I'm having a problem with Specification as stated in the title.
Here is my FishingLocation class:
#Entity
#Table(name = "tbl_fishing_location")
public class FishingLocation {
...
#JsonIgnore
#OneToMany(mappedBy = "id", fetch = FetchType.EAGER)
private List<Lake> lakeList;
...
}
And here is my Lake class:
#Entity
#Table(name = "tbl_lake")
public class Lake {
...
#JsonIgnore
#ManyToOne
#JoinColumn
private FishingLocation fishingLocation;
...
}
They both have a StaticMetamodel as follow:
#StaticMetamodel(FishingLocation.class)
public class FishingLocation_ {
public static volatile ListAttribute<FishingLocation, Lake> lakeList;
public static volatile SingularAttribute<FishingLocation, Long> id;
}
#StaticMetamodel(Lake.class)
public class Lake_ {
public static volatile SingularAttribute<Lake, Long> id;
public static volatile SingularAttribute<Lake, FishingLocation> fishingLocation;
}
I have created a Specification to filter FishingLocation as follow (the fishing method is not relevant):
public static Specification<FishingLocation> fishingMethodIdIn(Set<Long> fishingMethodIds) {
if (fishingMethodIds == null || fishingMethodIds.isEmpty()) {
return null;
}
return root.join(FishingLocation_.lakeList)
.join(Lake_.fishingMethodList)
.get(FishingMethod_.id).in(fishingMethodIds);
};
}
The problem is that when I run the program and send a request to filter, Hibernate showed me this SQL query:
select
fishingloc0_.id as id1_7_,
fishingloc0_.active as active2_7_,
...
from
tbl_fishing_location fishingloc0_
.... (some inner joins)
inner join
tbl_lake lakelist4_
on fishingloc0_.id=lakelist4_.id
....
It is supposed to be on fishingloc0_.id=lakelist4_.fishing_location_id. So where is the problem in my code? Huge thanks to anyone helping me with this problem.
Edit: This is just a small fraction of my code. My filter has multiple criteria and join many tables so I cannot just use premade function in FishingLocationRepository. I want to create dynamic query which should be build depend on what user chooses to filter by.
you don't need to call this all using custom queries the Hibernate will take care of everything with OneTO Many and ManyTOOne mappings.
You can get the required data by simply calling the getter methods of its class.
for example, in the case of Fishing Location, you can get all the lakes related to that location by calling new FishingLocation().getLakeList().
same for the other case. https://www.javatpoint.com/hibernate-many-to-one-example-using-annotation
I have a JPA-persisted object model that contains a many-to-one relationship: an Account has many Transactions. A Transaction has one Account.
Here's a snippet of the code:
#Entity
public class Transaction {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
#ManyToOne(cascade = {CascadeType.ALL},fetch= FetchType.EAGER)
private Account fromAccount;
....
#Entity
public class Account {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
#OneToMany(cascade = {CascadeType.ALL},fetch= FetchType.EAGER, mappedBy = "fromAccount")
private Set<Transaction> transactions;
I am able to create an Account object, add transactions to it, and persist the Account object correctly. But, when I create a transaction, using an existing already persisted Account, and persisting the the Transaction, I get an exception:
Caused by: org.hibernate.PersistentObjectException: detached entity passed to persist: com.paulsanwald.Account
at org.hibernate.event.internal.DefaultPersistEventListener.onPersist(DefaultPersistEventListener.java:141)
So, I am able to persist an Account that contains transactions, but not a Transaction that has an Account. I thought this was because the Account might not be attached, but this code still gives me the same exception:
if (account.getId()!=null) {
account = entityManager.merge(account);
}
Transaction transaction = new Transaction(account,"other stuff");
// the below fails with a "detached entity" message. why?
entityManager.persist(transaction);
How can I correctly save a Transaction, associated with an already persisted Account object?
The solution is simple, just use the CascadeType.MERGE instead of CascadeType.PERSIST or CascadeType.ALL.
I have had the same problem and CascadeType.MERGE has worked for me.
I hope you are sorted.
This is a typical bidirectional consistency problem. It is well discussed in this link as well as this link.
As per the articles in the previous 2 links you need to fix your setters in both sides of the bidirectional relationship. An example setter for the One side is in this link.
An example setter for the Many side is in this link.
After you correct your setters you want to declare the Entity access type to be "Property". Best practice to declare "Property" access type is to move ALL the annotations from the member properties to the corresponding getters. A big word of caution is not to mix "Field" and "Property" access types within the entity class otherwise the behavior is undefined by the JSR-317 specifications.
