I have the following domain objects:
public class Department {
private long departmentId;
}
public class Manager {
private long managerId;
}
public class Project {
private ProjectId compositeId;
#ManyToOne
private Department department;
#ManyToOne
private Manager manager;
}
public class ProjectId {
private long departmentId;
private long managerId;
}
Project is identified by a composite key (departmentId,managerId). The question is how should Project.setManager(..) or Project.setDepartment(..) be implemented? Is the implemention listed below the best practice?
public void setManager( Manager manager ) {
this.manager = manager;
this.compositeId.setManagerId( manager.getId() );
}
My understanding is that compositeId needs to be updated whenever an property is set.
A harder and related question is how should Project.setCompositeId(..) be implemented? Project wouldn't be able to update property manager nor department based on a composite id (long). Overwriting the compositeId without updating the properties would leave Project at an incongruous state.
I suggest the following:
#Entity
#IdClass(ProjectId.class)
public class Project {
#Id #Column(name="DEPARTMENT_ID")
private long departmentId;
#Id #Column(name="MANAGER_ID")
private long managerId;
#ManyToOne
#PrimaryKeyJoinColumn(name="DEPARTMENT_ID", referencedColumnName="DPT_ID")
private Department department;
#ManyToOne
#PrimaryKeyJoinColumn(name="MANAGER_ID", referencedColumnName="MGR_ID")
private Manager manager;
...
}
This mapping is very well explained in 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();
}
...
}
Reference
JPA Wikibook
Primary Keys through OneToOne and ManyToOne Relationships
Related
im relative new to Hibernate Mappings im trying to achieve this functionality between the class Post and Comentario without luck
Relational model
#Embeddable
public class PostPK implements Serializable {
#Column(name="idPost")
private int postID;
#Column(name="idUsuario")
private int userIDFK;
-------------------------------
#Entity
#Table(name="Post")
public class Post {
#EmbeddedId
private PostPK id;
#ManyToOne
#MapsId(value="userIDFK")
#JoinColumn(name="idUsuario")
private Usuario usuario;
#OneToMany(mappedBy="post")
private List<Comentario> comentarios;
#Column(name="titulo")
private String titulo;
-----------------------------------
#Embeddable
public class ComentarioPK implements Serializable{
#Column(name="idComentario")
private int comentarioId;
#Column(name="idPost")
private int postIdFK;
---------------------------
#Entity
#Table(name="Comentario")
public class Comentario {
#EmbeddedId
private ComentarioPK id;
#ManyToOne
#MapsId("postIdFK")
#JoinColumn(name="idPost",referencedColumnName="idPost")
private Post post;
#Column(name="texto")
private String texto;
without mapping comentario and its fields in Post its working fine but when i decide to map it i get this error
Unable to find column reference in the #MapsId mapping: idUsuario
is it not finding the idUsuario column in Comentario table? i dont want to add it , i can achieve joins in mysql but i dont know how to do it in Hibernate
#MapsId annotation is used to map the primary key fields of the parent entity with the child entity(with the same name).
In your case your are having composite primary key in your parent entity but in child entity you want to refer only one field of it.
PostPK has two fields : idPost and idUsuario. But in Comentario class when your are specifying ManyToOne relationship you are mentioning single column in #JoinColumn(which is idPost) and no field for idUsuario is available in your mapping. But as per the behavior of #MapsId annotation both the fields(idPost and idUsuario) are expected in Comentario class.
Thus, in your case #MapsId annotation won't work
I have a simple table (ActivityLog) and I want it to have a PK that is also a FK to another table (User).
It seems to be a common thing to have, and I tried to follow this wikibook
Primary Keys through OneToOne and ManyToOne Relationships. The example there involved a composite key. I need just a primitive key, so I ended up with:
#Entity
public class User {
#Id
private Long id;
// other stuff
}
#Entity
public class ActivityLog {
#Id
#OneToOne(optional = false)
#JoinColumn(name="user_id", referencedColumnName="id")
private User user;
// other stuff
}
Unfortunately i am getting:
Caused by: java.lang.IllegalArgumentException: This class [class com.example.ActivityLog] does not define an IdClass
at org.hibernate.metamodel.internal.AbstractIdentifiableType.getIdClassAttributes(AbstractIdentifiableType.java:183)
at org.springframework.data.jpa.repository.support.JpaMetamodelEntityInformation$IdMetadata.<init>(JpaMetamodelEntityInformation.java:253)
I tried to annotate ActivityLog with:
#IdClass(Long.class)
(even though from what I understand it is applicable only for composite keys), yet I am getting the exact same error.
Is my case different than what's on the mentioned wikibook?
Is Spring at fault here? (As suggested in this question? (no accepted answers)).
This should help:
#Entity
public class ActivityLog {
#Id
#Column(name = "user_id")
private Long id;
#OneToOne(optional = false)
#JoinColumn(name="user_id", referencedColumnName="id")
private User user;
// other stuff
}
Btw. I would expect, that you need more logs per user, so you would probably need some additional (generated) id anyway ...
