spring / hibernate: Generate UUID automatically for Id column - java

I am trying to persist a simple class using Spring, with hibernate/JPA and a PostgreSQL database.
The ID column of the table is a UUID, which I want to generate in code, not in the database.
This should be straightforward since hibernate and postgres have good support for UUIDs.
Each time I create a new instance and write it with save(), I get the following error:
o.h.j.JdbcSQLIntegrityConstraintViolationException: NULL not allowed for column "ID"; SQL statement: INSERT INTO DOODAHS (fieldA, fieldB) VALUES $1, $2) ...
This error indicates that it's expecting the ID column to be auto-populated (with some default value) when a row is inserted.
The class looks like this:
#lombok.Data
#lombok.AllArgsConstructor
#org.springframework.data.relational.core.mapping.Table("doodahs")
public class Doodah {
#org.springframework.data.annotation.Id
#javax.persistence.GeneratedValue(generator = "UUID")
#org.hibernate.annotations.GenericGenerator(name="UUID", strategy = "uuid2")
#javax.persistence.Column(nullable = false, unique = true)
private UUID id;
//... other fields
Things I have tried:
Annotate the field with #javax.persistence.Id (in addition to existing spring Id)
Annotate the field with #org.hibernate.annotations.Type(type = "pg-uuid")
Create the UUID myself - results in Spring complaining that it can't find the row with that id.
Specify strategy = "org.hibernate.id.UUIDGenerator"
Annotate class with #Entity
Replace spring #Id annotation with #javax.persistence.Id
I've seen useful answers here, here and here but none have worked so far.
NB the persistence is being handled by a class which looks like this:
#org.springframework.stereotype.Repository
public interface DoodahRepository extends CrudRepository<Doodah, UUID> ;
The DDL for the table is like this:
CREATE TABLE DOODAHS(id UUID not null, fieldA VARCHAR(10), fieldB VARCHAR(10));
Update
Thanks to Sve Kamenska, with whose help I finally got it working eventually. I ditched the JPA approach - and note that we are using R2DBC, not JDBC, so the answer didn't work straight away. Several sources (here, here, here, here, here and here) indicate that there is no auto Id generation for R2DBC. So you have to add a callback Bean to set your Id manually.
I updated the class as follows:
#Table("doodahs")
public class Doodah {
#org.springframework.data.annotation.Id
private UUID id;
I also added a Bean as follows:
#Bean
BeforeConvertCallback<Doodah> beforeConvertCallback() {
return (d, row, table) -> {
if (d.getId() == null){
d.id = UUID.randomUUID();
}
return Mono.just(d);
};
}
When a new object (with id = null, and isNew = true) is passed to the save() method, the callback method is invoked, and it sets the id.
Initially I tried using BeforeSaveCallback but it was being called too late in the process, resulting in the following exception:
JdbcSQLIntegrityConstraintViolationException: NULL not allowed for column "ID"....