Remove cascading from the child entity Transaction, it should be just:
#Entity class Transaction {
#ManyToOne // no cascading here!
private Account account;
}
(FetchType.EAGER can be removed as well as it's the default for #ManyToOne)
That's all!
Why? By saying "cascade ALL" on the child entity Transaction you require that every DB operation gets propagated to the parent entity Account. If you then do persist(transaction), persist(account) will be invoked as well.
But only transient (new) entities may be passed to persist (Transaction in this case). The detached (or other non-transient state) ones may not (Account in this case, as it's already in DB).
Therefore you get the exception "detached entity passed to persist". The Account entity is meant! Not the Transaction you call persist on.
You generally don't want to propagate from child to parent. Unfortunately there are many code examples in books (even in good ones) and through the net, which do exactly that. I don't know, why... Perhaps sometimes simply copied over and over without much thinking...
Guess what happens if you call remove(transaction) still having "cascade ALL" in that #ManyToOne? The account (btw, with all other transactions!) will be deleted from the DB as well. But that wasn't your intention, was it?
Don't pass id(pk) to persist method or try save() method instead of persist().
Removing child association cascading
So, you need to remove the #CascadeType.ALL from the #ManyToOne association. Child entities should not cascade to parent associations. Only parent entities should cascade to child entities.
#ManyToOne(fetch= FetchType.LAZY)
Notice that I set the fetch attribute to FetchType.LAZY because eager fetching is very bad for performance.
Setting both sides of the association
Whenever you have a bidirectional association, you need to synchronize both sides using addChild and removeChild methods in the parent entity:
public void addTransaction(Transaction transaction) {
transcations.add(transaction);
transaction.setAccount(this);
}
public void removeTransaction(Transaction transaction) {
transcations.remove(transaction);
transaction.setAccount(null);
}
Using merge is risky and tricky, so it's a dirty workaround in your case. You need to remember at least that when you pass an entity object to merge, it stops being attached to the transaction and instead a new, now-attached entity is returned. This means that if anyone has the old entity object still in their possession, changes to it are silently ignored and thrown away on commit.
You are not showing the complete code here, so I cannot double-check your transaction pattern. One way to get to a situation like this is if you don't have a transaction active when executing the merge and persist. In that case persistence provider is expected to open a new transaction for every JPA operation you perform and immediately commit and close it before the call returns. If this is the case, the merge would be run in a first transaction and then after the merge method returns, the transaction is completed and closed and the returned entity is now detached. The persist below it would then open a second transaction, and trying to refer to an entity that is detached, giving an exception. Always wrap your code inside a transaction unless you know very well what you are doing.
Using container-managed transaction it would look something like this. Do note: this assumes the method is inside a session bean and called via Local or Remote interface.
#TransactionAttribute(TransactionAttributeType.REQUIRED)
public void storeAccount(Account account) {
...
if (account.getId()!=null) {
account = entityManager.merge(account);
}
Transaction transaction = new Transaction(account,"other stuff");
entityManager.persist(account);
}
Probably in this case you obtained your account object using the merge logic, and persist is used to persist new objects and it will complain if the hierarchy is having an already persisted object. You should use saveOrUpdate in such cases, instead of persist.
My Spring Data JPA-based answer: I simply added a #Transactional annotation to my outer method.
Why it works
The child entity was immediately becoming detached because there was no active Hibernate Session context. Providing a Spring (Data JPA) transaction ensures a Hibernate Session is present.
Reference:
https://vladmihalcea.com/a-beginners-guide-to-jpa-hibernate-entity-state-transitions/
An old question, but came across the same issue recently . Sharing my experience here.
Entity
#Data
#Entity
#Table(name = "COURSE")
public class Course {
#Id
#GeneratedValue
private Long id;
}
Saving the entity (JUnit)
Course course = new Course(10L, "testcourse", "DummyCourse");
testEntityManager.persist(course);
Fix
Course course = new Course(null, "testcourse", "DummyCourse");
testEntityManager.persist(course);
Conclusion : If the entity class has #GeneratedValue for primary key (id), then ensure that you are not passing a value for the primary key (id)
If nothing helps and you are still getting this exception, review your equals() methods - and don't include child collection in it. Especially if you have deep structure of embedded collections (e.g. A contains Bs, B contains Cs, etc.).
