I have user entity:
#ToString
#Data
#Entity
#Table(name = "users")
#NamedEntityGraph(name = "UserWithItems",
attributeNodes = {
#NamedAttributeNode("items"),
#NamedAttributeNode("roles")
})
public class User {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Column(name = "name")
private String name;
#OneToMany(mappedBy = "user", cascade = CascadeType.ALL)
private List<Item> items;
#OneToMany(mappedBy = "user", cascade = CascadeType.ALL)
private List<Role> roles;
}
item:
#ToString(exclude = "user")
#Data
#Entity
#Table(name = "items")
public class Item {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Column(name = "name")
private String name;
#ManyToOne
#JoinColumn(name = "user_id", referencedColumnName = "id", nullable = false)
private User user;
}
role:
#ToString
#Data
#Entity
#Table(name = "roles")
public class Role {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Column(name = "name")
private String name;
#ManyToOne
#JoinColumn(name = "user_id", referencedColumnName = "id", nullable = false)
private User user;
}
I want load user with items and roles. I use #NamedEntityGraph. It is my repository:
#EntityGraph(value = "UserWithItems", type = EntityGraph.EntityGraphType.LOAD)
#Query("select u from User u where u.id = ?1 and u.name =?2")
User getOneById(Long id, String name);
But I get an error:
Caused by: org.hibernate.loader.MultipleBagFetchException: cannot simultaneously fetch multiple bags: [com.example.egerload.entity.User.roles, com.example.egerload.entity.User.items]
at org.hibernate.loader.BasicLoader.postInstantiate(BasicLoader.java:75) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
at org.hibernate.loader.hql.QueryLoader.<init>(QueryLoader.java:108) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:212) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
at org.hibernate.hql.internal.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:143) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:119) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
at org.hibernate.engine.query.spi.HQLQueryPlan.<init>(HQLQueryPlan.java:85) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
at org.hibernate.query.internal.AbstractProducedQuery.makeQueryParametersForExecution(AbstractProducedQuery.java:1350) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
at org.hibernate.query.internal.AbstractProducedQuery.doList(AbstractProducedQuery.java:1539) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
at org.hibernate.query.internal.AbstractProducedQuery.list(AbstractProducedQuery.java:1505) ~[hibernate-core-5.3.7.Final.jar:5.3.7.Final]
... 41 common frames omitted
You can split your "UserWithItems" #NamedEntityGraph into two #NamedEntityGraphs, resulting in two queries, as described in Hibernate throws MultipleBagFetchException - cannot simultaneously fetch multiple bags - answer of Vlad Mihalcea.
User
#ToString
#Data
#Entity
#Table(name = "users")
#NamedEntityGraphs(
{
#NamedEntityGraph(
name = "UserWithItems",
attributeNodes = {
#NamedAttributeNode("items")
}
),
#NamedEntityGraph(
name = "UserWithRoles",
attributeNodes = {
#NamedAttributeNode("roles")
}
),
}
)
public class User {
...
}
I assume you have a repository class. For example with extends JpaRepository. Use each NamedEntityGraph on an extra method. (I have omitted the name condition and #Query("..."). The id condition should be sufficient, since it is the user's identifier. #Query("...") is not needed.)
UserRepository
public interface UserRepository extends JpaRepository<User, Long> {
#EntityGraph(value = "UserWithItems", type = EntityGraph.EntityGraphType.LOAD)
Optional<User> getOneWithItemsById(Long id);
#EntityGraph(value = "UserWithRoles", type = EntityGraph.EntityGraphType.LOAD)
Optional<User> getOneWithRolesById(Long id);
....
}
Finally, you can call both methods in a service.
UserService
public interface UserService {
Optional<User> readById(Long id);
}
UserServiceImpl
#Service
public class UserServiceImpl implements UserService {
private final UserRepository userRepository;
public UserServiceImpl(UserRepository userRepository) {
this.userRepository = userRepository;
}
#Override
#Transactional
public Optional<User> readById(Long id) {
// Load user with items into persistence contex
userRepository.getOneWithItemsById(id);
// Load user with roles into persistence context
// (There is only one user instance by id within the persistence context)
return userRepository.getOneWithRolesById(id);
}
}
Related
User.java
#Entity
#Table(name = "users")
public class User implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "user_role_id", referencedColumnName = "id")
private UserRole userRole;
}
UserRole.java
#Data
#Entity
#Table(name = "user_roles")
public class UserRole implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
}
Client.java
#Data
#Entity
#Table(name = "clients")
public class Client implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
#OneToOne(cascade = CascadeType.ALL)
#JoinColumns({ #JoinColumn(name = "user_id", referencedColumnName = "id"),
#JoinColumn(name = "user_role_id", referencedColumnName = "user_role_id") })
private User user;
}
Error
org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'entityManagerFactory' defined in class path resource [org/springframework/boot/autoconfigure/orm/jpa/HibernateJpaConfiguration.class]: Invocation of init method failed; nested exception is org.hibernate.MappingException: Unable to find column with logical name: user_role_id in users
In RDBMS, users.(id, user_role_id) is unique so clients table can refer to that.
