I'm using toplink JPA in a webapp and I want to map just one table to a class hierarchy. I want to have one class that represents most of the data, and one class that inherits from that (so it gets all the fields of the superclass, plus a couple of other that hold large amounts of data). I don't want the large amounts of data all the time, don't want to hold them in request objects etc. I only want the large bits when someone has selected one of the summaries. I've setup the classes as follows (simplified as an example).
#Entity
#Table(name = "TRANSCRIPTS")
#MappedSuperclass //also tried without this - same error
public class Summary {
#Id
#Column(name = "id")
private long id;
#Column(name = "title")
private String title;
//rest of class etc.
}
#Entity
#Table(name = "TRANSCRIPTS")
public class Detail extends Summary {
#Id
#Column(name = "fullText")
private String fullText;
//rest of class etc.
}
When I try and get data using this hierarchy, I get an error along the lines of
Unknown column 'DTYPE'
So it's looking for a descriminator column. Which I haven't setup, because it's not that sort of relationship.
Is there a different way I can map this summary/detail relationship in JPA? Or should I give up on the class inheritance and have two separate unrelated classes, one representing summary data and one representing the full data (and redefining the summary fields).
Thanks.
DTYPE it is discriminator column that Toplink tries to access to choose between your entities,
If you add that column to your table schema, it will start working.
DTYPE is INTEGER typed column in database.
You could specify your own discriminator column using following code snippet:
#Entity
#DiscriminatorColumn(name="type",discriminatorType=DiscriminatorType.INTEGER)
#DiscriminatorValue("1")
class TestClass {}
Here is some documentation for you http://www.oracle.com/technetwork/middleware/ias/toplink-jpa-annotations-096251.html#CHDJHIAG
Related
I have a set of Java classes with the following UML diagram:
public class Invoice {
#Id
private long id;
...
}
public class InvoiceDetail {
#Id
private long id;
...
private String productName;
private int quantity;
private double price;
}
My purpose is using JPA annotations to establish the different relationships between them. There is a composition relationship between Invoice and InvoiceDetail, which is resolved using #Embedded and #Embeddable annotations for Invoice and InvoiceDetail respectively. However, a problem appears by establishing the relationships between InvoiceDetail, Class3 and Class4. In these relationships InvoiceDetail must be annotated as #Entity. However, when a class is annotated at the same time as #Entity and #Embeddable, the corresponding server will throw a runtime error during the deployment.
Basing on the information of this website, I have written the following possible solution:
#Entity
public class Invoice {
#Id
private long id;
...
#ElementCollection
#CollectionTable(name="INVOICEDETAIL", joinColumns=#JoinColumn(name="INVOICE_ID"))
private List<InvoiceDetail> invoiceDetails;
...
}
Would be this right in order to resolve my problem?
Thanks in advance.
Although without knowing what the classes really are it is hard to tell, I suppose that you have a design problem. The composition between Class1 and Class2 says that any Class2 instance only exists within the lifecycle of a corresponding Class1 instance. But on the other hand you have Class3 instances and Class4 instances which can / must have a relationship to a Class2 instance.
What I'm trying to say is that from my point of view the relationship between Class1 and Class2 should be a simple association and not a composition. Following this path Class2 would be an Entity in JPA and then you should have your problem solved.
I usually use #Embeddable for classes whose instances never exist by themselfes and #Entity for any class whose instances can exist without other instances. An address for example could be implemented either way but not on the same system. Address would be #Embeddable if I don't want to link addresses but it had to be #Entity if I want to make sure the same address isn't saved in more than one row.
[edit: added after classes 1 and 2 were renamed to Invoice and InvoiceDetails]
Having a composition between Invoice and InvoiceDetails makes perfect sense. But I still think you should avoid the need of double personality for InvoiceDetails. I can think of two solutions (both refactorings):
If you prefer having InvoiceDetails as #Embeddable you could change the associations of Class3 and Class4 to Invoice instead of InvoiceDetails. InvoiceDetails would still be traversable via the Invoice object.
If you prefer keeping the associations as is you could declare InvoiceDetails to be an entity. You could still achieve your composition with a cascading delete (see javax.persistence.CascadeType). As it seems that InvoiceDetails already has it's own table, this probably is the better option.
