Saving nested objects with spring data - using an id as a reference - java

Say you are creating a new entity of the type User, User has the nested object Billing given that you know that a Billing exists with the ID 1, is there a simple way with which you can form an association between a new User and an existing Billing?
Assume that fetching a Billing object to set to the user is an expensive operation, therefore the solution of fetching the entire Billing object and setting it to the user is not an option.
My question is, Is there a short hand method of saving this relationship between an entity and its nested counterpart, using spring data?
#Entity
public class User {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private int userId;
private String name;
#ManyToOne
#JoinColumn(name = "billing_id")
private Billing userBill;
// constructor
// getters and setters
}
For example in sudo code:
User bob = new User();
bob.billingId.id = 1;
userRepository.save(bob);

Absolutely.
JpaRepository.getOne(id) (as opposed to CrudRepository.findById) will call EntityManager.getReference(entityType, id) internally, which is the method designated to handle this exact use case (getting the reference to an entity, without loading its associated state).
To answer your question, what you want is: customer.setBilling(billingRepository.getOne(billingId)).

Related

Data doesn't update immediately after changes in tables

I've recently tried to implement Spring Security into my web store project to distinguish between single users. Websites are working properly except there is one issue which I can't track to resolve. I have object called Customer within User class. Customer object has fields like id, balance, etc., and User has OneToOne relationship to Customer, so I can have single object for credentials and foreign key to specifics of user - his first name, last name, balance, owned products, etc.
I also have Product class which has ManyToOne relationship with Customer. It has its' own id, productCost, etc.
I'm using Spring MVC to take care of proper URL dispatching. When some action is taken, I'm using #AuthenticationPrincipal annotation to get currently logged Customer (through foreign key in User) and modify data regarding Customer linked with that foreign key.
When I modify Customer data through #AuthenticationPrincipal in controller, changes are immediate and they show up on website. But when I try to modify data through some DAO, for example by searching for Customer through id or try to get Customer that owns Product from Product getter (ManyToOne has reference to owning Customer), changes are not immediate. Database updates itself immediately and properly, like in first case, but collections in code and website state are not changed until I logout and login again - that's when data is updated. I suspect it may be due to fact that updating UserDetails updates data directly for currently logged user but then - how may I achieve same effect for Customer found by id?
Snippets of code:
Users.java:
#Entity
#Table(name="users")
public class Users {
#Id
#Column(name="username")
private String username;
#Column(name="password")
private String password;
#Column(name="enabled")
private boolean isActive;
#OneToMany(mappedBy="user")
private Set<Authorities> authorities;
#OneToOne
#JoinColumn(name="customer_id")
private Customer customer;
Product.java:
#Entity
#Table(name="product")
public class Product {
#Id
#GeneratedValue(strategy=GenerationType.IDENTITY)
#Column(name="id")
private int id;
#Column(name="name")
private String productName;
#Column(name="description")
private String productDescription;
#Column(name="category")
private String productCategory;
#Column(name="cost")
private int productCost;
#ManyToOne(fetch=FetchType.EAGER)
#JoinColumn(name="owner_id")
private Customer productOwner;
Customer.java:
#Entity
#Table(name="customer")
public class Customer {
//Class fields
#Id
#GeneratedValue(strategy=GenerationType.IDENTITY)
#Column(name="id")
private int id;
#Column(name="balance")
private int balance;
#Column(name="first_name")
private String firstName;
#Column(name="last_name")
private String lastName;
#Column(name="email")
private String email;
#OneToMany(mappedBy="productOwner", fetch=FetchType.EAGER)
private List<Product> ownedProducts;
Piece of controller code:
#Autowired
CustomerService customerService;
#Autowired
ProductService productService;
/*(...)*/
#GetMapping("/showOffer/{offerId}")
public String getOffer(#PathVariable int offerId, Model theModel, #AuthenticationPrincipal MyUserDetails user) {
Product retrievedProduct = productService.findById(offerId);
if (user.getCustomer().getBalance() >= retrievedProduct.getProductCost())
{
Customer retrievedProductOwner = retrievedProduct.getProductOwner();
/* This is where changes aren't applied immediately and I need to logout and login to process them. */
retrievedProductOwner.setBalance(1000);
/* This is where changes are immediately shown and Java collections are updated: */
user.getCustomer().setBalance(user.getCustomer().getBalance()-retrievedProduct.getProductCost());
/* Code below is an attempt to force immediate changes by updating collections directly from database - but that approach doesn't work */
productService.delete(retrievedProduct.getId());
retrievedProduct.getProductOwner().getOwnedProducts().clear();
retrievedProduct.getProductOwner().setOwnedProducts(productService.listOwnerProducts(retrievedProduct.getProductOwner()));
}
else {
System.out.println("Insufficient funds!");
}
return "redirect:/home";
TL:DR
I use UserDetails object in controller and I am also using DAO for Customer used as foreign key in UserDetails. Using UserDetails directly updates data and everything works fine, using DAO doesn't make changes until I logout and login.
as far as i understand your changes are only commited when you log out .
just try to synchronize and commit any modification at the right time and it would be safer that you manage sessions and transactions at the same time so you don't get any sort of incoherence when you do that. then tell me about the results .
Check whether CTRL+F5 in your browser (force cache clearance) updates your data similarly to logging out and back in. If so, it's a question of cached information. (this and (3) may occur at the same time)
Alternatively ... or perhaps complementarly ... your data fetch reqeust may be called before the database update/commit operation is completed. If so, it should become evident if you run distinct update and show routines. i.e. turn A into B, then into C, and you'd get something like B when you're expecting C... A instead of B... etc.
Lastly, depending on how you set up your back end, it is possible that you only populate whatever form you use for the front end exactly once, instead of dynamically querying the database whenever you access that form.