In example of Account -> Transactions:
public class Account {
private Long id;
private String accountName;
private Set<Transaction> transactions;
#Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (!(obj instanceof Account))
return false;
Account other = (Account) obj;
return Objects.equals(this.id, other.id)
&& Objects.equals(this.accountName, other.accountName)
&& Objects.equals(this.transactions, other.transactions); // <--- REMOVE THIS!
}
}
In above example remove transactions from equals() checks. This is because hibernate will imply that you are not trying to update old object, but you pass a new object to persist, whenever you change element on the child collection.
Of course this solutions will not fit all applications and you should carefully design what you want to include in the equals and hashCode methods.
In your entity definition, you're not specifying the #JoinColumn for the Account joined to a Transaction. You'll want something like this:
#Entity
public class Transaction {
#ManyToOne(cascade = {CascadeType.ALL},fetch= FetchType.EAGER)
#JoinColumn(name = "accountId", referencedColumnName = "id")
private Account fromAccount;
}
EDIT: Well, I guess that would be useful if you were using the #Table annotation on your class. Heh. :)
Even if your annotations are declared correctly to properly manage the one-to-many relationship you may still encounter this precise exception. When adding a new child object, Transaction, to an attached data model you'll need to manage the primary key value - unless you're not supposed to. If you supply a primary key value for a child entity declared as follows before calling persist(T), you'll encounter this exception.
#Entity
public class Transaction {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
....
In this case, the annotations are declaring that the database will manage the generation of the entity's primary key values upon insertion. Providing one yourself (such as through the Id's setter) causes this exception.
Alternatively, but effectively the same, this annotation declaration results in the same exception:
#Entity
public class Transaction {
#Id
#org.hibernate.annotations.GenericGenerator(name="system-uuid", strategy="uuid")
#GeneratedValue(generator="system-uuid")
private Long id;
....
So, don't set the id value in your application code when it's already being managed.
Here is my fix.
Below is my Entity. Mark that the id is annotated with #GeneratedValue(strategy = GenerationType.AUTO), which means that the id would be generated by the Hibernate. Don't set it when entity object is created. As that will be auto generated by the Hibernate.
Mind you if the entity id field is not marked with #GeneratedValue then not assigning the id a value manually is also a crime, which will be greeted with IdentifierGenerationException: ids for this class must be manually assigned before calling save()
#Entity
#Data
#NamedQuery(name = "SimpleObject.findAll", query="Select s FROM SimpleObject s")
public class SimpleObject {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
#Column
private String key;
#Column
private String value;
}
And here is my main class.
public class SimpleObjectMain {
public static void main(String[] args) {
System.out.println("Hello Hello From SimpleObjectMain");
SimpleObject simpleObject = new SimpleObject();
simpleObject.setId(420L); // Not right, when id is a generated value then no need to set this.
simpleObject.setKey("Friend");
simpleObject.setValue("Bani");
EntityManager entityManager = EntityManagerUtil.getEntityManager();
entityManager.getTransaction().begin();
entityManager.persist(simpleObject);
entityManager.getTransaction().commit();
List<SimpleObject> simpleObjectList = entityManager.createNamedQuery("SimpleObject.findAll").getResultList();
for(SimpleObject simple : simpleObjectList){
System.out.println(simple);
}
entityManager.close();
}
}
When I tried saving that, it was throwing that
PersistentObjectException: detached entity passed to persist.
All I needed to fix was remove that id setting line for the simpleObject in the main method.
Maybe It is OpenJPA's bug, When rollback it reset the #Version field, but the pcVersionInit keep true. I have a AbstraceEntity which declared the #Version field. I can workaround it by reset the pcVersionInit field. But It is not a good idea. I think it not work when have cascade persist entity.
private static Field PC_VERSION_INIT = null;
static {
try {
PC_VERSION_INIT = AbstractEntity.class.getDeclaredField("pcVersionInit");
PC_VERSION_INIT.setAccessible(true);
} catch (NoSuchFieldException | SecurityException e) {
}
}
public T call(final EntityManager em) {
if (PC_VERSION_INIT != null && isDetached(entity)) {
try {
PC_VERSION_INIT.set(entity, false);
} catch (IllegalArgumentException | IllegalAccessException e) {
}
}
em.persist(entity);
return entity;
}
/**
* #param entity
* #param detached
* #return
*/
private boolean isDetached(final Object entity) {
if (entity instanceof PersistenceCapable) {
PersistenceCapable pc = (PersistenceCapable) entity;
if (pc.pcIsDetached() == Boolean.TRUE) {
return true;
}
}
return false;
}
You need to set Transaction for every Account.
foreach(Account account : accounts){
account.setTransaction(transactionObj);
}
Or it colud be enough (if appropriate) to set ids to null on many side.
