I'm writing 3 tables in the following relation:
Club class:
#Setter
#Getter
#Entity
#Table(name = "Club")
public class Club {
#Id
#GeneratedValue
private Long id;
private String name;
private String type;
private String mainPage;
private String logo;
#OneToMany(mappedBy="clubProductKey.club", cascade = CascadeType.ALL)
#JsonIgnoreProperties(value = "clubProductKey.club", allowSetters=true)
private Set<ClubProduct> clubProducts;
...
Product class:
#Setter
#Getter
#Entity
#Table(name = "Product")
public class Product {
#Id
#GeneratedValue
private Long id;
#OneToMany(mappedBy="clubProductKey.product", cascade = CascadeType.ALL)
#JsonIgnoreProperties(value = "clubProductKey.product", allowSetters=true)
private Set<ClubProduct> clubProducts;
...
ClubProduct class:
#Setter
#Getter
#Entity
#Table(name = "ClubProduct")
public class ClubProduct {
#EmbeddedId
private ClubProductKey clubProductKey;
...
ClubProductKey class:
#Setter
#Getter
#Embeddable
public class ClubProductKey implements Serializable {
#ManyToOne(cascade = {CascadeType.MERGE,CascadeType.REFRESH })
#JoinColumn(name = "club_id", referencedColumnName = "id")
#JsonIgnoreProperties(value = "clubProducts", allowSetters=true)
private Club club;
#ManyToOne(cascade = {CascadeType.MERGE,CascadeType.REFRESH })
#JoinColumn(name = "product_id", referencedColumnName = "id")
#JsonIgnoreProperties(value = "clubProducts", allowSetters=true)
private Product product;
...
ClubProductRepository class:
public interface ClubProductRepository extends JpaRepository<ClubProduct, ClubProductKey> {
public List<ClubProduct> findByClubProductKeyClub(Club club);
public List<ClubProduct> findByClubProductKeyProduct(Product product);
}
I try to save clubProduct like this:
#Service
public class ClubProductServiceImp implements ClubProductService {
#Autowired
private ClubProductRepository clubProductRepository;
...
ClubProduct savedClubProduct = clubProductRepository.save(clubProduct);
return savedClubProduct;
}
However I find that the clubProduct is not saved in the clubProducts list in the club or product entity, the list is null. Must I add lines like club.getClubProducts.add(clubProduct) or is there any other way to make it added automatically?
Thank you.
The #OnetoMany mapping in your Club class uses the attribute mappedby which means that it represents the owning side of the relation responsible for handling the mapping. However, we still need to have both sides in sync as otherwise, we break the Domain Model relationship consistency, and the entity state transitions are not guaranteed to work unless both sides are properly synchronized.
The answer is yes, you have to manage the java relations yourself so that the clubProducts gets persisted. You are using an instance of the repository class club to persist the data so , you should add a setter method like :
public void addClubProduct(ClubProduct clubProduct) {
if (clubProduct!= null) {
if (clubProduct== null) {
clubProduct= new ArrayList<ClubProduct>();
}
clubProducts.add(clubProduct);
clubProduct.setClubProduct(this);
}
}
also a method to remove it from the list and use these method in your code to set the values to the list properly before initiating save . Read related article
Related
I have the following entity classes:
#Embeddable
#Getter
#Setter
public class OrganizationCyclePlageKey implements Serializable {
#Column(name = "organization_id")
Long organizationId;
#Column(name = "cycle_plages_id")
Long cyclePlagesId;
...
equals() and hashCode() methods come here
#Entity
#Table(name = "organization_cycle_plages")
#Getter
#Setter
public class OrganizationCyclePlage {
#EmbeddedId
private OrganizationCyclePlageKey id;
#ManyToOne
#MapsId("organizationId")
#JoinColumn(name = "organization_id")
Organization organization;
#ManyToOne
#MapsId("cyclePlagesId")
#JoinColumn(name = "cycle_plages_id")
CyclePlage cyclePlage;
...
other attributes
}
#Entity
#Getter
#Setter
public class CyclePlage extends AbstractEntity {
#OneToMany(mappedBy = "cyclePlage")
private Set<OrganizationCyclePlage> organizationCyclePlages;
...
