Let's say I have two entities:
#Entity
public class Phone {
#Id
private Long id;
private String number;
}
#Entity
public class Person {
#Id
private Long id;
private String name;
}
The relationship between a person and a phone is one to one.
How could I access only the phone's number in the Person entity mapped by the phone's id
#Entity
public class Person {
#Id
private Long id;
private String name;
// ???
private String phoneNumber;
}
The reason for not mapping the whole entity is because in some more realistic entities there are too many properties.
I don't think you can, but something like this might be acceptable:
public class Person {
#OneToOne
#JoinColumn(name = "phone_id")
private Phone phone;
public String getPhoneNumber() {
return phone.getNumber();
}
}
Although you have mapped the whole object, not just the single property, you have only exposed the single property you want. The other stuff is hidden.
Alternatively, do it at the DB layer using a View:
create view person_with_phone as
select p.id, p.name,f.number
from person p
join phone f on f.id=p.phone_id
and then have an entity class to match the view. You'll need to turn off schema creation in your JPA implementation.
Related
I use h2 in memory db and I don't want to create duplicate locations in my DataBase. Only when I use createItem and input location column id manualy it write it to the same location. Otherwise even if the country city gps coordinates are the same app write it to other location with it's id.
I tried to understand but It's not working
I got these entities.
#Entity
#Table(name = "item_System_items")
public class Item {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
private String title;
private String description;
private BigDecimal price;
private Integer stock;
#ManyToOne
#JoinColumn(name = "location_id")
#Cascade(value={org.hibernate.annotations.CascadeType.ALL})
private Location location;
And
#Entity
#Table(name = "item_System_locations")
public class Location {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Integer id;
private String country;
private String city;
private String street;
private String gpsCoordinates;
SETTERS AND GETTERS IS THERE I JUST NOT POST THEM HERE
Controller
#RestController
#RequestMapping("/items")
public class ItemsController {
#Autowired
private ItemsService service;
#PostMapping
#ResponseStatus(value=HttpStatus.CREATED)
public int createItem(#RequestBody Item item) {
return service.createItem(item);
}
Service
#Service
#Transactional
public class ItemsService {
#Autowired
private ItemJPARepository repository;
public int createItem(Item item) {
return repository.save(item).getId();
}
I expect after re-coding app doesn't make new location if the column values are the same.
Thank you people!
If you really help me I would be so happy!
There is nothing in your Entity definitions to tell the ORM about the constraint you want.
You can add #UniqueConstraint to the #Table in your Location entity and specify which column(s) must all be in a unique combination (example given in the linked documentation):
#Table(
name="EMPLOYEE",
uniqueConstraints=
#UniqueConstraint(columnNames={"EMP_ID", "EMP_NAME"})
)
This will add a check in the database, if the ORM is managing your database schema, which will throw an exception when violated.
So in this example scenario I have an attendance DTO, and a worker DTO, workers in this context are separated by department, and a worker can only ever be inside of one department. It is important to note that Worker {id='123-123-123', department='a'} is different to Worker {id='123-123-123', department='b'}, despite them both sharing the same Id.
I have the following class setup to try and separate functions by id and department
public class IdAndDepartmentPK implements Serializable {
private String id;
private String department;
public IdAndDepartmentPK() {}
...
}
This key class is shared between DTOs that require both the Worker's id and department, below are the two DTOs that are causing a problem.
#Entity
#IdClass(IdAndDepartmentPK.class)
public class AttendencetDto {
#Id private String id; // This is a departmentally unique attendenceId
#Id private String department;
#Column private String workerId;
#JoinColumns({
#JoinColumn(name = "workerId"),
#JoinColumn(name = "department")
})
#ManyToOne(fetch = FetchType.EAGER)
private WorkerDto workerDto;
....
}
#Entity
#IdClass(IdAndDepartmentPK.class)
public class WorkerDto {
#Id
private String id;
#Id
private String department;
...
}
WorkerDto does not need to have knowledge of AttendencetDto, but AttendencetDto, does need to have access to WorkerDto and the other data it contains.
Hibernate complains that fields like workerId should be mapped with insert="false" update="false", but if I was to do this then I wouldn't be able to persist those values to the database.