// list of existing accounts
List<Account> accounts = new ArrayList<>(transactionObj.getAccounts());
foreach(Account account : accounts){
account.setId(null);
}
transactionObj.setAccounts(accounts);
// just persist transactionObj using EntityManager merge() method.
cascadeType.MERGE,fetch= FetchType.LAZY
Resolved by saving dependent object before the next.
This was happened to me because I was not setting Id (which was not auto generated). and trying to save with relation #ManytoOne
#OneToMany(mappedBy = "xxxx", cascade={CascadeType.MERGE, CascadeType.PERSIST, CascadeType.REMOVE})
worked for me.
In my case I was committing transaction when persist method was used.
On changing persist to save method , it got resolved.
If above solutions not work just one time comment the getter and setter methods of entity class and do not set the value of id.(Primary key)
Then this will work.
Another reason I have encountered this issue is having Entities that aren't versioned by Hibernate in a transaction.
Add a #Version annotation to all mapped entities
#Entity
public class Customer {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private UUID id;
#Version
private Integer version;
#OneToMany(cascade = CascadeType.ALL)
#JoinColumn(name = "orders")
private CustomerOrders orders;
}
#Entity
public class CustomerOrders {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private UUID id;
#Version
private Integer version;
private BigDecimal value;
}
This error comes from the JPA Lifecycle.
To solve, no need to use specific decorator. Just join the entity using merge like that :
entityManager.merge(transaction);
And don't forget to correctly set up your getter and setter so your both side are sync.
So I stumbled across this Question and Answers because I got the same Error but a very basic object with just Strings and Integers.
But in my case I was trying to set a Value to a Field which was annotated with #Id.
So if you are using #Id it seems that you can't create a new Object on a Class and set an Id by yourself and persist it to Database. You should then leave the Id blank. I wasn't aware and maybe this helps anyone else.
The problem here is lack of control.
When we use the CrudRepository/JPARepository save method we loose the transactional control.
To overcome this issue we have Transaction Management
I prefer the #Transactional mechanism
imports
import javax.transaction.Transactional;
Entire Source Code:
package com.oracle.dto;
import lombok.*;
import javax.persistence.*;
import java.util.Date;
import java.util.List;
#Entity
#Data
#ToString(exclude = {"employee"})
#EqualsAndHashCode(exclude = {"employee"})
public class Project {
#Id
#GeneratedValue(strategy = GenerationType.AUTO,generator = "ps")
#SequenceGenerator(name = "ps",sequenceName = "project_seq",initialValue = 1000,allocationSize = 1)
#Setter(AccessLevel.NONE)
#Column(name = "project_id",updatable = false,nullable = false)
private Integer pId;
#Column(name="project_name",nullable = false,updatable = true)
private String projectName;
#Column(name="team_size",nullable = true,updatable = true)
private Integer teamSize;
#Column(name="start_date")
private Date startDate;
#ManyToMany(cascade = CascadeType.ALL)
#JoinTable(name="projectemp_join_table",
joinColumns = {#JoinColumn(name = "project_id")},
inverseJoinColumns = {#JoinColumn(name="emp_id")}
)
private List<Employee> employees;
}
package com.oracle.dto;
import lombok.*;
import javax.persistence.*;
import java.util.List;
#Entity
#Data
#EqualsAndHashCode(exclude = {"projects"})
#ToString(exclude = {"projects"})
public class Employee {
#Id
#GeneratedValue(strategy = GenerationType.AUTO,generator = "es")
#SequenceGenerator(name = "es",sequenceName = "emp_seq",allocationSize = 1,initialValue = 2000)
#Setter(AccessLevel.NONE)
#Column(name = "emp_id",nullable = false,updatable = false)
private Integer eId;
#Column(name="fist_name")
private String firstName;
#Column(name="last_name")
private String lastName;
#ManyToMany(mappedBy = "employees")
private List<Project> projects;
}
package com.oracle.repo;
import com.oracle.dto.Employee;
import org.springframework.data.jpa.repository.JpaRepository;
public interface EmployeeRepo extends JpaRepository<Employee,Integer> {
}
package com.oracle.repo;
import com.oracle.dto.Project;