Last time, I was using insertable = false, updatable = false on user_role_id, but when I want to add records of new client, I always need to add user_role_id manually user.setUserRoleId(userRole.getId()) after user.setUserRole(userRole) and I think that is bad practice of ORM (it should be added automatically when I set user.setUserRole(userRole))
#Column(name = "user_role_id", insertable = false, updatable = false)
private Integer userRoleId;
What should I do so the relation can be mapped in Spring JPA? and what is the best practice?
In other words, this is also mean how to reference to foreign key generated logical name column?
OK! Please try following configuration:
Below is a important code part and under this link you may find repository with working example
UserRole.java
#Data
#Entity
#Table(name = "user_roles")
public class UserRole implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "role_id")
private Integer roleId;
}
User.java
#Data
#Entity
#Table(name = "users")
public class User implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "user_id")
private Integer userId;
#ManyToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
#JoinColumn(name = "user_role_id", referencedColumnName = "role_id")
private UserRole userRole;
}
Client.java
#Data
#Entity
#Table(name = "clients")
public class Client implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "client_id")
private Integer clientId;
#OneToOne(fetch = FetchType.LAZY)
#JoinColumns(
value = {
#JoinColumn(name = "client_role_id", referencedColumnName = "user_role_id"),
#JoinColumn(name = "client_user_id", referencedColumnName = "user_id"),
}
,
foreignKey = #ForeignKey(
name = "FK_user_with_role",
foreignKeyDefinition = "FOREIGN KEY (client_user_id, client_role_id)\n" +
" REFERENCES users \n" +
" (user_id, user_role_id) \n" +
" ON UPDATE CASCADE\n" +
" ON DELETE CASCADE")
)
private User user;
}
Please note that beside adding a foreignKey in the Client implementation, you MUST keep the sequence of #JoinColum annotations.. I don't know what is the reason behind, but if you flip those lines you'll still get your error as it was before :)
EDIT: I've added another answer which fits best in my opinion. I'm leaving this one as well to see the other steps I tried.
Though the solution is not elegant and not using JPA as requested. Just in case anything in here would be helpful
If I understand the main issue correctly - you want to bind Client entity with Role entity via User entity, by first setting User's Role and then transfer that "property" by using only UserId instead setting additionally RoleId while creating Client.
Basically after playing for a while with your model I think the main issue is to assign data to each other within a #Transactional methods. That seems to be caused ba Lazy fetch strategy.
My proposal for solution that binds all your Entities according expectations differs only from yours with ommiting the RoleId JoinColumn in Clients table. I have checked that when calling a service that would have #Transactional methods, you can assign a Role to the User and User to the Client with simple user.setRole(roleEntity) followed by client.setUser(userEntity).
All the data is then consistent. No need to call further like getters and setters as you mentioned in the second part of your question. Question is if for any reason you need to have RoleId as well in your Clients Table, then this soultion would have to be enhanced by additional column?