I checked my JPA applications and haven't found any occurence of the same class being #Entity and #Embeddable. Honestly, I doubt if this is possible at all because the official javadoc of #Embeddable says:
Specifies a class whose instances are stored as an intrinsic part of an owning entity and share the identity of the entity.
As #Entity has it's own identity, you would try to declare the same object having two identities - and this can't work.
[/edit]
[edit2: adding code for solution proposal #2]
This code should work with some assumptions (see below). This is the implementation of bi-directional navigation for a 1:n-relationship.
#Entity
public class Invoice {
#Id
private long id;
#OneToMany(mappedBy="invoice", cascade = CascadeType.ALL)
private List<InvoiceDetail> details;
}
#Entity
public class InvoiceDetails {
#Id
private long id;
#ManyToOne
#JoinColumn(name="invoice_id")
private Invoice invoice;
}
Assumptions: Tables are named like the entities, the foreign key column for invoice_details table is named "invoice_id" and both tables have a primary key column named "id". Note that the mappedBy-value "invoice" refers to the entity field while the name-value "invoice_id" refers to the database table.
Be cautious when deleting an Invoice object whose InvoiceDetails still are referenced by your Class3 or Class4 instances - you have to release these references first.
For information about JPA refer to these resources:
The Java EE 7 Tutorial: Persistence
Wikibooks: Java Persistence
Javadoc of Package javax.persistence
[/edit]
I have the following three simplified classes that represent a relationship I need to map in a legacy database I'm working with.
class One
{
#Id
String id
}
class Two
{
#Id
#ManyToOne
#JoinColumn(name = 'ONE', referencedColumn = 'ID')
One one
}
class Three
{
#Id
#ManyToOne
#JoinColumn(name = 'TWO', referencedColumn = 'ONE')
Two two
}
The relationship between class One and Two works fine, however, the relationship between class Two and Three doesn't work. Hibernate returns a "Unable to find column with logical name: ONE in org.hibernate.mapping.Table(TWO) and its related supertables and secondary tables" error. I've debugged into the hibernate code to where this error gets thrown and the org.hibernate.mapping.Table object that defines the TWO table doesn't contain any column details. It not treating the Many To One join in the Two class as a column.
Is it possible to make inheritance in JPA/Hibernate without id?
My use case:
I have multiple entities and I want every time change is being done, timestamp being recorded. So I created AbstractResource and want each derived class inherit properties and logic (to avoid writing same stuff over and over again in each class).
My problem that hibernate wants an ID to entity, and I do not care about id, since my only concern is additional properties. And each entity can have whatever id it wants (String, int, long, different name, etc.).
I tried with Embeddable, but looks like hibernate does not support inheritance for Embeddable. Do you have any ideas, how my task can be achieved?
My parent class from which "Audited" entities are derived:
#Embeddable
#EntityListeners(AbstractResourceListener.class)
#Inheritance(strategy = InheritanceType.SINGLE_TABLE)
public class AbstractResource {
private long modifyTimestamp;
#Column(name = "_MODIFY_TIMESTAMP", nullable = true)
public long getModifyTimestamp() {
return modifyTimestamp;
}
public void setModifyTimestamp(long modifyTimestamp) {
this.modifyTimestamp = modifyTimestamp;
}
}
#MappedSuperclass is an annotation for super classes that you can extend and use in audit. Please see example
I need to create a relation in Hibernate, linking three tables: Survey, User and Group.
The Survey can be visible to a User or to a Group, and a Group is form of several Users.
My idea was to create a superclass for User and Group, and create a ManyToMany relationship between that superclass and Survey.
My problem is that Group, is not map to a table, but to a view, so I can't split the fields of Group among several tables -which would happen if I created a common superclass-.
I thought about creating a common interface, but mapping to them is not allowed.
I will probably end up going for a two relations solution (Survey-User and Survey-Group), but I don't like too much that approach.
I thought as well about creating a table that would look like:
Survey Id | ElementId | Type
ElementId would be the Group or UserId, and the type... the type of it.
Does anyone know how to achieve it using hibernate annotations? Any other ideas?
Thanks a lot
I posted a very similar answer yesterday. To summarize, you can't use a mapped superclass because a mapped superclass is not an entity and can't be part of an association (which is what you want) but you can use an abstract Entity with a TABLE_PER_CLASS inheritance strategy to obtain a similar result.