Maybe this is an overly simple question, but I am getting an exception when I try to delete a user entity.
The user entity:
#Entity
#Table(name = "users")
public class User
{
#Transient
private static final int SALT_LENGTH = 32;
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private int id;
#NotNull
private String firstName;
#NotNull
private String lastName;
#Column(unique = true, length = 254)
#NotNull
private String email;
// BCrypt outputs 60 character results.
#Column(length = 60)
private String hashedPassword;
#NotNull
private String salt;
private boolean enabled;
#CreationTimestamp
#Temporal(TemporalType.TIMESTAMP)
#Column(updatable = false)
private Date createdDate;
And I have an entity class which references a user with a foreign key. What I want to happen is that when the user is deleted, any PasswordResetToken objects that reference the user are also deleted. How can I do this?
#Entity
#Table(name = "password_reset_tokens")
public class PasswordResetToken
{
private static final int EXPIRATION_TIME = 1; // In minutes
private static final int RESET_CODE_LENGTH = 10;
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private int id;
private String token;
#OneToOne(targetEntity = User.class, fetch = FetchType.EAGER)
#JoinColumn(nullable = false, name = "userId")
private User user;
private Date expirationDate;
The exception I am getting boils down to Cannot delete or update a parent row: a foreign key constraint fails (`heroku_bc5bfe73a752182`.`password_reset_tokens`, CONSTRAINT `FKk3ndxg5xp6v7wd4gjyusp15gq` FOREIGN KEY (`user_id`) REFERENCES `users` (`id`))
I'd like to avoid adding a reference to PasswordResetToken in the parent entity, becaue User shouldn't need to know anything about PasswordResetToken.
It is not possible on JPA level without creating a bidirectional relation. You need to specify cascade type in User class. User should be owner of the relation and it should provide the information on how to deal with related PasswordResetToken.
But if you cannot have a bidirectional relation I would recommend you to setup relation directly in schema generation SQL script.
If you create your schema via SQL script and not via JPA autogeneration (I believe all serious projects must follow this pattern) you can add ON DELETE CASCADE constraint there.
It will look somehow like this:
CREATE TABLE password_reset_tokens (
-- columns declaration here
user_id INT(11) NOT NULL,
CONSTRAINT FK_PASSWORD_RESET_TOKEN_USER_ID
FOREIGN KEY (user_id) REFERENCES users (id)
ON DELETE CASCADE
);
Here is the documentation on how to use DB migration tools with spring boot. And here is the information on how to generate schema script from hibernate (that will simplify the process of writing your own script).
Parent Entity:
#OneToOne
#JoinColumn(name = "id")
private PasswordResetToken passwordResetToken;
Child Entity:
#OneToOne(mappedBy = "PasswordResetToken", cascade = CascadeType.ALL, orphanRemoval = true)
private User user;
If you want the Password entity to be hidden from the client, you can write a custom responses and hide it. Or if you want to ignore it by using #JsonIgnore
If you don't want the reference in the Parent Entity (User), then you have to override the default method Delete() and write your logic to find and delete the PasswordResetToken first and then the User.
You can use Entity listener and Callback method #PreRemove to delete an associated 'Token' before the 'User'.
#EntityListeners(UserListener.class)
#Entity
public class User {
private String name;
}
#Component
public class UserListener {
private static TokenRepository tokenRepository;
#Autowired
public void setTokenRepository(TokenRepository tokenRepository) {
PersonListener.tokenRepository = tokenRepository;
}
#PreRemove
void preRemove(User user) {
tokenRepository.deleteByUser(user);
}
}
where deleteByPerson is very simple method of your 'Token' repository:
public interface TokenRepository extends JpaRepository<Token, Long> {
void deleteByUser(User user);
}
Pay attention on static declaration of tokenRepository - without this Spring could not inject TokenRepository because, as I can understand, UserListener is instantiated by Hybernate (see additional info here).
Also as we can read in the manual,
a callback method must not invoke EntityManager or Query methods!
But in my simple test all works OK.
Working example and test.
First of all, sorry for my english. This is my first post.
Im developing an app for a friend that uses JPA (EclipseLink) and i can´t figure out how to make a ManyToMany reflexive relationship with an extra column describing the relationship.
I tried the solution of this post (without using a bidirectional relation):
#Entity
#Table(name="relationships")
public class Relationship implements Serializable {
private static final long serialVersionUID = 1L;
#ManyToOne()
#JoinColumn(name = "associated_id_one")
private Person associatedPersonOne;
#ManyToOne()
#JoinColumn(name = "associated_id_two")
private Person associatedPersonTwo;
#Column(name="description")
private String description;
//Getters, Setters and constructor
...
}
But JPA gives me this error: The entity has no primary key attribute defined.
I know i have to specify a primary key but i don´t now how to make a composite primary key with those two foreign key.
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).