JPA 2.0 many-to-many with extra column
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I'm trying to do a ManyToMany relationship in JPA 2.0 (JBoss 7.1.1) with an extra column (in bold, below) in the relationship, like:
Employer EmployerDeliveryAgent DeliveryAgent
(id,...) (employer_id, deliveryAgent_id, **ref**) (id,...)
I wouldn't like to have duplicate attributes, so I would like to apply the second solution presented in http://giannigar.wordpress.com/2009/09/04/mapping-a-many-to-many-join-table-with-extra-column-using-jpa/ . But I can't get it to work, I get several errors like:
- Embedded ID class should not contain relationship mappings (in fact the spec says so);
- In attribute 'employerDeliveryAgent', the "mapped by" value 'pk.deliveryAgent' cannot be resolved to an attribute on the target entity;
- In attribute 'employerDeliveryAgent', the "mapped by" value 'pk.employer' cannot be resolved to an attribute on the target entity;
- Persistent type of override attribute "pk.deliveryAgent" cannot be resolved;
- Persistent type of override attribute "pk.employer" cannot be resolved;
Many people on that link said that it worked fine, so I suppose something is different in my environment, perhaps JPA or Hibernate version. So my question is: how do I achieve such scenario with JPA 2.0 (Jboss 7.1.1 / using Hibernate as JPA implementation)? And to complement that question: should I avoid using composite keys and instead use plain generated id and a unique constraint?
Thanks in advance.
Obs.: I didn't copy my source code here because it is essentially a copy of the one at the link above, just with different classes and attributes names, so I guess it is not necessary.
hibernate java-ee orm jboss7.x jpa-2.0
add a comment |
I'm trying to do a ManyToMany relationship in JPA 2.0 (JBoss 7.1.1) with an extra column (in bold, below) in the relationship, like:
Employer EmployerDeliveryAgent DeliveryAgent
(id,...) (employer_id, deliveryAgent_id, **ref**) (id,...)
I wouldn't like to have duplicate attributes, so I would like to apply the second solution presented in http://giannigar.wordpress.com/2009/09/04/mapping-a-many-to-many-join-table-with-extra-column-using-jpa/ . But I can't get it to work, I get several errors like:
- Embedded ID class should not contain relationship mappings (in fact the spec says so);
- In attribute 'employerDeliveryAgent', the "mapped by" value 'pk.deliveryAgent' cannot be resolved to an attribute on the target entity;
- In attribute 'employerDeliveryAgent', the "mapped by" value 'pk.employer' cannot be resolved to an attribute on the target entity;
- Persistent type of override attribute "pk.deliveryAgent" cannot be resolved;
- Persistent type of override attribute "pk.employer" cannot be resolved;
Many people on that link said that it worked fine, so I suppose something is different in my environment, perhaps JPA or Hibernate version. So my question is: how do I achieve such scenario with JPA 2.0 (Jboss 7.1.1 / using Hibernate as JPA implementation)? And to complement that question: should I avoid using composite keys and instead use plain generated id and a unique constraint?
Thanks in advance.
Obs.: I didn't copy my source code here because it is essentially a copy of the one at the link above, just with different classes and attributes names, so I guess it is not necessary.
hibernate java-ee orm jboss7.x jpa-2.0
add a comment |
I'm trying to do a ManyToMany relationship in JPA 2.0 (JBoss 7.1.1) with an extra column (in bold, below) in the relationship, like:
Employer EmployerDeliveryAgent DeliveryAgent
(id,...) (employer_id, deliveryAgent_id, **ref**) (id,...)
I wouldn't like to have duplicate attributes, so I would like to apply the second solution presented in http://giannigar.wordpress.com/2009/09/04/mapping-a-many-to-many-join-table-with-extra-column-using-jpa/ . But I can't get it to work, I get several errors like:
- Embedded ID class should not contain relationship mappings (in fact the spec says so);
- In attribute 'employerDeliveryAgent', the "mapped by" value 'pk.deliveryAgent' cannot be resolved to an attribute on the target entity;
- In attribute 'employerDeliveryAgent', the "mapped by" value 'pk.employer' cannot be resolved to an attribute on the target entity;
- Persistent type of override attribute "pk.deliveryAgent" cannot be resolved;
- Persistent type of override attribute "pk.employer" cannot be resolved;
Many people on that link said that it worked fine, so I suppose something is different in my environment, perhaps JPA or Hibernate version. So my question is: how do I achieve such scenario with JPA 2.0 (Jboss 7.1.1 / using Hibernate as JPA implementation)? And to complement that question: should I avoid using composite keys and instead use plain generated id and a unique constraint?
Thanks in advance.
