
#Entity
#AttributeOverride(name="id", column=#Column(name="oid"))
public class OtherElement extends Element {
}

Jpa 2.0 - EntityManager.find (SomeEntity.class,PK) need to fill Descriminator value to key

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.

javax.persistence annotations and inheritance

I have 4 persistent classes which all have the same fields (exactly) the only 3 difference between them is 1) the class name, 2) the table name and 3) the data. i am aware that this might seem strange to some but trust me there is a good reason which i won't go into here.
now, i'm using hibernate annotations to configure my class which should work like so:
#Entity
#Table(name = "store")
public class Store
{
#Id
#Column(name = "unique_id")
protected String id;
#Column
protected String category;
...
}
.. and this does work, for a single stand-alone class, however there are many fields to map and i'd like to do it all in one hit for all four similar classes, ie:
public class StoreBase
{
#Id
#Column(name = "unique_id")
protected String id;
#Column
protected String category;
...
}
#Entity
#Table(name = "store1")
public class Store1 extends StoreBase
{}
#Entity
#Table(name = "store2")
public class Store2 extends StoreBase
{}
#Entity
#Table(name = "store3")
public class Store3 extends StoreBase
{}
#Entity
#Table(name = "store4")
public class Store4 extends StoreBase
{}
however when attempting this i get the following exception:
Caused by: org.hibernate.AnnotationException: No identifier specified for entity: package.entities.Store1
at org.hibernate.cfg.AnnotationBinder.bindClass(AnnotationBinder.java:672)
at org.hibernate.cfg.AnnotationConfiguration.processArtifactsOfType(AnnotationConfiguration.java:546)
at org.hibernate.cfg.AnnotationConfiguration.secondPassCompile(AnnotationConfiguration.java:291)
at org.hibernate.cfg.Configuration.buildSessionFactory(Configuration.java:1292)
at org.hibernate.cfg.AnnotationConfiguration.buildSessionFactory(AnnotationConfiguration.java:867)
i'm guessing this is because the super class is not being searched for the identifier?
is there a way to utilise inheritance in this context?
thanks, paul.
#MappedSuperclass
public class StoreBase
See docs for more info.
Have a look at #MappedSuperclass.

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