import org.springframework.data.jpa.repository.JpaRepository;
public interface ProjectRepo extends JpaRepository<Project,Integer> {
}
package com.oracle.services;
import com.oracle.dto.Employee;
import com.oracle.dto.Project;
import com.oracle.repo.ProjectRepo;
import org.springframework.beans.factory.annotation.Autowired;
import org.springframework.stereotype.Component;
import javax.transaction.Transactional;
import java.util.Date;
import java.util.LinkedList;
import java.util.List;
#Component
public class DBServices {
#Autowired
private ProjectRepo repo;
#Transactional
public void performActivity(){
Project p1 = new Project();
p1.setProjectName("Bank 2");
p1.setTeamSize(20);
p1.setStartDate(new Date(2020, 12, 22));
Project p2 = new Project();
p2.setProjectName("Bank 1");
p2.setTeamSize(21);
p2.setStartDate(new Date(2020, 12, 22));
Project p3 = new Project();
p3.setProjectName("Customs");
p3.setTeamSize(11);
p3.setStartDate(new Date(2010, 11, 20));
Employee e1 = new Employee();
e1.setFirstName("Pratik");
e1.setLastName("Gaurav");
Employee e2 = new Employee();
e2.setFirstName("Ankita");
e2.setLastName("Noopur");
Employee e3 = new Employee();
e3.setFirstName("Rudra");
e3.setLastName("Narayan");
List<Employee> empList1 = new LinkedList<Employee>();
empList1.add(e2);
empList1.add(e3);
List<Employee> empList2 = new LinkedList<Employee>();
empList2.add(e1);
empList2.add(e2);
List<Project> pl1=new LinkedList<Project>();
pl1.add(p1);
pl1.add(p2);
List<Project> pl2=new LinkedList<Project>();
pl2.add(p2);pl2.add(p3);
p1.setEmployees(empList1);
p2.setEmployees(empList2);
e1.setProjects(pl1);
e2.setProjects(pl2);
repo.save(p1);
repo.save(p2);
repo.save(p3);
}
}
I'm trying to use JPA (with Hibernate) to save 2 entities. Spring data is providing the interface but I don't think it matters here.
I have a main entity called 'Model'. This model has many 'Parameter' entities linked. I'm writing a method to save a model and its parameters.
This is the method:
private void cascadeSave(Model model) {
modelRepository.save(model);
for (ParameterValue value : model.getParameterValues()) {
parameterValueRepository.save(value);
}
}
This is the problem:
When I load a Model that already existed before, add some new parameters to it and then call this method to save both of them something strange happens:
Before the first save (modelRepository.save) this is what the model's data looks like when debugging:
The model has 2 parameters, with filled in values (name and model are filled).
Now, after saving the model the first save in my method, this happens. Note that the object reference is a different one so Hibernate must have done something magical and recreated the values instead of leaving them alone:
For some reason hibernate cleared all the attributes of the parameters in the set.
Now when the saving of the new parameters happens in the following code it fails because of not null constraints etc.
My question: Why does hibernate clear all of the fields?
Here are the relevant mappings:
ParameterValue
#Entity
#Table(name = "tbl_parameter_value")
#Inheritance(strategy = InheritanceType.SINGLE_TABLE)
#DiscriminatorColumn(name = "PARAMETER_TYPE")
public abstract class ParameterValue extends AbstractBaseObject {
#Column(nullable = false)
#NotBlank
private String name;
private String stringValue;
private Double doubleValue;
private Integer intValue;
private Boolean booleanValue;
#Enumerated(EnumType.STRING)
private ModelType modelParameterType;
#Column(precision = 7, scale = 6)
private BigDecimal bigDecimalValue;
#Lob
private byte[] blobValue;
ParameterValue() {
}
ParameterValue(String name) {
this.name = name;
}
ModelParameterValue
#Entity
#DiscriminatorValue(value = "MODEL")
public class ModelParameterValue extends ParameterValue {
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "model_id", foreignKey = #ForeignKey(name = "FK_VALUE_MODEL"))
private Model model;
ModelParameterValue() {
super();
}
ModelParameterValue(String name) {
super(name);
}
Model
#Entity
#Table(name = "tbl_model")
public class Model extends AbstractBaseObject implements Auditable {
#OneToMany(fetch = FetchType.LAZY, mappedBy = "model")
private Set<ModelParameterValue> parameterValues = new HashSet<>();
EDIT
I was able to reproduce this with a minimal example.