Get generated id for related JPA entity

In my project I have something like this:
#Entity
public class Refund {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#ManyToOne(fetch = FetchType.LAZY)
private Payment payment;
// ... other fields...
}
#Entity
public class Payment {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#OneToMany(orphanRemoval = true, cascade = CascadeType.ALL, mappedBy = "payment")
private List<Refund> refunds;
// ... other fields...
}
Payment is the aggregate root and when I need to add a new Refund I use this logic:
Payment payment = // fetch or create a Payment
Refund refund = // create + populate new Refund
refund.setPayment(payment);
payment.getRefunds().add(refund);
paymentRepository.save(payment); // <= here I'm using a Spring Data JPA repository
After saving payment I'd like to retrieve the id of the newly created Refund, but the id field is not populated in refund instance.
NOTES
The method implementing the save logic is annotated with #Transactional.
I also tried to use repo saveAndFlush method but nothing changes...
Is there a way to achieve this?
You need to make sure to update the payment object upon the call to the save method. What you need to do is to update to the following:
payment = paymentRepository.save(payment);
After this, you should be able to see the IDs set, etc.
The id is available on the value returned from save.
In your case, the merge operation is cascaded to the collection of Refunds, and the updated refund will be available in the collection from returned payment.
Having said that, if your goal is only to add a Refund to the DB, you may prefer to save the Refund instead - it is the owning side of the relation.
See javadoc for CrudRepository.save
<S extends T> 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.
paymentRepository.save(payment)
should return the saved instance payment.
From there you can access your refunds and get the id.

Can I use only Entity's id instead of fetching an entity from DB when there is a relationship in an Entity?

