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.
JPA newbie here. Here's my question:
Say we have an entity like this:
#Entity
#Table(name="thingies")
public class Thingy implements Serializable {
private Long thingyId;
private String thingyName;
private Integer thingyPrice;
// Constructor, getters, setters
}
Mapped to a table like this:
create table thingies (
thingy_id serial primary key,
thingy_name text,
thingy_price smallint
);
Is there a way to make the provider aware of the attribute naming policy instead of needing to explicitly provide a #Column annotation on each getter? In other words, can we automatically map all underscored names to the corresponding camelcased names without using #Column?
(I know I can quote names, that is not an answer to my question above.)
You need to add the following property to your persistence.xml file:
<property name="hibernate.ejb.naming_strategy" value="org.hibernate.cfg.ImprovedNamingStrategy"/>
And all camelcasedproperty names will bemapped to underscored column names.
You should use #Column.
#Column(name="thingy_name")
private String thingyName;
Becouse 'thingy_name' is not equal 'thingyName'. JPA dont know what you want mapped. But in table thingyname and in entity thingyName is equal for JPA.
Use adnotation #Column, it does not take a lot of time. But you will get a explicit documentation.
in JPA2 when we are using Embed-able (Basic Type like String.. etc ) object in Entity using with #ElementCollection and #CollectionTable annotation , the new table is created , but in new table how to declare primary-key contraint in column ? following is my code
public class Employee {
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
private int id;
private String name;
private String salary;
#Transient
private String phnNum;
#Enumerated(EnumType.STRING)
private EmployeeType type;
#ElementCollection
#CollectionTable(name="vacations" , joinColumns=#JoinColumn(name="Emp_Id"))
private Collection<Vacation> vacationBooking;
#ElementCollection
private Set<String> nickNames;
...................
with this code the "vacation" and "employee_nickname" two tables are created in schema. but i want to declare the one primary-key column in both table . what i do for this?
It looks like a primary key per se is not supported by JPA 2.0:
From Wikibooks:
The JPA 2.0 specification does not provide a way to define the Id in the Embeddable. However, to delete or update an element of the ElementCollection mapping, some unique key is normally required. Otherwise, on every update the JPA provider would need to delete everything from the CollectionTable for the Entity, and then insert the values back. So, the JPA provider will most likely assume that the combination of all of the fields in the Embeddable are unique, in combination with the foreign key (JoinColumn(s)). This however could be inefficient, or just not feasible if the Embeddable is big, or complex.
Some JPA providers may allow the Id to be specified in the Embeddable, to resolve this issue. Note in this case the Id only needs to be unique for the collection, not the table, as the foreign key is included. Some may also allow the unique option on the CollectionTable to be used for this. Otherwise, if your Embeddable is complex, you may consider making it an Entity and use a OneToMany instead.
Do you mean that you want to assign 'id' from Employee table as foreign key to the Vacation table?
In that case, you should use #OneToMany instead of #ElementCollection
I have the following existing DB schema, which I'd like to recreate with Java and plain JPA annotations (using hibernate as provider, so hibernate specific annotations would work as a last resort):
CREATE TABLE users (
user_id NUMBER NOT NULL -- pk
);
CREATE TABLE userdata_keys (
userdata_key_id NUMBER NOT NULL, -- pk
key VARCHAR2(128) NOT NULL
);
CREATE TABLE users_userdata (
user_id NUMBER NOT NULL, -- fk users.user_id
userdata_key_id NUMBER NOT NULL, -- fk userdata_keys.userdata_key_id
value VARCHAR2(256)
);
I've thus created the following classes and annotations:
class User {
#Id
Long id;
#OneToMany
Set<Userdata> userdata;
}
class UserdataKey {
#Id
Long id;
String key;
}
class Userdata {
String value;
#EmbeddedId
UserdataId userdataId;
}
#Embeddable
class UserdataId {
User user;
UserdataKey userdataKey;
}
I left out columnName attributes and other attributes of the entities here.
It does however not quite work as intended. If I do not specify a mappedBy attribute for User.userdata, hibernate will automatically create a table USERS_USERS_USERDATA, but as far as I've seen does not use it. It does however use the table which I specified for the Userdata class.
Since I'm rather new to Java and hibernate as well, all I do to test this currently is looking at the DB schema hibernate creates when persisting a few sample entries.
As a result, I'm entirely puzzled as to whether I'm doing this the right way at all. I read the hibernate documentation and quite a bunch of Google results, but none of them seemed to deal with what I want to do (composite key with "subclasses" with their own primary key).
The mappedBy attribute is mandatory at one of the sides of every bidirectional association. When the association is a one-to-many, the mappedBy attribute is placed ot the one- side (i.e. on the User's userdata field in your case).
That's because when an association is bidirectional, one side of the association is always the inverse of the other, so there's no need to tell twice to Hibernate how the association is mapped (i.e. which join column or join table to use).
If you're ready to recreate the schema, I would do it right (and easier), and use a surrogate auto-generated key in users_userdata rather than a composite one. This will be much easier to handle, in all the layers of your application.
I want to wrap id in custom class. Like this
#Entity
#Table(name = "USERS")
public class User {
#EmbeddedId
UserId id;
}
#Embeddable
public class UserId implements Serializable {
private Long value;
}
The issue in auto generation value for UserId. What I should do to make #GeneratedValue on value be workable?
BTW, It would be great if id would be initialized automatically itself.
As far as I know Hibernate only generates values for a field marked as the #Id. I found this post and Hardy's answer supports this.
We have tried to do similar and managed it via a pre-insert listener. It was fairly complex and non-ideal though. Also you might find different behaviour on different database palatforms. Using Oracle sequences would mean that you need to assign the value pre-insert (Hibernate does a select to get the value and then an insert) but with MySQL the auto incrementing field would assign the value and hibernate does an insert to generate the auto generated value and then select to find out what the value was.