If you replace everything spring data does this is what happened under the hood (em is a JPA EntityManager):
public Model simpleTest() {
Model model = new Model("My Test Model");
em.persist(model);
model.addParameter(new Parameter("Param 1"));
em.merge(model);
for (Parameter child : model.getParameters()) {
em.persist(child);
}
return model;
}
When the merge is executed, all of the attributes of the parameters are set to null. They are actually just replaced with completely new parameters.
I guess you are using Spring Data Jpa as your modelRepository. This indicates following consequences.
Spring Repository Save
S save(S entity)
Saves a given entity. Use the returned
instance for further operations as the save operation might have
changed the entity instance completely.
So it is normal behaviour which you had encountered.
Code should be changed to :
model = modelRepository.save(model);
for (ParameterValue value : model.getParameterValues()) {
parameterValueRepository.save(value);
}
EDIT:
I think that your saving function is broken in sense, that you do it backwards. Either you can use CascadeType on your relation or you have to save children first.
Cascade
Cascade works like that "If you save Parent, save Children, if you update Parent, update Children ..."
So we can put cascade on your relation like that :
#Entity
#Table(name = "tbl_model")
public class Model extends AbstractBaseObject implements Auditable {
#OneToMany(fetch = FetchType.LAZY, mappedBy = "model", cascade = CascadeType.ALL)
private Set<ModelParameterValue> parameterValues = new HashSet<>();
and then only save like this
private void cascadeSave(Model model) {
modelRepository.save(model);
//ParamValues will be saved/updated automaticlly if your model has changed
}
Bottom-Up save
Second option is just to save params first and then model with them.
private void cascadeSave(Model model) {
model.setParameterValues(
model.getParameterValues().stream()
.map(param -> parameterValueRepository.save(param))
.collect(Collectors.toSet())
);
modelRepository.save(model);
}
I haven't checked second code in my compiler but the idea is to first save children (ParamValues), put it into Model and then save Model : )
Lets say we have User entity class. User can be friends with other users. How can i map this self-reference collection field without creating a new entity called Connection or creating multiple entries in the database?
#Entity
public class User {
...
#ManyToMany
private Collection<User> friends;
...
}
USER_ID-FRIEND_ID
1 - 2
2 - 1 (duplicate... I don't need it)
Following is snapshot from my code for ElementEntity:
#OneToMany(mappedBy = "parent", fetch = FetchType.LAZY)
private List<ElementEntity> children;
#JoinColumn(name = "ParentId", referencedColumnName = "ElementId")
#ManyToOne(fetch = FetchType.LAZY)
private ElementEntity parent;
Where on database there are fields:
ElementId - primary key;
ParentId relation with parent
You can't - you need both records in the database.
Actually, for friendship relations, I'd say that a graph database like neo4j is the proper thing to use. There you have the two users and simply add an edge "friends".
At least you will need a relational table.
So you have a USER table and a FRIENDS:
user_id friend_id
1 2
But #Bozho answer is way better than mine (neo4j).
Well, in fact you can.
You can use annotations like #PreUpdate, #PrePersists, #PostUpdate and so to convert manually the elements of a collection. This way your entity can render then them way you want while in database you only store a raw text.
A more pausible alternative will be to use #Convert annotation, available since jpa 2.1 (#UserType in hibernate). It tells jpa to convert the field into another type everytime it read/save in database.
For it you should use #Convert anotation, specifying and AttributeConverter object.
For example
public class Parent {
#Id
private Integer id;
#Convert(converter = FriendConverter.class)
private Set<Parent>friends;
}
And converter class like the following:
#Component
public class FriendConverter implements AttributeConverter<List, String>{
#Autowired
private SomeRepository someRepository;
#Override
public String convertToDatabaseColumn(List attribute) {
StringBuilder sb = new StringBuilder();
for (Object object : attribute) {
Parent parent = (parent) object;
sb.append(parent.getId()).append(".");
}
return sb.toString();
}
#Override
public List convertToEntityAttribute(String dbData) {
String[] split = dbData.split(".");
List<Parent>friends = new ArrayList<>();
for (String string : split) {
Parent parent = someRepository.findById(Integer.valueOf(string));
friends.add(accion);
}
return friends;
}
}
It is a dummy implementation but it gives you the idea.
As a personal comment, I do recommend to map the relationship as it should. In the future it will avoid you problems. AttributeConverter comes in handy when working with enums