I am using Spring Data JPA with Hibernate.
Lets say I have the following entity defined:
#Entity
#Table(name = "foods")
public class Food {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "food_id")
private Long foodId;
#ManyToOne(fetch = FetchType.EAGER)
#JoinColumn(name = "food_type_id")
#NotNull
private FoodType foodType;
...
}
#Entity
#Table(name = "food_types")
public class FoodType {
public static final Integer PERISHABLE;
public static final Integer NON_PERISHABLE;
#Id
#Column(name = "food_type")
private Integer foodTypeId;
private String name;
...
}
Every time when I want to create a Food entity and save it to the database, currently code looks like this:
Food food = new Food();
FoodType foodType = foodTypeRepository.findById(FoodType.PERISHABLE); // Call to DB to get Entity
food.setFoodType(foodType);
....
foodRepository.save(food);
If we consider FoodType to be constant in the DB. Can I use it like this:
Food food = new Food();
FoodType foodType = new FoodType();
foodType.setFoodTypeId(FoodType.PERISHABLE); // No Call to DB
food.setFoodType(foodType);
....
foodRepository.save(food);
I have tested it and yes I can use it that way, hibernate will save the Food entity, but are there any downsides, pitfalls, etc... I am not seeing.
PS. This is just a simple example illustrating the idea, it is part of old legacy project which I cannot modify to remove constant from DB, and use an enum instead.
To avoid extra call to DB you should use:
FoodType foodType = foodTypeRepository.getOne(FoodType.PERISHABLE);
under the hood it calls EntityManager.getReference that obtain a reference to an entity without having to load its data as opposed to the foodTypeRepository.findById that lead to call EntityManager.find that obtain an entity along with its data.
See also this section of the hibernate documentation.
P.S. You can not use:
Food food = new Food();
FoodType foodType = new FoodType();
foodType.setFoodTypeId(FoodType.PERISHABLE);
as in this case hibernate consider foodType as a transient entity (not associated with a persistence context) and will try to save it as a new record if you have a proper cascading on your #ManyToOne association.
P.S.S. As it's mentioned in the documentation the method JpaRepository#getOne(ID) is deprecated and you should use JpaRepository#getById(ID) instead.
You do not need to fetch the entity associated with FoodType.PERISHABLE in order to set the relation on a Food entity to it and I'm not aware of any side effects or pitfalls of using FoodType.PERISHABLE directly as long it is a valid FoodType id.
As others mentioned, you could also use JpaRepository#getById(ID id) and that's probably the more canonical way of addressing this problem:
T getById(ID id) Returns a reference to the entity with the given
identifier. Depending on how the JPA persistence provider is
implemented this is very likely to always return an instance and throw
an EntityNotFoundException on first access. Some of them will reject
invalid identifiers immediately.

How to create an entity object in Spring Boot when the entire table is the key?

I'm creating an entity object that I will use to query a database with. However, I'm not sure how to build it since all of the columns are the key for the table. Typically I'd create an Entity object with the parameters that aren't the key and make a nested object within that one as the key. That entity object key would hold the values needed to build the key. But, in this case, all 5 values of the table are used for the key.
This is what I have currently, but I know that it's wrong.
Would I just have to make an #EmbeddedId key object within this one and push all of the values to that while practically leaving this base entity object empty? Or is there a better way to do this?
#Entity
#Table(name = "<insert>", schema="<insert>")
public class Market {
#Id
#Column(name="MKT_ID")
private String marketId;
#Column(name="DLR_CD")
private String dealerCd;
#Column(name="SER_NO_PFX")
private String serialNoPrefix;
#Column(name="SER_NO_BDY")
private String serialNoBody;
#Column(name="USER_ID")
private String userId;
}

Using primary key of one table as foreign key to another table [duplicate]