Update
There are, at least, 2 types of Spring Data: JPA and JDBC.
The issue happens because you are mixing the 2 of them.
So, in order to fix, there are 2 solutions.
Solution 1 - Use Spring Data JDBC only.
Pom.xml dependency
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-data-jdbc</artifactId>
</dependency>
Generate ID.
Spring Data JDBC assumes that ID is generated on database level (like we already figured that out from log). If you try to save an entity with pre-defined id, Spring will assume that it is existing entity and will try to find it in the database and update. That is why you got this error in your attempt #3.
In order to generate UUID, you can:
Leave it to DB (it looks like Postgre allows to do it)
or Fill it in BeforeSaveCallback (more details here https://spring.io/blog/2021/09/09/spring-data-jdbc-how-to-use-custom-id-generation)
#Bean BeforeSaveCallback<Doodah> beforeSaveCallback() {
return (doodah, mutableAggregateChange) -> {
if (doodah.id == null) {
doodah.id = UUID.randomUUID();
}
return doodah;
};
}
Solution 2 - Use Spring Data JPA only
Pom.xml dependency
<dependency>
<groupId>org.springframework.boot</groupId>
<artifactId>spring-boot-starter-data-jpa</artifactId>
</dependency>
Generate ID.
Here you can, actually, use the approach with the UUID auto-generation, like you wanted to do initially
Use javax.persistence #Entity annotation instead of springdata #Table on the class-level
and Use #javax.persistence.Id and #javax.persistence.GeneratedValue with all defaults on id-field.
#javax.persistence.Id
#javax.persistence.GeneratedValue
private UUID id;
Other notes:
Specification of generator and strategy is not required, since it will generate based on the type of the id field (UUID in this case).
Specification of Column(nullable = false, unique = true) is not required either, since putting #Id annotation already assumes these constraints.
Initial answer before update
The main question: how do you save the entity? As id-generation is handled by JPA provider, Hibernate in this case. It is done during save method of em or repository. In order to create entities and ids Hibernate is looking for javax.persistence annotations, while you have Spring-specific, so I am wandering how do you save them.
And another question here: the error you provided INSERT INTO DOODAHS (fieldA, fieldB) VALUES $1, $2 shows that there is no id field in the insert-query at all. Did you just simplified the error-message and removed ID from it? Or this is original error and your code does not even "see" field ID? In that case the issue in not related to the id-generation, but rather is related to the question why your code does not see this field.

Related

org.springframework.orm.jpa.JpaSystemException: identifier of an instance of com.cc.domain.User was altered from 90 to null; [duplicate]