}
#Entity
#DynamicUpdate
#Getter
#Setter
public class Organization extends AbstractEntity {
#OneToMany(mappedBy = "organization")
private Set<OrganizationCyclePlage> organizationCyclePlages = new HashSet<>();
...
}
SpringBoot app starts up normally without errors.
But when I try to save an instance of OrganizationCyclePlage:
OrganizationCyclePlage ocp = new OrganizationCyclePlage();
ocp.setOrganization(organization);
ocp.setCyclePlage(cyclePlage);
organizationCyclePlageRepository.save(ocp);
it raises the error when calling organizationCyclePlageRepository.save(ocp):
org.hibernate.PropertyAccessException: Could not set field value [361] value by reflection : [class com.XXXX.OrganizationCyclePlageKey.cyclePlagesId] setter of com.XXXX.OrganizationCyclePlageKey.cyclePlagesId
What's wrong with these relations?
I had to add the constructor into the OrganizationCyclePlageKey class to init the foreign keys values as well a default constructor via #NoArgsConstructor annotation:
public OrganizationCyclePlageKey(Long organizationId, Long cyclePlagesId) {
this.organizationId = organizationId;
this.cyclePlagesId = cyclePlagesId;
}
and init the OrganizationCyclePlageKey instance in the OrganizationCyclePlage class:
public class OrganizationCyclePlage {
private OrganizationCyclePlageKey id = new OrganizationCyclePlageKey();
...
}
I have two entities SuperAlbumEntity and AlbumEntity reflecting the same table "albums".
SuperAlbumEntity:
#Entity
#Table(name = "albums")
#Inheritance(strategy = InheritanceType.SINGLE_TABLE)
public class SuperAlbumEntity {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Integer id;
//other fields
}
AlbumEntity:
#EqualsAndHashCode(callSuper = true)
#Entity
#Table(name = "albums")
public class AlbumEntity extends SuperEntity{
//some fields
#Column(name = "country")
private String country;
#OneToMany(fetch = FetchType.EAGER)
#JoinColumn(name = "country_name", referencedColumnName = "country")
private Set<CountryEntity> countrySet = new HashSet<>();
}
AlbumEntity has #OneToMany mapping to CountryEntity:
#Entity
#Table(name = "countries")
public class CountryEntity implements Serializable {
#Id
String id;
String country_name;
//other fields
}
Running my application I get the folowing error:
...
Caused by: org.hibernate.AnnotationException: referencedColumnNames(country) of CountryEntity.countrySet referencing AlbumEntity not mapped to a single property
...
What's interesting is that if I move country field from SuperAlbumEntity to AlbumEntity everything just works fine...
Can someone explain me why I get this error?
I'm not sure but I think is connected with the type of inherence that you used it. Try to modify your superclass to something like this:
SuperAlbumEntity:
#MappedSuperclass
public abstract class SuperAlbumEntity {
}
AlbumEntity:
#Entity
#Inheritance(strategy=InheritanceType.JOINED)
#Table(name = "albums")
public class AlbumEntity extends SuperEntity {
#OneToMany(fetch = FetchType.EAGER)
#JoinColumn(name = "country_name", referencedColumnName = "country")
private Set<CountryEntity> countrySet = new HashSet<>();
}
I have a problem with JPA inheritance. The database model is also specially built. It contains several tables with the same attributes (the tables were intentionally cut by country) and all these tables connect to another table (OneToOne).
Here is an example of the data model:
usa_user, germany_user, austria_user. All these tables have the same attributes (id, name, address). Now the address was also built up according to the countries e.g. usa_address, germany_address, austria_address.
Now I don't know or have the problem that I have been mapping them correctly for a long time. I have the following:
// All Lombok Getter, Setter Args,...
#MappedSuperclass
public abstract Address {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#JsonIgnore
private Long id;
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name = "user_id", referencedColumnName = "id")
#JsonIgnore
private User user;
private String name;
private String addr_num;
...
}
// All Lombok Getter, Setter Args,...