I receive following error when I save the object using Hibernate
object references an unsaved transient instance - save the transient instance before flushing
You should include cascade="all" (if using xml) or cascade=CascadeType.ALL (if using annotations) on your collection mapping.
This happens because you have a collection in your entity, and that collection has one or more items which are not present in the database. By specifying the above options you tell hibernate to save them to the database when saving their parent.
I believe this might be just repeat answer, but just to clarify, I got this on a #OneToOne mapping as well as a #OneToMany. In both cases, it was the fact that the Child object I was adding to the Parent wasn't saved in the database yet. So when I added the Child to the Parent, then saved the Parent, Hibernate would toss the "object references an unsaved transient instance - save the transient instance before flushing" message when saving the Parent.
Adding in the cascade = {CascadeType.ALL} on the Parent's reference to the Child solved the problem in both cases. This saved the Child and the Parent.
Sorry for any repeat answers, just wanted to further clarify for folks.
#OneToOne(cascade = {CascadeType.ALL})
#JoinColumn(name = "performancelog_id")
public PerformanceLog getPerformanceLog() {
return performanceLog;
}
Introduction
When using JPA and Hibernate, an entity can be in one of the following 4 states:
New - A newly created object that hasn’t ever been associated with a Hibernate Session (a.k.a Persistence Context) and is not mapped to any database table row is considered to be in the New or Transient state.
To become persisted we need to either explicitly call the persist method or make use of the transitive persistence mechanism.
Persistent - A persistent entity has been associated with a database table row and it’s being managed by the currently running Persistence Context.
Any change made to such an entity is going to be detected and propagated to the database (during the Session flush-time).
Detached - Once the currently running Persistence Context is closed all the previously managed entities become detached. Successive changes will no longer be tracked and no automatic database synchronization is going to happen.
Removed - Although JPA demands that managed entities only are allowed to be removed, Hibernate can also delete detached entities (but only through a remove method call).
Entity state transitions
To move an entity from one state to the other, you can use the persist, remove or merge methods.
Fixing the problem
The issue you are describing in your question:
object references an unsaved transient instance - save the transient instance before flushing
is caused by associating an entity in the state of New to an entity that's in the state of Managed.
This can happen when you are associating a child entity to a one-to-many collection in the parent entity, and the collection does not cascade the entity state transitions.
So, you can fix this by adding cascade to the entity association that triggered this failure, as follows:
The #OneToOne association
#OneToOne(
mappedBy = "post",
orphanRemoval = true,
cascade = CascadeType.ALL)
private PostDetails details;
Notice the CascadeType.ALL value we added for the cascade attribute.
The #OneToMany association
#OneToMany(
mappedBy = "post",
orphanRemoval = true,
cascade = CascadeType.ALL)
private List<Comment> comments = new ArrayList<>();
Again, the CascadeType.ALL is suitable for the bidirectional #OneToMany associations.
Now, in order for the cascade to work properly in a bidirectional, you also need to make sure that the parent and child associations are in sync.
The #ManyToMany association
#ManyToMany(
mappedBy = "authors",
cascade = {
CascadeType.PERSIST,
CascadeType.MERGE
}
)
private List<Book> books = new ArrayList<>();
In a #ManyToMany association, you cannot use CascadeType.ALL or orphanRemoval as this will propagate the delete entity state transition from one parent to another parent entity.
Therefore, for #ManyToMany associations, you usually cascade the CascadeType.PERSIST or CascadeType.MERGE operations. Alternatively, you can expand that to DETACH or REFRESH.
This happens when saving an object when Hibernate thinks it needs to save an object that is associated with the one you are saving.
I had this problem and did not want to save changes to the referenced object so I wanted the cascade type to be NONE.
The trick is to ensure that the ID and VERSION in the referenced object is set so that Hibernate does not think that the referenced object is a new object that needs saving. This worked for me.