UserRole.java
#Entity
#Table(name = "user_roles")
public class UserRole implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "role_id")
private Integer roleId;
//getters and setters and toString
}
User.java
#Entity
#Table(name = "users")
public class User implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "user_id")
private Integer userId;
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "user_role_id", referencedColumnName = "role_id")
private UserRole userRole;;
//getters and setters and toString;
}
Client.java
#Entity
#Table(name = "clients")
public class Client implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "client_id")
private Integer clientId;
#OneToOne(fetch = FetchType.LAZY)
#JoinColumns({
#JoinColumn(name = "client_user_id", referencedColumnName = "user_id"),
})
private User user;
#Column(name = "client_role_id")
private Integer roleId;
#PrePersist
#PreUpdate
private void prePersist(){
try {
roleId = getUser().getUserRole().getRoleId();
} catch (NullPointerException e){
roleId = null;
}
}
//getters and setters and toString
}
UserService.java
#Service
public class UserService {
UserRepo userRepo;
public UserService(UserRepo userRepo) {
this.userRepo = userRepo;
}
#Transactional
public void save(User user) {
userRepo.save(user);
}
#Transactional
public User getReferenceById(int i) {
return userRepo.getReferenceById(i);
}
}
ClientService.java
#Service
public class ClientService {
private ClientRepo clientRepo;
private UserService userService;
public ClientService(ClientRepo clientRepo, UserService userService) {
this.clientRepo = clientRepo;
this.userService = userService;
}
#Transactional
public Client save(Client client){
return clientRepo.save(client);
}
#Transactional
public Client getReferenceById(int i) {
return clientRepo.getReferenceById(i);
}
#Transactional
public void printClient(Client client){
client = clientRepo.getReferenceById(client.getClientId());
System.out.println(client);
}
#Transactional
public void bindUserToClient(int userId, int clientId) {
Client entity = clientRepo.findById(clientId).orElseGet(Client::new);
entity.setUser(userService.getReferenceById(userId));
}
#Transactional
public void printClient(int i) {
clientRepo.findById(i).ifPresentOrElse(this::printClient, EntityNotFoundException::new);
}
}
This configuration after running this commandLineRunner:
#Configuration
public class Config {
#Bean
#Transactional
public CommandLineRunner commandLineRunner(
#Autowired UserRoleRepo roleRepo,
#Autowired UserService userService,
#Autowired ClientService clientService
) {
return args -> {
for (int i = 0; i < 5; i++) {
roleRepo.save(new UserRole());
}
for (int i = 5; i > 0; i--) {
User user = new User();
user.setUserRole(roleRepo.getReferenceById(i));
userService.save(user);
}
Client client = new Client();
client.setUser(userService.getReferenceById(2));
client = clientService.save(client);
clientService.printClient(client);
client = new Client();
client.setClientId(1);
clientService.printClient(client);
int userId = 5;
clientService.bindUserToClient(userId, 1);
clientService.printClient(1);
};
}
}
gave me correct output in the console:
Client{id=1, user=User{id=2, userRole=UserRole{id=4}}}
Client{id=1, user=User{id=2, userRole=UserRole{id=4}}}
Client{id=1, user=User{id=5, userRole=UserRole{id=1}}}
WORKAROUND
I tried to reach the goal by use of Spring JPA but could'nt.
The workaround that keeps the referential integrity was by creating a constrains through DB like below and add #PrePersist and #PreUpdate annotated method which is updating the client's roleId as intended.
create table clients
(
client_id integer not null,
client_user_id integer,
client_role_id integer,
primary key (client_id)
);
create table user_roles
(
role_id integer generated by default as identity,
primary key (role_id)
);
create table users
(
user_id integer generated by default as identity,
user_role_id integer,
primary key (user_id),
CONSTRAINT User_Role UNIQUE (user_id, user_role_id)
);
alter table users
add constraint FK_role_id foreign key (user_role_id) references user_roles (role_id);
alter table clients
add constraint FK_user_id foreign key (client_user_id, client_role_id) references users (user_id, user_role_id) on update cascade ;
Thanks to that I could for instance update userRole in user entity, and the change was reflected in the clients table as well without any further actions
#Getter #Setter #NoArgsConstructor #AllArgsConstructor
#Table(name = "my_users")
public class MyUsers {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(nullable = false)
private Long id;
#Column(nullable = false, unique = true)
private String userName;
private String password;
#ManyToMany
private List<MyUsers> connections;
}
This is my MyUsers Model Class. I am using Hibernate and MySQL.
#ManyToMany
private List<MyUsers> connections;
This ManyToMany relationship is automatically creating the table 'my_users_connections' with 'my_users_id' and 'connections_id' colums. How can I add extra columns to this auto mapped table?
It's not ideal solution...