Something like this (not tested):
#Entity
#Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstract class AbstractEntity {
#Id #GeneratedValue(strategy = GenerationType.TABLE)
private Long id;
#ManyToMany(mappedBy="entities")
private Set<Survey> surveys = new HashSet<Survey>();
...
}
#Entity
public class User extends AbstractEntity {
...
}
#Entity
public class Group extends AbstractEntity {
...
}
#Entity
public class Survey {
#Id #GeneratedValue
private Long id;
#ManyToMany
private Set<AbstractEntity> entities = new HashSet<AbstractEntity>();
...
}
References
Annotations, inheritance and interfaces
using MappedSuperclass in relation one to many
Polymorphic association to a MappedSuperclass throws exception
You can use the table per concrete class inheritance strategy, hibernate will replicate all properties for each subclass, this will work with a view.
I would also suggest the composite pattern for users/groups (which is close to your first option).
http://en.wikipedia.org/wiki/Composite_pattern
This is possible. Such an 'inherited properties' approach can be achieved by defining the superclass as a MappedSuperclass.
EDIT:
There is also some alternatives listed in section 2.2.4 in the hibernate annotations reference doc, section 2.2.4.4 covers MappedSuperclass.
I'm currently using Eclipselink, but I know now days most JPA implementations have been pretty standardized. Is there a native way to map a JPA entity to a view? I am not looking to insert/update, but the question is really how to handle the #Id annotation. Every entity in the JPA world must have an ID field, but many of the views I have created do not conform to this. Is there native support for this in the JPA or do I need to use hacks to get it to work? I've searched a lot and found very little information about doing this.
While using the #Id annotation with fields of directly supported types is not the only way to specify an entity's identity (see #IdClass with multiple #Id annotations or #EmbeddedId with #Embedded), the JPA specification requires a primary key for each entity.
That said, you don't need entities to use JPA with database views. As mapping to a view is no different from mapping to a table from an SQL perspective, you could still use native queries (createNativeQuery on EntityManager) to retrieve scalar values instead.
I've been looking into this myself, and I've found a hack that I'm not 100% certain works but that looks promising.
In my case, I have a FK column in the view that can effectively function as a PK -- any given instance of that foreign object can only occur once in the view. I defined two objects off of that one field: one is designated the ID and represents the raw value of the field, and the other is designated read-only and represents the object being referred to.
#Id
#Column(name = "foreignid", unique = true, nullable = false)
public Long getForeignId() {
...
#OneToOne
#JoinColumn(name = "foreignid", insertable=false, updatable=false)
public ForeignObject getForeignObject() {
...
Like I said, I'm not 100% sure on this one (and I'll just delete this answer if it turns out not to work), but it got my code past a particular crash point.
Dunno if it applies to your specific situation, though. And there's an excellent chance that after 11 months, you no longer care. :-) What the hell, that "Necromancer" badge doesn't just earn itself....
In my view I have a "unique" id, so I mapped it as the Entity id.
It works very well:
#Entity
#Table(name="table")
#NamedQuery(name="Table.findAll", query="SELECT n FROM Table n")
public class Table implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#Column(name="column_a")
private int columnA;
JPA - 2.5.4
CREATE MATERIALIZED VIEW IF NOT EXISTS needed_article as select product_id, count(product_id) as count from product_article group by product_id;
CREATE MATERIALIZED VIEW IF NOT EXISTS available_article as select product_id, count(product_id) as count from article a inner join product_article p
on a.id = p.article_id and a.stock >= p.amount_of group by product_id;
CREATE UNIQUE INDEX productId_available_article ON available_article (product_Id);
CREATE UNIQUE INDEX productId_needed_article ON needed_article (product_Id);
Entity.java
#Entity
#Immutable // hibernate import
#Getter
#Setter
public class NeededArticle {
#Id
Integer productId;
Integer count;
}
Repository.java
#Repository
public interface AvailableProductRepository extends CrudRepository<AvailableArticle, Integer> {
#Query("select available.productId from AvailableArticle available, NeededArticle needed where available.productId = needed.productId and available.count = needed.count")
List<Integer> availableProduct();