org.hibernate.HibernateException: identifier of an instance
of org.cometd.hibernate.User altered from 12 to 3
in fact, my user table is really must dynamically change its value, my Java app is multithreaded.
Any ideas how to fix it?
Are you changing the primary key value of a User object somewhere? You shouldn't do that. Check that your mapping for the primary key is correct.
What does your mapping XML file or mapping annotations look like?
You must detach your entity from session before modifying its ID fields
In my case, the PK Field in hbm.xml was of type "integer" but in bean code it was long.
In my case getters and setter names were different from Variable name.
private Long stockId;
public Long getStockID() {
return stockId;
}
public void setStockID(Long stockID) {
this.stockId = stockID;
}
where it should be
public Long getStockId() {
return stockId;
}
public void setStockId(Long stockID) {
this.stockId = stockID;
}
In my case, I solved it changing the #Id field type from long to Long.
In my particular case, this was caused by a method in my service implementation that needed the spring #Transactional(readOnly = true) annotation. Once I added that, the issue was resolved. Unusual though, it was just a select statement.
Make sure you aren't trying to use the same User object more than once while changing the ID. In other words, if you were doing something in a batch type operation:
User user = new User(); // Using the same one over and over, won't work
List<Customer> customers = fetchCustomersFromSomeService();
for(Customer customer : customers) {
// User user = new User(); <-- This would work, you get a new one each time
user.setId(customer.getId());
user.setName(customer.getName());
saveUserToDB(user);
}
In my case, a template had a typo so instead of checking for equivalency (==) it was using an assignment equals (=).
So I changed the template logic from:
if (user1.id = user2.id) ...
to
if (user1.id == user2.id) ...
and now everything is fine. So, check your views as well!
It is a problem in your update method. Just instance new User before you save changes and you will be fine. If you use mapping between DTO and Entity class, than do this before mapping.
I had this error also. I had User Object, trying to change his Location, Location was FK in User table. I solved this problem with
#Transactional
public void update(User input) throws Exception {
User userDB = userRepository.findById(input.getUserId()).orElse(null);
userDB.setLocation(new Location());
userMapper.updateEntityFromDto(input, userDB);
User user= userRepository.save(userDB);
}
Also ran into this error message, but the root cause was of a different flavor from those referenced in the other answers here.
Generic answer:
Make sure that once hibernate loads an entity, no code changes the primary key value in that object in any way. When hibernate flushes all changes back to the database, it throws this exception because the primary key changed. If you don't do it explicitly, look for places where this may happen unintentionally, perhaps on related entities that only have LAZY loading configured.
In my case, I am using a mapping framework (MapStruct) to update an entity. In the process, also other referenced entities were being updates as mapping frameworks tend to do that by default. I was later replacing the original entity with new one (in DB terms, changed the value of the foreign key to reference a different row in the related table), the primary key of the previously-referenced entity was already updated, and hibernate attempted to persist this update on flush.
I was facing this issue, too.
The target table is a relation table, wiring two IDs from different tables. I have a UNIQUE constraint on the value combination, replacing the PK.
When updating one of the values of a tuple, this error occured.
This is how the table looks like (MySQL):
CREATE TABLE my_relation_table (
mrt_left_id BIGINT NOT NULL,
mrt_right_id BIGINT NOT NULL,
UNIQUE KEY uix_my_relation_table (mrt_left_id, mrt_right_id),
FOREIGN KEY (mrt_left_id)
REFERENCES left_table(lef_id),
FOREIGN KEY (mrt_right_id)
REFERENCES right_table(rig_id)
);
The Entity class for the RelationWithUnique entity looks basically like this:
#Entity
#IdClass(RelationWithUnique.class)
#Table(name = "my_relation_table")
public class RelationWithUnique implements Serializable {
...
#Id
#ManyToOne
#JoinColumn(name = "mrt_left_id", referencedColumnName = "left_table.lef_id")
private LeftTableEntity leftId;
#Id
#ManyToOne
#JoinColumn(name = "mrt_right_id", referencedColumnName = "right_table.rig_id")
private RightTableEntity rightId;
...
I fixed it by
// usually, we need to detach the object as we are updating the PK
// (rightId being part of the UNIQUE constraint) => PK
// but this would produce a duplicate entry,
// therefore, we simply delete the old tuple and add the new one
final RelationWithUnique newRelation = new RelationWithUnique();
newRelation.setLeftId(oldRelation.getLeftId());
newRelation.setRightId(rightId); // here, the value is updated actually
entityManager.remove(oldRelation);
entityManager.persist(newRelation);
Thanks a lot for the hint of the PK, I just missed it.
Problem can be also in different types of object's PK ("User" in your case) and type you ask hibernate to get session.get(type, id);.
In my case error was identifier of an instance of <skipped> was altered from 16 to 32.
Object's PK type was Integer, hibernate was asked for Long type.
In my case it was because the property was long on object but int in the mapping xml, this exception should be clearer
If you are using Spring MVC or Spring Boot try to avoid:
#ModelAttribute("user") in one controoler, and in other controller
model.addAttribute("user", userRepository.findOne(someId);
This situation can produce such error.
This is an old question, but I'm going to add the fix for my particular issue (Spring Boot, JPA using Hibernate, SQL Server 2014) since it doesn't exactly match the other answers included here:
I had a foreign key, e.g. my_id = '12345', but the value in the referenced column was my_id = '12345 '. It had an extra space at the end which hibernate didn't like. I removed the space, fixed the part of my code that was allowing this extra space, and everything works fine.
Faced the same Issue.
I had an assosciation between 2 beans. In bean A I had defined the variable type as Integer and in bean B I had defined the same variable as Long.
I changed both of them to Integer. This solved my issue.
I solve this by instancing a new instance of depending Object. For an example
instanceA.setInstanceB(new InstanceB());
instanceA.setInstanceB(YOUR NEW VALUE);
In my case I had a primary key in the database that had an accent, but in other table its foreign key didn't have. For some reason, MySQL allowed this.
It looks like you have changed identifier of an instance
of org.cometd.hibernate.User object menaged by JPA entity context.
In this case create the new User entity object with appropriate id. And set it instead of the original User object.
Did you using multiple Transaction managers from the same service class.
Like, if your project has two or more transaction configurations.
If true,
then at first separate them.
I got the issue when i tried fetching an existing DB entity, modified few fields and executed
session.save(entity)
instead of
session.merge(entity)
Since it is existing in the DB, when we should merge() instead of save()
you may be modified primary key of fetched entity and then trying to save with a same transaction to create new record from existing.

How to save entities with manually assigned identifiers using Spring Data JPA?