#MappedSuperclass
public abstract User {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#JsonIgnore
private Long id;
#OneToOne(mappedBy = "user", cascade = CascadeType.ALL, fetch = FetchType.LAZY, optional = false)
#JsonIgnore
private Address address;
private String name;
}
#Entity
#Table(name = "usa_user")
public class UsaUser extends User {}
#Entity
#Table(name = "austria_user")
public class AustriaUser extends User {}
#Entity
#Table(name = "germany_user")
public class GermanyUser extends User {}
#Entity
#Table(name = "usa_address")
public class UsaAddress extends Address {}
#Entity
#Table(name = "austria_address")
public class AustriaAddress extends Address {}
#Entity
#Table(name = "germany_address")
public class GermanyAddress extends Address {}
But unfortunately this does not work. Every time I start it JPA notices that it can't map the Entities Address - User (which is understandable because they are not entities but abstract classes). What would be the best way to solve this? I want to avoid that I have to list the attributes in all these entities because it would be redundant.
The goal is to find out how I can use a #MappedSuperclass in a #MappedSuperclass.
MappedSuperclass is not queryable and thus also not joinable. You need to map this as an abstract entity with the table per class inheritance strategy. Just switch to #Entity on the Address and User and add #Inheritance(TABLE_PER_CLASS).
I'm trying to have Spring Data JPA issue one query using joins to eagerly get a graph of entities:
#Entity
#NamedEntityGraph(name = "PositionKey.all",
attributeNodes = {#NamedAttributeNode("positionKey.account"),
#NamedAttributeNode("positionKey.product")
})
#Data
public class Position {
#EmbeddedId
private PositionKey positionKey;
}
#Embeddable
#Data
public class PositionKey implements Serializable {
#ManyToOne
#JoinColumn(name = "accountId")
private Account account;
#ManyToOne
#JoinColumn(name = "productId")
private Product product;
}
Here's my Spring Data repo:
public interface PositionRepository extends JpaRepository<Position, PositionKey> {
#EntityGraph(value = "PositionKey.all", type = EntityGraphType.LOAD)
List<Position> findByPositionKeyAccountIn(Set<Account> accounts);
}
This produces the following exception:
java.lang.IllegalArgumentException: Unable to locate Attribute with the the given name [positionKey.account] on this ManagedType
I want all of the accounts and products to be retrieved in one join statement with the positions. How can I do this / reference the embedded ID properties?
I would suggest refactoring the entity this way if it possible
#Entity
#NamedEntityGraph(name = "PositionKey.all",
attributeNodes = {#NamedAttributeNode("account"),
#NamedAttributeNode("product")
})
#Data
public class Position {
#EmbeddedId
private PositionKey positionKey;
#MapsId("accountId")
#ManyToOne
#JoinColumn(name = "accountId")
private Account account;
#MapsId("productId")
#ManyToOne
#JoinColumn(name = "productId")
private Product product;
}
#Embeddable
#Data
public class PositionKey implements Serializable {
#Column(name = "accountId")
private Long accountId;
#Column(name = "productId")
private Long productId;
}
Such an EmbeddedId is much easier to use. For instance, when you are trying to get an entity by id, you do not need to create a complex key containing two entities.
I had entities User, BlockUnit, Block and Unit as Follows.
User has ManyToMany relation with blockunit.
#Entity
public class User {
#ManyToMany
private Set<BlockUnit> blockUnits;
}
#Entity
public class BlockUnit {
#Id
private Long id;
#OneToOne(targetEntity = Block.class)
#JoinColumn(name = "block_id")
private Block block;
#OneToOne(targetEntity = Unit.class)
#JoinColumn(name = "unit_id")
private Unit unit;
#ManyToMany(fetch = FetchType.LAZY)
private Set<User> users = new HashSet<>();
}
BlockUnit has OneToOne relation with Block and Unit.
#Entity
public class Block {
#Id
private Long id;
}
#Entity
public class Unit {
#Id
private Long id;
}
How should I create the JPA criteria specification for above in order to select user's having Block ID and Unit ID?
Thank you very much in advance!