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 have four entities to map together, "Association", "Account", "Transaction" and "TransactionEvent". The id of Association is a simple integer id. Account and Transaction each have embedded id's consisting of a mapping to an Association and a number.
TransactionEvent should have an embedded id consisting of one Account and one Association. Now, each of those are mapped to an Association, and I want it to be the same Association for one TransactionEvent.
JPA Annotations is used for the Hibernate mapping, but I cannot make this work. I have tried forcing the same column name for the Association key, but Hibernate complains about repeated columns.
Is this possible to solve, or am I not thinking straight?
Here are the annotated classes, but I trimmed away getters/setters and non-id columns, annotations from the javax.persistence namespace:
#Entity
public class Association implements Serializable {
#Id #GeneratedValue(strategy = GenerationType.AUTO)
private long id;
}
#Embeddable
public class AccountPK implements Serializable {
#ManyToOne(optional=false)
private Association association;
#Column(nullable=false)
private int number;
}
#Embeddable
public class TransactionPK implements Serializable {
#ManyToOne
private Association association;
#GeneratedValue(strategy=GenerationType.AUTO)
private long number;
}
#Embeddable
public class AccountEventPK implements Serializable {
#ManyToOne(optional=false)
#JoinColumns({
#JoinColumn(name="association_id", referencedColumnName="association_id"),
#JoinColumn(name="account_number", referencedColumnName="number")
})
private Account account;
#ManyToOne(optional=false)
#JoinColumns({
#JoinColumn(name="association_id", referencedColumnName="association_id"),
#JoinColumn(name="transaction_number", referencedColumnName="number")
})
private Transaction transaction;
}
Actual Account, Transaction and AccountEvent entities are on the form
#Entity
public class Account implements Serializable {
#EmbeddedId
private AccountPK id;
}
I don't have much experience with placing associations directly in the embedded id component since this is not supported by JPA but is Hibernate specific.
As an alternative my suggestion would be to use the approach described in the Composite Primary Keys section of the JPA wikibook:
(...) JPA 1.0 requires that all #Id
mappings be Basic mappings, so if
your Id comes from a foreign key
column through a OneToOne or
ManyToOne mapping, you must also
define a Basic #Id mapping for the
foreign key column. The reason for
this is in part that the Id must be a
simple object for identity and caching
purposes, and for use in the IdClass
or the EntityManager find() API.
Because you now have two mappings for
the same foreign key column you must
define which one will be written to
the database (it must be the Basic
one), so the OneToOne or ManyToOne
foreign key must be defined to be
read-only. This is done through
setting the JoinColumn attributes
insertable and updatable to false,
or by using the
#PrimaryKeyJoinColumn instead of the
#JoinColumn.
A side effect of having two mappings
for the same column is that you now
have to keep the two in synch. This is
typically done through having the set
method for the OneToOne attribute
also set the Basic attribute value to
the target object's id. This can
become very complicated if the target
object's primary key is a
GeneratedValue, in this case you
must ensure that the target object's
id has been assigned before relating
the two objects.
(...)
Example ManyToOne id annotation
...
#Entity
#IdClass(PhonePK.class)
public class Phone {
#Id
#Column(name="OWNER_ID")
private long ownerId;
#Id
private String type;
#ManyToOne
#PrimaryKeyJoinColumn(name="OWNER_ID", referencedColumnName="EMP_ID")
private Employee owner;
...
public void setOwner(Employee owner) {
this.owner = owner;
this.ownerId = owner.getId();
}
...
}
This looks like to be what you're looking for (and maybe less complicated). I'd try to implement this solution (incrementally).