I'm updating an existing code that handles the copy or raw data from one table into multiple objects within the same database.
Previously, every kind of object had a generated PK using a sequence for each table.
Something like that :
#Id
#Column(name = "id")
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
In order to reuse existing IDs from the import table, we removed GeneratedValue for some entities, like that :
#Id
#Column(name = "id")
private Integer id;
For this entity, I did not change my JpaRepository, looking like this :
public interface EntityRepository extends JpaRepository<Entity, Integer> {
<S extends Entity> S save(S entity);
}
Now I'm struggling to understand the following behaviour, within a spring transaction (#Transactional) with the default propagation and isolation level :
With the #GeneratedValue on the entity, when I call entityRepository.save(entity) I can see with Hibernate show sql activated that an insert request is fired (however seems to be only in the cache since the database does not change)
Without the #GeneratedValue on the entity, only a select request is fired (no insert attempt)
This is a big issue when my Entity (without generated value) is mapped to MyOtherEntity (with generated value) in a one or many relationship.
I thus have the following error :
ERROR: insert or update on table "t_other_entity" violates foreign key constraint "other_entity_entity"
Détail : Key (entity_id)=(110) is not present in table "t_entity"
Seems legit since the insert has not been sent for Entity, but why ? Again, if I change the ID of the Entity and use #GeneratedValue I don't get any error.
I'm using Spring Boot 1.5.12, Java 8 and PostgreSQL 9
You're basically switching from automatically assigned identifiers to manually defined ones which has a couple of consequences both on the JPA and Spring Data level.
Database operation timing
On the plain JPA level, the persistence provider doesn't necessarily need to immediately execute a single insert as it doesn't have to obtain an identifier value. That's why it usually delays the execution of the statement until it needs to flush, which is on either an explicit call to EntityManager.flush(), a query execution as that requires the data in the database to be up to date to deliver correct results or transaction commit.
Spring Data JPA repositories automatically use default transactions on the call to save(…). However, if you're calling repositories within a method annotated with #Transactional in turn, the databse interaction might not occur until that method is left.
EntityManager.persist(…) VS. ….merge(…)
JPA requires the EntityManager client code to differentiate between persisting a completely new entity or applying changes to an existing one. Spring Data repositories w ant to free the client code from having to deal with this distinction as business code shouldn't be overloaded with that implementation detail. That means, Spring Data will somehow have to differentiate new entities from existing ones itself. The various strategies are described in the reference documentation.
In case of manually identifiers the default of inspecting the identifier property for null values will not work as the property will never be null by definition. A standard pattern is to tweak the entities to implement Persistable and keep a transient is-new-flag around and use entity callback annotations to flip the flag.
#MappedSuperclass
public abstract class AbstractEntity<ID extends SalespointIdentifier> implements Persistable<ID> {
private #Transient boolean isNew = true;
#Override
public boolean isNew() {
return isNew;
}
#PrePersist
#PostLoad
void markNotNew() {
this.isNew = false;
}
// More code…
}
isNew is declared transient so that it doesn't get persisted. The type implements Persistable so that the Spring Data JPA implementation of the repository's save(…) method will use that. The code above results in entities created from user code using new having the flag set to true, but any kind of database interaction (saving or loading) turning the entity into a existing one, so that save(…) will trigger EntityManager.persist(…) initially but ….merge(…) for all subsequent operations.
I took the chance to create DATAJPA-1600 and added a summary of this description to the reference docs.

How do I stop spring data JPA from doing a SELECT before a save()?

