I'm trying to use an #Embeddable inside annother #Embeddable, which is supported according to the specs, however, it looks as though the JPA annotation in de Code class below is ignored. The enum is (de)serialized by ordinal instead of string representation. I can think of several workarounds, but that's not the issue. I'm trying to understand what I'm doing wrong.
I'm using hibernate 4.2.5 with Spring 3.2.4
#Entity
public class ExampleEntity implements Serializable {
#Id
#GeneratedValue
private Long id;
#ElementCollection(fetch = EAGER)
#CollectionTable(name="example_embeddable",
joinColumns=#JoinColumn(name="example_entity_id"))
private Set<ExampleEmbeddable> embeddables = new HashSet<>();
// This enum is (de)serialized as a string just fine
#Enumerated(EnumType.STRING)
private AnotherEnum another;
private String stuff;
}
#Embeddable
public class ExampleEmbeddable implements Serializable {
#Embedded
private Code code;
// This enum is (de)serialized as a string just fine
#Enumerated(EnumType.STRING)
private StatusEnum status;
}
#Embeddable
public class Code implements Serializable {
public enum Version {
RED,
BLUE,
GREEN
}
private String code;
//Why is this enum is (de)serialized by ordinal?
#Enumerated(EnumType.STRING)
private Version version;
}
Related
I'm trying to do a JPA mapping for an existing database. I can't really change the existing structure.
I managed to make it works, but Intellij is telling me that some column doesn't exist even if it works. So I don't know if there's a better way to do this or if it's Intellij that doesn't support all the use cases.
I simplified my mapping and table for the question.
The 2 tables with primary composite keys are:
Table_A
some_id_a
some_seq_a
Table B
some_id_a
some_seq_a
some_seq_b
And my mapping is:
#Data
#Entity(name="Table_A")
public class TableA {
#EmbeddedId
private Key key;
#OneToMany
#JoinColumn(name = "someIdA")
#JoinColumn(name = "someSeqA")
private List<TableB> tableBs;
#Data
#Embeddable
public static final class Key implements Serializable {
private String someIdA;
private long someSeqA;
}
}
#Data
#Entity(name="Table_B")
public class TableB {
#EmbeddedId
private Key key;
#Data
#Embeddable
public static final class Key implements Serializable {
private String someIdA;
private long someSeqA;
private long someSeqB;
}
}
So like I said it works but I have an error in Intellij saying that the #JoinColumn(name ="someIdA") #JoinColumn(name = "someSeqA") don't exist and is expecting something like #JoinColumn(name ="some_id_a") #JoinColumn(name = "some_seq_a").
Using it the way Intellij is telling me, JPA has en error that says: Table [table_b] contains physical column name [some_id_a] referred to by multiple logical column names: [some_id_a], [someIdA].
My mapping is ok despite Intellij but is there's a better alternative ?
Thanks
You can use a "derived identity" and map your classes like this:
#Data
#Entity(name="Table_A")
public class TableA {
#EmbeddedId
private Key key;
#OneToMany(mappedBy = "tableA")
private List<TableB> tableBs;
#Data
#Embeddable
public static final class Key implements Serializable {
private String someIdA;
private long someSeqA;
}
}
#Data
#Entity(name="Table_B")
public class TableB {
#EmbeddedId
private Key key;
#MapsId("tableAKey") // maps tableAKey attribute of embedded id
#JoinColumns({
#JoinColumn(name="some_id_a", referencedColumnName="some_id_a"),
#JoinColumn(name="some_seq_a", referencedColumnName="some_seq_a")
})
#ManyToOne
private TableA tableA;
#Data
#Embeddable
public static final class Key implements Serializable {
private TableA.Key tableAKey; // corresponds to PK type of TableA
private long someSeqB;
}
}
Derived identities are discussed (with examples) in the JPA 2.2 spec in section 2.4.1.