#Getter
#Setter
#NoArgsConstructor
#AllArgsConstructor
#Entity
#Table(name = "my_users")
public class MyUsers implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(nullable = false)
private Long myUsersId;
#Column(nullable = false, unique = false)
private String userName;
private String password;
#ManyToMany(cascade = CascadeType.ALL)
#JoinTable(name = "my_users_connections",
joinColumns = { #JoinColumn(name = "my_users_id") },
inverseJoinColumns = { #JoinColumn(name = "connections_id") })
private List<MyUsers> connections;
}
Create embedded id MyUsersConnectionsPK:
#Data
#Embeddable
public class MyUsersConnectionsPK implements Serializable {
#Column(name = "my_users_id")
private Long myUsersId;
#Column(name = "connections_id")
private Long connectionsId;
}
Create MyUsersConnections, which represent ManyToMany
#Data
#Entity
#Table(name = "my_users_connections")
public class MyUsersConnections implements Serializable {
#EmbeddedId
private MyUsersConnectionsPK id;
#ManyToOne
#MapsId("my_users_id")
#JoinColumn(name = "my_users_id")
private MyUsers myUsersId;
#ManyToOne
#MapsId("connections_id")
#JoinColumn(name = "connections_id")
private MyUsers connectionsId;
#Column(name = "extra_column")
private String extraColumn;
}
Create JPA repository
#Repository
public interface MyUsersConnectionsRepository extends JpaRepository<MyUsersConnections, MyUsersConnectionsPK> {
List<MyUsersConnections> findMyUsersConnectionsByMyUsersIdMyUsersId(Long id);
}
And simple sample for using:
#Service
public class Test {
#Autowired
private MyUsersConnectionsRepository myUsersConnectionsRepository;
#Autowired
private MyUsersRepository myUsersRepository;
public void test() {
MyUsers myUsers = new MyUsers();
myUsers.setUserName("user name");
myUsers.setPassword("password");
MyUsers myUsers2 = new MyUsers();
myUsers2.setUserName("user name 2");
myUsers2.setPassword("password 2");
myUsers.setConnections(Collections.singletonList(myUsers2));
myUsers = myUsersRepository.saveAndFlush(myUsers);
List<MyUsersConnections> myUsersConnections = myUsersConnectionsRepository.findMyUsersConnectionsByMyUsersIdMyUsersId(myUsers.getMyUsersId());
MyUsersConnections item = myUsersConnections.get(0);
item.setExtraColumn("Extra column");
myUsersConnectionsRepository.saveAndFlush(item);
}
}
I use OneToOne in the spring data JPA and I want to delete a record from the Address table without touching the user. But I can't.
If I remove User, in this case Address is removed, that's good.
But how can you delete an Address without touching the User?
https://github.com/myTestPercon/TestCascade
User.Java
#Entity
#Table(name = "user", schema = "testCascade")
public class User implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id")
private Long id;
#Column(name = "name")
private String name;
#OneToOne(mappedBy = "user", cascade = CascadeType.ALL)
private Address address;
// Getter and Setter ...
}
Address.java
#Entity
#Table(name = "address", schema = "testCascade")
public class Address implements Serializable {
#Id
private Long id;
#Column(name = "city")
private String city;
#OneToOne
#MapsId
#JoinColumn(name = "id")
private User user;
// Getter and Setter ...
}
DeleteController.java
#Controller
public class DeleteController {
#Autowired
ServiceJpa serviceJpa;
#GetMapping(value = "/deleteAddressById")
public String deleteAddressById () {
serviceJpa.deleteAddressById(4L);
return "redirect:/home";
}
}
You got your mapping wrong thats all is the problem .
try the below and see
User.java
#Entity
#Table(name = "user", schema = "testCascade")
public class User implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id")
private Long id;
#Column(name = "name")
private String name;
#OneToOne(cascade=CascadeType.ALL)
#JoinColumn(name="foriegn key column in user table for address example.. address_id")
private Address address;
// Getter and Setter ...
}
Address.java
#Entity
#Table(name = "address", schema = "testCascade")
public class Address implements Serializable {
#Id
private Long id;
#Column(name = "city")
private String city;
//name of the address variable in your user class
#OneToOne(mappedBy="address",
cascade={CascadeType.DETACH, CascadeType.MERGE, CascadeType.PERSIST,
CascadeType.REFRESH})
private User user;
// Getter and Setter ...
}
In order to solve this problem, you need to read the hibernate Documentation Hibernate Example 162, Example 163, Example 164.
And also I recommend to look at this is Using #PrimaryKeyJoinColumn annotation in spring data jpa
This helped me in solving this problem.