We are writing a new app against an existing database. I'm using Spring Data JPA, and simply doing a
MyRepository.save()
on my new entity, using
MyRepository extends CrudRepository<MyThing, String>
I've noticed in the logs that hibernate is doing a Select before the insert, and that they are taking a long time, even when using the indexes.
I've searched for this here, and the answers I've found usually are related to Hibernate specifically. I'm pretty new to JPA and it seems like JPA and Hibernate are pretty closely intertwined, at least when using it within the context of Spring Data. The linked answers suggest using Hibernate persist(), or somehow using a session, possibly from an entityManager? I haven't had to do anything with sessions or entityManagers, or any Hibernate API directly. So far I've gotten simple inserts done with save() and a couple #Query in my Repositories.
Here is the code of Spring SimpleJpaRepository you are using by using Spring Data repository:
#Transactional
public <S extends T> S save(S entity) {
if (entityInformation.isNew(entity)) {
em.persist(entity);
return entity;
} else {
return em.merge(entity);
}
}
It does the following:
By default Spring Data JPA inspects the identifier property of the given entity. If the identifier property is null, then the entity will be assumed as new, otherwise as not new.
Link to Spring Data documentation
And so if one of your entity has an ID field not null, Spring will make Hibernate do an update (and so a SELECT before).
You can override this behavior by the 2 ways listed in the same documentation. An easy way is to make your Entity implements Persistable (instead of Serializable), which will make you implement the method "isNew".
If you provide your own id value then Spring Data will assume that you need to check the DB for a duplicate key (hence the select+insert).
Better practice is to use an id generator, like this:
#Entity
public class MyThing {
#Id
#GeneratedValue(generator = "uuid2")
#GenericGenerator(name = "uuid2", strategy = "uuid2")
private UUID id;
}
If you really must insert your own id and want to prevent the select+insert then implement Persistable, e.g.
#Entity
public class MyThing implements Persistable<UUID> {
#Id
private UUID id;
#Override
public UUID getId() {
return id;
}
//prevent Spring Data doing a select-before-insert - this particular entity is never updated
#Override
public boolean isNew() {
return true;
}
}
I created a custom method in the #Repository:
public void persistAll(Iterable<MyThing> toPersist) {
toPersist.forEach(thing -> entityManager.persist(thing));
}
If you provide your own ID value then Spring Data will assume that you need to check the DB for a duplicate key (hence the select+insert).
One option is to use a separate autogenerated ID column as Primary key but this option seems redundant. Because if you already have a Business/Natural ID that is unique then it is easier to make this as the #ID column instead of having a separate ID column.
So how to solve the problem?
The solution is to use #javax.persistence.Version on a new versionNumber column in all the tables. If you have a parent and child table then use #Version column in all the entity classes.
Add a column in the Entity class like this:
#javax.persistence.Version
#Column(name = "data_version")
private Long dataVersion;
add column in SQL file:
"data_version" INTEGER DEFAULT 0
Then I see that Spring data does not do Select before doing Insert.