I have superclass:
#MappedSuperclass
public abstract class BaseEntity {
#Id #GeneratedValue
private Long id;
#Version
private long version;
}
and two subclasses:
#Entity
#Table(name = "\"user\"")
public class User extends BaseEntity {
private String username;
#org.hibernate.annotations.Type(type = "yes_no")
private boolean isAdmin;
// constructor/getters/setters etc.
}
#Entity
public class Product extends BaseEntity {
public String name;
public BigDecimal price;
// constructor/getters/setters etc.
}
I can query for all subclasses using code:
entityManager.unwrap(Session.class)
.createCriteria(BaseEntity.class)
.list()
.forEach(x -> System.out.println(x));
how I can get the same results via JPA (without unwrap, is it possible?). I tried using createQuery("from BaseEntity") but get BaseEntity not mapped exception.
EDIT: I know that this will result in two SELECT statement. And it must be MappedSuperclass - I would like to not change that.
I want to save Tesis class using entityManager.persist() method but I get following error.
Caused by: javax.validation.UnexpectedTypeException: HV000030: No validator could be found for type: thymeleafexamples.layouts.acenta.Acenta.
#Entity
public class Tesis {
public Tesis(){
}
public Tesis(String adi, Acenta acenta) {
this.adi = adi;
this.acenta = acenta;
}
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
#NotEmpty
private String adi;
#NotEmpty
#ManyToOne(fetch = FetchType.EAGER,cascade=CascadeType.ALL)
#JoinColumn(name="acenta_id")
private Acenta acenta;
//GETTERS AND SETTERS
}
/////////////////////////////////////////////////////////////////////
#SuppressWarnings("serial")
#Entity
public class Acenta implements java.io.Serializable {
public Acenta(String adi) {
this.adi = adi;
}
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
#NotEmpty
private String adi;
#OneToMany(mappedBy="acenta")
private Set<Tesis> tesiss;
}
According to Hibernate Validator's API org.hibernate.validator.constraints.NotEmpty annotation
"asserts that the annotated string, collection, map or array is not
null or empty"
Based on the above definition Acenta type is not a valid type to check. You may consider using javax.validation.constraints.NotNull annotation instead as it's valid for all types and moreover not vendor specific.
I have a problem with abstract class where we implement an interface.
Now the interface is in our implementation of the abstract class other than the other implementation.
I'll show you the code here:
#MappedSuperclass
public abstract class AbstractOrder {
#Id
#GeneratedValue
private Long idOrder;
#Temporal(TemporalType.TIMESTAMP)
#Column(name="datum")
private Date date = new Date();
#OneToMany(targetEntity=IOrderLine.class)
private List<IOrderLine> orderLines = new ArrayList<IOrderLine>();
private String refOrder;
...
}
#Entity
#Table(name="supplyorders")
public class SupplyOrder extends AbstractOrder implements Comparable<SupplyOrder>, Serializable {
#ManyToOne
private Supplier supplier;
#Enumerated(EnumType.STRING)
private SupplyOrderStatus status = SupplyOrderStatus.TOBESUPPLIED;
#ElementCollection
private Set<CustomerOrder> customerOrders = new HashSet<CustomerOrder>();
...
}
#Entity
#Table(name="customerorders")
public class CustomerOrder extends AbstractOrder implements commparable<CustomerOrder>,Serializable {
#ManyToOne
private Customer customer;
#ManyToOne
private Place place;
#ManyToOne
private User vendor;
private double deposit;
#Enumerated(EnumType.STRING)
private OrderStatus status = OrderStatus.CREATED;
}
#MappedSuperclass
public interface IOrderLine {
double getSubTotal();
int getQuantity();
Furniture getFurniture();
}
#Entity
#Table(name="supplyorderlines")
public class SupplyOrderLine implements IOrderLine, Serializable {
#Id
#GeneratedValue
private Long id;
#OneToMany
private List<CustomerOrderLine> customerOrderLines = new ArrayList<CustomerOrderLine>();
...
}
and of course a class CustormerOrderLine that implements IOrderLine.