I essentially want to have those fields available whilst also having the WorkerDto available, is this possible?
You should remove #Column private String workerId; because you already map it by relation to WorkerDto.
If you want to create relation between that you should use setWorkerDto method in your AttendencetDto and just save. After transaction ends you will have your relation in DB.
I have a class that looks something like this:
#Entity
public class EdgeInnovation {
#Id
public long id;
#ManyToOne
public NodeInnovation destination;
#ManyToOne
public NodeInnovation origin;
}
and another one that looks something like this:
#Entity
public class NodeInnovation {
#Id
public long id;
#OneToOne
public EdgeInnovation replacedEdge;
}
and so each table map to the other, so one entity will refer to other entities that will refer to more entities and so on, so that in the end there will be many entities that will be fetched from the database. Is there any way to only get the value (integer/long) of the key and not the entity it refers to? something like this:
#ManyToOne(referToThisTable="NodeInnovation")
#Entity
public class EdgeInnovation {
#Id
public long id;
#ManyToOne(referToTable="NodeInnovation")
public Long destination;
#ManyToOne(referToTable="NodeInnovation")
public Long origin;
}
and
#Entity
public class NodeInnovation {
#Id
public long id;
#OneToOne(referToTable="EdgeInnovation")
public Long replacedEdge;
}
Here's an example. I want the stuff in green, I get all the stuff in red along with it. This wastes memory and time reading from disk.
You would just map the foreign keys as basic mappings instead of Relationships:
#Entity
public class EdgeInnovation {
#Id
public long id;
#Column(name="DESTINATION_ID")
public Long destination;
#Column(name="ORIGIN_ID")
public Long origin;
}
Or you can have access to both the ID and the referenced entity within EdgeInnovation, but you'll need to decide which you want to use to set the mapping:
#Entity
public class EdgeInnovation {
#Id
public long id;
#Column(name="DESTINATION_ID", updatable=false, insertable=false)
public Long destination_id;
#ManyToOne
public NodeInnovation destination;
#Column(name="ORIGIN_ID", updatable=false, insertable=false)
public Long origin_id;
#ManyToOne
public NodeInnovation origin;
}
In the above example, the origin_id is read-only while the origin reference is used to set the foreign key in the table. Any changes though should be made to both fields to keep the object mappings in synch with each other.
Another alternative is to use the provider's native code to find if the reference is lazy and wasn't triggered, and then get the foreign key value. If it has been triggered, you can just use the reference to get the ID value, since it won't cause a query to fetch anything. This is something you would have to look into EclipseLink's source code for though.
Sorry, I cant comment so I put it here ,
I think it should be like that
#Entity
public class EdgeInnovation {
#Id
public long id;
#ManyToOne
public NodeInnovation destination;
#ManyToOne
public NodeInnovation origin;
}
And the other class is :
#Entity
public class NodeInnovation {
#Id
public long id;
#OneToMany(mappedBy="origin")
public List<EdgeInnovation> replacedEdges;
}
If I'm getting the situation wrong sorry, (Could you draw your classes with the relations so I can get it straight?)
Why not use a new construction in JPA and a custom constructor in NodeInnovation? Basically, create a transient property in NodeInnovation for use when you only want the EdgeInnovation id:
#Entity
public class NodeInnovation {
#Id #GeneratedValue private Long id;
private Integer type;
#OneToOne
private EdgeInnovation replacedEdge;
#Transient
private Long replacedEdgeId;
public NodeInnovation() {}
public NodeInnovation(Long id, Integer type, Long replacedEdgeId ) {
this.id = id;
this.type = type;
this.replacedEdgeId = replacedEdgeId;
}
...
}
Use it like so:
NodeInnovation n = em.createQuery("select new NodeInnovation(n.id, n.type, n.replacedEdge.id) from NodeInnovation n where n.id = 20", NodeInnovation.class).getSingleResult();
You didn't say how you were selecting NodeInnovation, whether directly or through a join, but either way the trick is the new NodeInnovation in the JPQL or CriteriaBuilder query.