Look through all of the relationships in the class you are saving to work out the associated objects (and the associated objects of the associated objects) and ensure that the ID and VERSION is set in all objects of the object tree.
Or, if you want to use minimal "powers" (e.g. if you don't want a cascade delete) to achieve what you want, use
import org.hibernate.annotations.Cascade;
import org.hibernate.annotations.CascadeType;
...
#Cascade({CascadeType.SAVE_UPDATE})
private Set<Child> children;
In my case it was caused by not having CascadeType on the #ManyToOne side of the bidirectional relationship. To be more precise, I had CascadeType.ALL on #OneToMany side and did not have it on #ManyToOne. Adding CascadeType.ALL to #ManyToOne resolved the issue.
One-to-many side:
#OneToMany(cascade = CascadeType.ALL, mappedBy="globalConfig", orphanRemoval = true)
private Set<GlobalConfigScope>gcScopeSet;
Many-to-one side (caused the problem)
#ManyToOne
#JoinColumn(name="global_config_id")
private GlobalConfig globalConfig;
Many-to-one (fixed by adding CascadeType.PERSIST)
#ManyToOne(cascade = CascadeType.PERSIST)
#JoinColumn(name="global_config_id")
private GlobalConfig globalConfig;
This occurred for me when persisting an entity in which the existing record in the database had a NULL value for the field annotated with #Version (for optimistic locking). Updating the NULL value to 0 in the database corrected this.
This isn't the only reason for the error. I encountered it just now for a typo error in my coding, which I believe, set a value of an entity which was already saved.
X x2 = new X();
x.setXid(memberid); // Error happened here - x was a previous global entity I created earlier
Y.setX(x2);
I spotted the error by finding exactly which variable caused the error (in this case String xid). I used a catch around the whole block of code that saved the entity and printed the traces.
{
code block that performed the operation
} catch (Exception e) {
e.printStackTrace(); // put a break-point here and inspect the 'e'
return ERROR;
}
Don't use Cascade.All until you really have to. Role and Permission have bidirectional manyToMany relation. Then the following code would work fine
Permission p = new Permission();
p.setName("help");
Permission p2 = new Permission();
p2.setName("self_info");
p = (Permission)crudRepository.save(p); // returned p has id filled in.
p2 = (Permission)crudRepository.save(p2); // so does p2.
Role role = new Role();
role.setAvailable(true);
role.setDescription("a test role");
role.setRole("admin");
List<Permission> pList = new ArrayList<Permission>();
pList.add(p);
pList.add(p2);
role.setPermissions(pList);
crudRepository.save(role);
while if the object is just a "new" one, then it would throw the same error.
beside all other good answers, this could happen if you use merge to persist an object and accidentally forget to use merged reference of the object in the parent class. consider the following example
merge(A);
B.setA(A);
persist(B);
In this case, you merge A but forget to use merged object of A. to solve the problem you must rewrite the code like this.
A=merge(A);//difference is here
B.setA(A);
persist(B);
If your collection is nullable just try: object.SetYouColection(null);
This issue happened to me when I created a new entity and an associated entity in a method marked as #Transactional, then performed a query before saving. Ex
#Transactional
public someService() {
Entity someEntity = new Entity();
AssocaiatedEntity associatedEntity = new AssocaitedEntity();
someEntity.setAssociatedEntity(associatedEntity);
associatedEntity.setEntity(someEntity);
// Performing any query was causing hibernate to attempt to persist the new entity. It would then throw an exception
someDao.getSomething();
entityDao.create(someEntity);
}
To fix, I performed the query before creating the new entity.
To add my 2 cents, I got this same issue when I m accidentally sending null as the ID. Below code depicts my scenario (and OP didn't mention any specific scenario).
Employee emp = new Employee();
emp.setDept(new Dept(deptId)); // --> when deptId PKID is null, same error will be thrown
// calls to other setters...
em.persist(emp);
Here I m setting the existing department id to a new employee instance without actually getting the department entity first, as I don't want to another select query to fire.
In some scenarios, deptId PKID is coming as null from calling method and I m getting the same error.
So, watch for null values for PK ID
It can also happen when you are having OneToMany relation and you try to add the child entity to the list in parent entity, then retrieve this list through parent entity (before saving this parent entity), without saving child entity itself, e.g.:
Child childEntity = new Child();
parentEntity.addChild(childEntity);
parentEntity.getChildren(); // I needed the retrieval for logging, but one may need it for other reasons.
parentRepository.save(parentEntity);
The error was thrown when I saved the parent entity. If I removed the retrieval in the previous row, then the error was not thrown, but of course that's not the solution.
The solution was saving the childEntity and adding that saved child entity to the parent entity, like this:
Child childEntity = new Child();
Child savedChildEntity = childRepository.save(childEntity);
parentEntity.addChild(savedChildEntity);
parentEntity.getChildren();
parentRepository.save(parentEntity);
If you're using Spring Data JPA then addition #Transactional annotation to your service implementation would solve the issue.
I also faced the same situation. By setting following annotation above the property made it solve the exception prompted.
The Exception I faced.
Exception in thread "main" java.lang.IllegalStateException: org.hibernate.TransientObjectException: object references an unsaved transient instance - save the transient instance before flushing: com.model.Car_OneToMany
To overcome, the annotation I used.
#OneToMany(cascade = {CascadeType.ALL})
#Column(name = "ListOfCarsDrivenByDriver")
private List<Car_OneToMany> listOfCarsBeingDriven = new ArrayList<Car_OneToMany>();
What made Hibernate throw the exception:
This exception is thrown at your console because the child object I attach to the parent object is not present in the database at that moment.
By providing #OneToMany(cascade = {CascadeType.ALL}) , it tells Hibernate to save them to the database while saving the parent object.
i get this error when i use
getSession().save(object)
but it works with no problem when I use
getSession().saveOrUpdate(object)
For the sake of completeness: A
org.hibernate.TransientPropertyValueException
with message
object references an unsaved transient instance - save the transient instance before flushing
will also occur when you try to persist / merge an entity with a reference to another entity which happens to be detached.
One other possible reason: in my case, I was attempting to save the child before saving the parent, on a brand new entity.
The code was something like this in a User.java model:
this.lastName = lastName;
this.isAdmin = isAdmin;
this.accountStatus = "Active";
this.setNewPassword(password);
this.timeJoin = new Date();
create();
The setNewPassword() method creates a PasswordHistory record and adds it to the history collection in User. Since the create() statement hadn't been executed yet for the parent, it was trying to save to a collection of an entity that hadn't yet been created. All I had to do to fix it was to move the setNewPassword() call after the call to create().
this.lastName = lastName;
this.isAdmin = isAdmin;
this.accountStatus = "Active";
this.timeJoin = new Date();
create();
this.setNewPassword(password);
There is another possibility that can cause this error in hibernate. You may set an unsaved reference of your object A to an attached entity B and want to persist object C. Even in this case, you will get the aforementioned error.
There are so many possibilities of this error some other possibilities are also on add page or edit page. In my case I was trying to save a object AdvanceSalary. The problem is that in edit the AdvanceSalary employee.employee_id is null Because on edit I was not set the employee.employee_id. I have make a hidden field and set it. my code working absolutely fine.
#Entity(name = "ic_advance_salary")
#Table(name = "ic_advance_salary")
public class AdvanceSalary extends BaseDO{
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id")
private Integer id;
#ManyToOne(fetch = FetchType.EAGER)
#JoinColumn(name = "employee_id", nullable = false)
private Employee employee;
#Column(name = "employee_id", insertable=false, updatable=false)
#NotNull(message="Please enter employee Id")
private Long employee_id;
#Column(name = "advance_date")
#DateTimeFormat(pattern = "dd-MMM-yyyy")
#NotNull(message="Please enter advance date")
private Date advance_date;
#Column(name = "amount")
#NotNull(message="Please enter Paid Amount")
private Double amount;
#Column(name = "cheque_date")
#DateTimeFormat(pattern = "dd-MMM-yyyy")
private Date cheque_date;
#Column(name = "cheque_no")
private String cheque_no;
#Column(name = "remarks")
private String remarks;
public AdvanceSalary() {
}
public AdvanceSalary(Integer advance_salary_id) {
this.id = advance_salary_id;
}
public Integer getId() {
return id;
}
public void setId(Integer id) {