Now for the supplyOrder they have supplyOrderLines in them and the customerOrder has the CustomerOrderLine in them.
The fault we get is that the abstract class doesn't know what implementation to take of the interface IOrderLine.
How can I override the field of orderLines from the abstract class in the implementation class and point to the implementation of the IOrderLine with annotations?
Thx in advance.
Chillworld
In Java you cannot instantiate an interface.
You can only instantiate actual classes that implement interfaces.
There's actually no such thing as IOrderLine.class.
You probably want to declare your orderLines field in your sub-classes.
In the sub-classes you can declare the field with a concrete class (that will map to a real database table).
If you need/want to use the abstract class to refer to your line items generically (this seems like a really good idea), you could use an abstract method that returns an interface.
Here's an example:
#MappedSuperclass
public abstract class AbstractOrder {
#Id
#GeneratedValue
private Long idOrder;
#Temporal(TemporalType.TIMESTAMP)
#Column(name="datum")
private Date date = new Date();
// This abstract method will be implemented by sub-classes
public abstract List<IOrderLine> getOrderLines();
}
Then you can add the fields in your sub-classes and implement the abstract method that returns them:
#Entity
#Table(name = "supplyorderlines")
public class SupplyOrderLine implements IOrderLine, Serializable {
#Id
#GeneratedValue
private Long id;
#OneToMany(targetEntity = SupplyOrderLine.class)
private List<SupplyOrderLine> customerOrderLines;
#Override
public List<IOrderLine> getOrderLines() {
return customerOrderLines;
}
}
If this wasn't about JPA entities, you could also do something like this:
public abstract class PojoClass {
private Long idOrder;
private Date date = new Date();
private List<? extends IOrderLine> orderLines = new ArrayList<? extends IOrderLine>();
}
However, I don't think this is an option in a JPA entity.
That's because your entity class needs to map to a concrete class and database table.
I am trying to do mapping in JPA.
#Entity
public class Auction {
#Id
private Integer auctionId;
#OneToMany(mappedBy="auctionId")
#MapKey(name="auctionParamId")
private Map<AuctionParam, AuctionParamValue> values;
}
#Entity
public class AuctionParam {
#Id
private Integer auctionParamId;
private String description;
}
#Entity
public class AuctionParamValue {
#EmbeddedId
private AuctionParamValuePK pk;
private String value;
}
#Embeddable
public class AuctionParamValuePK {
#ManyToOne
#JoinColumn(name="auctionId")
private Auction auction;
#ManyToOne
#JoinColumn(name="auctionParamId")
private AuctionParam auctionParam;
}
Showing an error:-
.Error-Details:-Exception Description:
Entity [class
com.eaportal.domain.AuctionParamValue]
uses [class
com.eaportal.domain.AuctionParamValuePK]
as embedded id class
whose access-type
has been determined as [FIELD].
But
[class
com.eaportal.domain.AuctionParamValuePK]
does not define any [FIELD]. It is
likely that you have not provided
sufficient metadata in your id class
[class
com.eaportal.domain.AuctionParamValuePK].
If you come up with a solution please let me know.
Thanks in Advance
Tushar
You cannot use an EmbeddedId with relationships. Use an IdClass.
#Entity
#IdClass(AuctionParamValuePK.class)
public class AuctionParamValue {
#Id
#ManyToOne
#JoinColumn(name="auctionId")
private Auction auction;
#Id
#ManyToOne
#JoinColumn(name="auctionParamId")
private AuctionParam auctionParam;
#Basic
private String value;
}
public class AuctionParamValuePK {
private int auction;
private int auctionParam;
}
I think there are some errors in your Auction class. This is how I think it should look
#Entity
public class Auction {
#Id
private Integer auctionId;
#OneToMany(mappedBy="auction") // not auctionId
#MapKey(name="auctionParam") // not auctionParamId
private Map<AuctionParam, AuctionParamValue> values;
}
(The annotation values have to correspond with fields (or properties), not with columns)