#GeneratedValue polymorphic abstract superclass over MySQL

In a Spring MVC application using Hibernate and MySQL, I have an abstract superclass BaseEntity that manages the values of the IDs for all the other entities in the model. The id field uses #GeneratedValue. I am encountering a problem whenever my code tries to save any of the subclasses that extend BaseEntity. The problem comes with the choice of GenerationType for the #GeneratedValue.
At every place in my code where a subclass of BaseEntity tries to save to the underlying MySQL database, I get the following error:
ERROR SqlExceptionHelper - Table 'docbd.hibernate_sequences' doesn't exist
I have read many postings about this on SO and on google, but they either deal with other databases (not MySQL) or they do not deal with abstract superclasses. I cannot solve the problem by using GenerationType.IDENTITY because I am using an abstract superclass to manage id fields for all entities in the model. Similarly, I cannot use GenerationType.SEQUENCE because MySQL does not support sequences.
So how do I solve this problem?
Here is the code for BaseEntity.java:
#Entity
#Inheritance(strategy = InheritanceType.TABLE_PER_CLASS)
public abstract class BaseEntity {
#Id
#GeneratedValue(strategy = GenerationType.TABLE)
protected Integer id;
public void setId(Integer id) {this.id = id;}
public Integer getId() {return id;}
public boolean isNew() {return (this.id == null);}
}
Here is an example of the code for one of the entities that extends BaseEntity:
#Entity
#Table(name = "ccd")
public class CCD extends BaseEntity{
//other stuff
}
Here is the DDL:
CREATE TABLE IF NOT EXISTS ccd(
id int(11) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY,
#other stuff
)engine=InnoDB;SHOW WARNINGS;
Here is the JPQL code in the DAO:
#Override
#Transactional
public void saveCCD(CCD ccd) {
if (ccd.getId() == null) {
System.out.println("[[[[[[[[[[[[ about to persist CCD ]]]]]]]]]]]]]]]]]]]]");
this.em.persist(ccd);
this.em.flush();
}
else {
System.out.println("]]]]]]]]]]]]]]]]]] about to merge CCD [[[[[[[[[[[[[[[[[[[[[");
this.em.merge(ccd);
this.em.flush();
}
}
EDIT:
The reason I cannot use #MappedSuperClass in this situation is that I need to have ManyToOne relationships that allow for multiple subtypes to be used interchangeably. Look at the AccessLog class below as an example. It has an actor_entity and a target_entity. There can be many types of actor entities and many types of target entities, but they all inherit from BaseEntity. This inheritance enables the underlying accesslogs data table in MySQL to just have one actor_entity_id field and just one target_entity_id field instead of having to have several fields for each. When I change #Entity above BaseEntity to #MappedSuperClass, a different error gets thrown indicating that AccessLog cannot find BaseEntity. BaseEntity needs #Entity annotation in order for AccessLog to have polymorphic properties.
#Entity
#Table(name = "accesslogs")
public class AccessLog extends BaseEntity{
#ManyToOne
#JoinColumn(name = "actorentity_id")
private BaseEntity actor_entity;
#ManyToOne
#JoinColumn(name = "targetentity_id")
private BaseEntity target_entity;
#Column(name="action_code")
private String action;
//getters, setters, & other stuff
}
SECOND EDIT:
As per JBNizet's suggestion, I created a hibernate_sequences table as follows:
CREATE TABLE IF NOT EXISTS hibernate_sequences(
sequence_next_hi_value int(11) UNSIGNED NOT NULL AUTO_INCREMENT PRIMARY KEY
)engine=InnoDB;SHOW WARNINGS;
But now I am getting the following error:
Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'sequence_name' in 'where clause'
Here is the hibernate sql causing the error, followed by the next 2 lines of the stack trace:
Hibernate: select sequence_next_hi_value from hibernate_sequences where sequence_name = 'BaseEntity' for update
ERROR MultipleHiLoPerTableGenerator - HHH000351: Could not read or init a hi value
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: Unknown column 'sequence_name' in 'where clause'
How do I resolve this?
What a mess... AUTO_INCREMENT is MySQL's hidden sequence. The radical problem is that MySQL can not insert and return the PK at the same time, but Hibernate need this while INSERTing a new Entity.
The Problems you run into:
If Hibernate save a new Entity, he try to immerdentelly set the id to the new EntityBean. Therefore hibernate must read what ID will the Database use before hibernate save the new Tuple to the Table.
If you have multiple Servers who access the database, you shall let hibernate's session-factory decide to use the built-in sequence(AUTO-INCREMENT) or let hibernate decide (GenerationType.AUTO/GenerationType.IDENTITY) how large the open range of reserved PK's is (Job of a DB-Architect). (We have about 20 servers to one Database, so on a good-used table we use a PK-distance of +100). If only one server have access to the database GenerationType.TABLE shall be correct.
Hibernate must calculate the next id by yourself using max(*)+1 but:
What if two requests ask for max(*)+1 at the same time/with the same result? Right: The last try to insert will fail.
So you need to have a Table LAST_IDS in the database who stores the last Table-PK's. If you like to add one, you must do this steps:
Start read-optimistic transaction.
SELECT MAX(address_id) FROM LAST_IDS
store the maximum in a java-variable i.e.: $OldID.
$NewID = $OldID + 1. (+100 in pessimistic-lock)
UPDATE LAST_IDS SET address_id= $newID WHERE address_id= $oldID?
commit the read-optimistic transaction.
if commit was successfull, store $newID to setID() in the HibernateBean you like to save.
Finally let Hibernate call the insert.
This is the only way i know.
BTW: Hibernate-Entitys shall only use inheritance if the Database support inheritance between tables like PostgreSQL or Oracle.
Because you use the TABLE identifier generator you need to have that table created. If you are not using the enhanced identifier generators, chances are you are going to use the MultipleHiLoPerTableGenerator.
The MultipleHiLoPerTableGenerator can use one table for all table identifier generators.
My suggestion is to grab the table ddl from your integration tests, in case you use hbmddl to build the test schema. If you use flyway or liquibase for testing, you can add a maven plugin to generate the ddl schema.
Once you have the schema, you need to take the exact create table command and make add it to your MySQL database.