I am aware I am quite late but some people might look for an answer to the same question - in your JPA repository you could do something like this:
#Query("SELECT new java.lang.Integer(model.id) FROM #{#entityName} model WHERE model.relationModeFieldlName.id IN :relationModelIds")
List<Integer> findIdByRelationModelIdIn(#Param("relationModelIds") List<Long> relationModelIds);
i'm curious about how HQL would assert equality between an entity instances.
Let's say I have a Entity called Person
#Entity
public class Person{
#Id
private Long id;
private String name;
}
and Department
#Entity
public class Department {
#Id
private Long id;
#ManyToOne
private Person person;
}
then it's fine if I do the following statement:
Query query = getSession().createQuery("from Department d where d.person = ?");
query.setProperty(0,new Person(1L));
but, what if I have an Embedded entity and no pk defined? like
#Embeddable
public class Adress {
private String email;
private String street;
private Long identifier;
}
#Entity
public class Person{
#Id
private Long id;
private String name;
#Embedded
private Address address;
}
would have any way so I could tell JPA to make it work:
Query query = getSession().createQuery("from Person p where p.address = ?");
query.setProperty(0,new Address(1L));
even though it's not exactly a primary key?
For sure i know i'd work if I tried p.adress.identifier, and then passed just the Long value, but the point is, can I tell JPA provider how it's gonna kind of 'implement' equality my way?
Thank you all
No, it is not supported and it would be difficult in general or would not make sense in some situations, like when there are collections in the Embeddable.
If you find that you need this often though, consider converting such Embeddables to custom user types. Then you can perform comparisons the way you described.
I am trying to figure out the best way to accomplish a relationship in hibernate. I have a Customer object. Each customer has a technical contact, a billing contact, and a sales contact. Each type of contact has the exact same data structure (phone, email, address, etc).
My first thought was to create a Contact table, and then have three columns in the Customer table - sales_contact, billing_contact, technical_contact. That would make three distinct foreign key one-to-one relationships between the same two tables. However, I have found that this is very difficult to map in Hibernate, at least using annotations.
Another thought was to make it a many to many relationship, and have a type flag in the mapping table. So, any Customer can have multiple Contacts (though no more than three, in this case) and any Contact can belong to multiple Customers. I was not sure how to map that one either, though. Would tere be a type field on the map table? Would this attribute show up on the Contact java model object? Would the Customer model have a Set of Contact objects. or three different individual Contact objects?
So I am really looking for two things here - 1. What is the best way to implement this in the database, and 2. How do I make Hibernate map that using annotations?
It can be as simple as :
#Entity
public class Contact {
#Id
private String id;
private String phome;
private String email;
private String address;
// ... Getters and Setters
}
#Entity
public class Customer {
#Id
#GeneratedValue
private String id;
#ManyToOne
#JoinColumn(name = "ID")
private Contact billingContact;
#ManyToOne
#JoinColumn(name = "ID")
private Contact salesContact;
#ManyToOne
#JoinColumn(name = "ID")
private Contact technicalContact;
public Customer() {
}
// ... Getters and Setters
}
Now, if you want to make the difference between a BillingContact and a SalesContact at the object level, you can make Contact abstract, and implement it with each type of contact. You will have to annotate the parent class with #Inheritance to specify the inheritance strategy of your choice (SINGLE_TABLE sounds appropriate here, it will use a technical discriminator column - see http://docs.jboss.org/hibernate/annotations/3.5/reference/en/html_single/#d0e1168).
How about using #OneToOne and just naming the #JoinColumn differently for each type:
#Entity
public class Contact {
#Id
private String id;
private String phone;
private String email;
private String address;
// ... Getters and Setters
}
#Entity
public class Customer {
#Id
#GeneratedValue
private String id;
#OneToOne(cascade=CascadeType.ALL)
#JoinColumn(name="billingContact_ID")
private Contact billingContact;
#OneToOne(cascade=CascadeType.ALL)
#JoinColumn(name="salesContact_ID")
private Contact salesContact;
#OneToOne(cascade=CascadeType.ALL)
#JoinColumn(name="technicalContact_ID")
private Contact technicalContact;
public Customer() {
}
// ....
}
For each row in Customer table should create three rows in Contact table