this.id = id;
}
public Employee getEmployee() {
return employee;
}
public void setEmployee(Employee employee) {
this.employee = employee;
}
public Long getEmployee_id() {
return employee_id;
}
public void setEmployee_id(Long employee_id) {
this.employee_id = employee_id;
}
}
I think is because you have try to persist an object that have a reference to another object that is not persist yet, and so it try in the "DB side" to put a reference to a row that not exists
Case 1:
I was getting this exception when I was trying to create a parent and saving that parent reference to its child and then some other DELETE/UPDATE query(JPQL). So I just flush() the newly created entity after creating parent and after creating child using same parent reference. It Worked for me.
Case 2:
Parent class
public class Reference implements Serializable {
#Id
#Column(precision=20, scale=0)
private BigInteger id;
#Temporal(TemporalType.TIMESTAMP)
private Date modifiedOn;
#OneToOne(mappedBy="reference")
private ReferenceAdditionalDetails refAddDetails;
.
.
.
}
Child Class:
public class ReferenceAdditionalDetails implements Serializable{
private static final long serialVersionUID = 1L;
#Id
#OneToOne
#JoinColumn(name="reference",referencedColumnName="id")
private Reference reference;
private String preferedSector1;
private String preferedSector2;
.
.
}
In the above case where parent(Reference) and child(ReferenceAdditionalDetails) having OneToOne relationship and when you try to create Reference entity and then its child(ReferenceAdditionalDetails), it will give you the same exception. So to avoid the exception you have to set null for child class and then create the parent.(Sample Code)
.
.
reference.setRefAddDetails(null);
reference = referenceDao.create(reference);
entityManager.flush();
.
.
In my case , issue was completely different. I have two classes let's say c1 & c2. Between C1 & C2 dependency is OneToMany. Now if i am saving C1 in DB it was throwing above error.
Resolution of this problem was to get first C2's id from consumer request and find C2 via repository call.Afterwards save c2 into C1 object .Now if i am saving C1, it's working fine.
I was facing the same error for all PUT HTTP transactions, after introducing optimistic locking (#Version)
At the time of updating an entity it is mandatory to send id and version of that entity. If any of the entity fields are related to other entities then for that field also we should provide id and version values, without that the JPA try to persist that related entity first as a new entity
Example: we have two entities --> Vehicle(id,Car,version) ; Car(id, version, brand); to update/persist Vehicle entity make sure the Car field in vehicle entity has id and version fields provided
Simple way of solving this issue is save the both entity.
first save the child entity and then save the parent entity.
Because parent entity is depend on child entity for the foreign key value.
Below simple exam of one to one relationship
insert into Department (name, numOfemp, Depno) values (?, ?, ?)
Hibernate: insert into Employee (SSN, dep_Depno, firstName, lastName, middleName, empno) values (?, ?, ?, ?, ?, ?)
Session session=sf.openSession();
session.beginTransaction();
session.save(dep);
session.save(emp);
One possible cause of the error is the inexistence of the setting of the value of the parent entity ; for example for a department-employees relationship you have to write this in order to fix the error :
Department dept = (Department)session.load(Department.class, dept_code); // dept_code is from the jsp form which you get in the controller with #RequestParam String department
employee.setDepartment(dept);
I faced this exception when I did not persist parent object but I was saving the child. To resolve the issue, with in the same session I persisted both the child and parent objects and used CascadeType.ALL on the parent.
My problem was related to #BeforeEach of JUnit. And even if I saved the related entities (in my case #ManyToOne), I got the same error.
The problem is somehow related to the sequence that I have in my parent.
If I assign the value to that attribute, the problem is solved.
Ex.
If I have the entity Question that can have some categories (one or more) and entity Question has a sequence:
#GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "feedbackSeq")
#Id
private Long id;
I have to assign the value question.setId(1L);
Just make Constructor of your mapping in your base class.
Like if you want One-To-One relation in Entity A, Entity B.
if your are taking A as base class, then A must have a Constructor have B as a argument.

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