Hibernate zeroToOne

I am trying to establish a relationship between 2 entities which would be zero-to-one. That is, the Parent can be saved without the associated Child entity and also along with the assoicated Child.
Following are the 2 Entity classes...
Employee (Parent)
public class Employee {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
#Column(name="EMP_NAME")
private String name;
#PrimaryKeyJoinColumn
#OneToOne(cascade = {CascadeType.ALL})
private EmployeeInfo info;
#Column(name="EMP_ENUM")
private Integer enumId;
EmployeeInfo (Child)
public class EmployeeInfo {
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
private Long id;
#Column(name="EMPLOYEE_EMAIL")
private String email;
With such kind of a relation and id column of the only Parent (Employee) table set to AUTO INCREMENT in MySql DB, the problem is that while saving a Parent->Child object graph, I get the following exception
org.springframework.orm.hibernate3.HibernateJdbcException: JDBC exception on Hibernate data access: SQLException for SQL [insert into EMP_INFO
Caused by: java.sql.SQLException: Field 'id' doesn't have a default value
I tried setting the Child Table's Id property to AUTO INCREMENT in the DB , and the persistence of such a Parent->Child object graph is successful.
However, the problem described here surfaces, because I have a scenario in which I would like to save the parent (Employee) object without the associated EmpInfo object, and hence do NOT want to have AUTO INCREMENT on the Child's id column.
One solution could be not use the PrimaryKeyJoinColumn, but use a particular JoinColumn, but that adds an unnecessary column to my existing Table.
Has anyone come across such a problem? If yes, any pointers would be much helpful.
Finally, I got it working thanks to Pascal and some googling from my side. Apparently, I cannot use the Native key generator for such relationships where the parent can exist without the child (optional = true).
The thing that worked finally was the following, leaving me the downside of having to deal with Hibernate specific annotation (#GenericGenerator) and also having to make-do with bi-directional relationships instead of the unidirectional that I wanted.
Employee (Parent) class remains unchanged as above. It has AUTO INCREMENT on the Id column.
As for the child class (EmployeeInfo) it changed to the following, and again WITHOUT having the AUTO INCREMENT set on the Id column.
#Table(name="EMP_INFO")
#Entity
public class EmployeeInfo {
#Id
#GeneratedValue(generator="foreign")
#GenericGenerator(name="foreign", strategy = "foreign", parameters={
#Parameter(name="property", value="verifInfo")})
private Long id;
#OneToOne(optional=false)
#JoinColumn (name="id")
private Employee emp;
#Column(name="EMPLOYEE_EMAIL")
private String email;
This helped me achieve what I wanted but on the downside, GenericGenerator is not a JPA annotation, it is a hibernate annotation, and sadly I have to make do with that as of now because JPA does not currently support this(or any similar) annotation.
Anyway, it helps to get through such cases :-)
I have a scenario in which I would like to save the parent (Employee) object without the associated EmpInfo object.
The optional attribute of a OneToOne is true by default, which is what you want.
However, you are somehow misusing the #PrimaryKeyJoinColumn here (well, it actually depends on what you really want to achieve but your current combination of annotations is not correct).
IF you want to map a OneToOne with a shared primary-key, use the #PrimaryKeyJoinColumn. But in that case, don't use a GeneratedValue on EmployeeInfo and set the id manually or, if you don't want to set it manually, use the Hibernate specific foreign generator that I already mentioned in your previous question. Check also the related question mentioned below.
And IF you do not want to use a shared primary key (like in your current code since you're trying to get the id generated by the database), then do not use the PrimaryKeyJoinColumn.
You have to make a choice.
References
JPA 1.0 specification:
9.1.32 PrimaryKeyJoinColumn Annotation
Related question
JPA Hibernate One-to-One relationship.

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