And also you need to specify the parameter orphanRemoval = true
User.java
#Entity(name = "User")
#Table(name = "user", schema = "testother")
public class User implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id")
private Long id;
#Column(name = "name")
private String name;
#OneToOne(mappedBy = "user", cascade = CascadeType.ALL, orphanRemoval = true)
private Address address;
public void addAddress(Address address) {
address.setUser( this );
this.address = address;
}
public void removeAddress() {
if ( address != null ) {
address.setUser( null );
this.address = null;
}
}
// Getter and Setter
}
Address.java
#Entity(name = "Address")
#Table(name = "address", schema = "testother")
public class Address implements Serializable {
#Id
private Long id;
#Column(name = "city")
private String city;
#OneToOne
#MapsId
#JoinColumn(name = "id")
private User user;
// Getter and Setter
}
DeleteController .java
#Controller
public class DeleteController {
#Autowired
ServiceJpa serviceJpa;
#GetMapping(value = "/deleteUser")
public String deleteUser () {
User user = serviceJpa.findUserById(2L).get();
user.removeAddress();
serviceJpa.saveUser(user);
return "/deleteUser";
}
}
Or make a custom SQL query.
#Repository
public interface DeleteAddress extends JpaRepository<Address, Long> {
#Modifying
#Query("delete from Address b where b.id=:id")
void deleteBooks(#Param("id") Long id);
}
public class Address {
#Id
private Long id;
#MapsId
#JoinColumn(name = "id")
private User user;
}
Rename #JoinColumn(name = "id") to #JoinColumn(name = "user_id")
You can't say that the column that will point to user will be the id of the Address
In my Spring Boot project I have an entity class User
#NoArgsConstructor
#AllArgsConstructor
#Getter
#Setter
#Builder
#Entity
#Table(name = "applicationusers")
public class User {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
[...]
#Singular
#ManyToMany(
cascade = {
CascadeType.PERSIST,
CascadeType.MERGE
},
fetch = FetchType.EAGER)
#Fetch(value = FetchMode.SUBSELECT)
#JoinTable(
name = "applicationusers_roles",
joinColumns = { #JoinColumn(name = "applicationuser_id") },
inverseJoinColumns = { #JoinColumn(name = "role_id") }
)
private Set<Role> roles;
}
and a second entity class role
#NoArgsConstructor
#AllArgsConstructor
#Getter
#Setter
#Builder
#Entity
#Table(name = "roles")
public class Role {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Column(nullable = false, unique = true)
#Length(min = 3, max = 15)
private String name;
#ManyToMany(fetch = FetchType.EAGER, mappedBy = "roles")
#Fetch(value = FetchMode.SUBSELECT)
#Singular
private Set<User> applicationUsers;
}
In one of my tests, I try to store a Role with a User to the repository like this
#Test
public void createRoleWithUsers() {
User newUser = User.builder()
.name("name")
.password("1234567")
.email("hello#world.net")
.enabled(true)
.build();
User savedUser = userRepository.save(newUser);
Set<User> users = new HashSet<>();
users.add(savedUser);
Role role = Role.builder()
.name("TestRole")
.applicationUsers(users)
.build();
Role createdRole = roleRepository.save(role);
Role foundRole = roleRepository.findRoleById(createdRole.getId()).get();
[...]
}
Debugging this code, I found out, that createdRole has the users set as expected but foundRole doesn't.
How can I get the users in foundRole too?
Please tell me if you need the repository-code - it's a very simple interface so I just skipped it.
Simplified response, read the fine manual: 6.3.10. Configuring Fetch- and LoadGraphs.
#NoArgsConstructor
#AllArgsConstructor
#Getter
#Setter
#Builder
#Entity
#Table(name = "applicationusers")
public class User {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Singular
#ManyToMany
private Set<Role> roles;
}
and
#NoArgsConstructor
#AllArgsConstructor
#Getter
#Setter
#Builder
#Entity
#Table(name = "roles")
public class Role {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#ManyToMany(mappedBy = "roles")
#Singular
private Set<User> applicationUsers;
}
The manual specifies:
public interface RoleRepository extends JpaRepository<Role, Long>{
#Query("select r from Role r where r.id = :id")
#EntityGraph(attributePaths = {"applicationUsers"})
Role findByIdFetchUsers(#Param("id") Long id);
}
The second System.out.println will give you a org.hibernate.LazyInitializationException.
#Override
public void run(String... args) throws Exception {
User u = save();
Role r2 = roleRepository.findByIdFetchUsers(u.getId());
System.out.println("R: " + r2 + " : " + r2.getApplicationUsers());
Role r1 = roleRepository.findById(u.getId()).get();
System.out.println("R: " + r1 + " : " + r1.getApplicationUsers());
}
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<>();
}