Obs.: I didn't copy my source code here because it is essentially a copy of the one at the link above, just with different classes and attributes names, so I guess it is not necessary.
hibernate java-ee orm jboss7.x jpa-2.0
I'm trying to do a ManyToMany relationship in JPA 2.0 (JBoss 7.1.1) with an extra column (in bold, below) in the relationship, like:
Employer EmployerDeliveryAgent DeliveryAgent
(id,...) (employer_id, deliveryAgent_id, **ref**) (id,...)
I wouldn't like to have duplicate attributes, so I would like to apply the second solution presented in http://giannigar.wordpress.com/2009/09/04/mapping-a-many-to-many-join-table-with-extra-column-using-jpa/ . But I can't get it to work, I get several errors like:
- Embedded ID class should not contain relationship mappings (in fact the spec says so);
- In attribute 'employerDeliveryAgent', the "mapped by" value 'pk.deliveryAgent' cannot be resolved to an attribute on the target entity;
- In attribute 'employerDeliveryAgent', the "mapped by" value 'pk.employer' cannot be resolved to an attribute on the target entity;
- Persistent type of override attribute "pk.deliveryAgent" cannot be resolved;
- Persistent type of override attribute "pk.employer" cannot be resolved;
Many people on that link said that it worked fine, so I suppose something is different in my environment, perhaps JPA or Hibernate version. So my question is: how do I achieve such scenario with JPA 2.0 (Jboss 7.1.1 / using Hibernate as JPA implementation)? And to complement that question: should I avoid using composite keys and instead use plain generated id and a unique constraint?
Thanks in advance.
Obs.: I didn't copy my source code here because it is essentially a copy of the one at the link above, just with different classes and attributes names, so I guess it is not necessary.
hibernate java-ee orm jboss7.x jpa-2.0
hibernate java-ee orm jboss7.x jpa-2.0
asked May 23 '14 at 20:03
RenanRenan
5841527
5841527
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
First of all You need to generate a EmployerDeliveryAgentPK
class because It has a multiple PK:
@Embeddable
public class EmployerDeliveryAgentPK implements Serializable {
@Column(name = "EMPLOYER_ID")
private Long employer_id;
@Column(name = "DELIVERY_AGENT_ID")
private Long deliveryAgent_id;
}
Next, You need to create a EmployerDeliveryAgent
class. This class represent many to many table of Employer
and DeliveryAgent
:
@Entity
@Table(name = " EmployerDeliveryAgent")
public class EmployerDeliveryAgent implements Serializable {
@EmbeddedId
private EmployerDeliveryAgentPK id;
@ManyToOne
@MapsId("employer_id") //This is the name of attr in EmployerDeliveryAgentPK class
@JoinColumn(name = "EMPLOYER_ID")
private Employer employer;
@ManyToOne
@MapsId("deliveryAgent_id")
@JoinColumn(name = "DELIVERY_AGENT_ID")
private DeliveryAgent deliveryAgent;
}
After that, in Employer
class You need to add:
@OneToMany(mappedBy = "deliveryAgent")
private Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
And in DeliveryAgent
class You need to add:
@OneToMany(mappedBy = "employer")
private Set<EmployerDeliveryAgent> employer = new HashSet<EmployerDeliveryAgent>();
This is all! Good luck!!
Thank you very much, I wasted two days tryng to figure out this problem.
– Paolo
Mar 28 '17 at 9:58
I think in theDeliveryAgent
class we should mapEmployerDeliveryAgent
class instead ofEmployer
class. So, the code in theDeliveryAgent
class should look likeprivate Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
, same as in theEmployer
class.
– tylerdurden
Oct 4 '17 at 7:10
@tylerdurden You are rigth! I have already modified it!
– Erik Lucio
Oct 4 '17 at 7:21
Is it required to overrideequals()
andhashcode()
? (assumimng I will notintend to put instances of persistent classes in a Set AND intend to use reattachment of detached instances
)
– user3529850
Oct 29 '18 at 21:13
add a comment |
Both answers from Eric Lucio and Renan helped, but there use of the ids in the association table is redundant. You have both the associated entities and their ids in the class. This is not required. You can simple map the associated entity in the association class with the @Id
on the associated entity field.
@Entity
public class Employer {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@OneToMany(mappedBy = "employer")
private List<EmployerDeliveryAgent> deliveryAgentAssoc;
// other properties and getters and setters
}
@Entity
public class DeliveryAgent {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@OneToMany(mappedBy = "deliveryAgent")
private List<EmployerDeliveryAgent> employerAssoc;
// other properties and getters and setters
}
The association class
@Entity
@Table(name = "employer_delivery_agent")
@IdClass(EmployerDeliveryAgentId.class)
public class EmployerDeliveryAgent {
@Id
@ManyToOne
@JoinColumn(name = "employer_id", referencedColumnName = "id")
private Employer employer;
@Id
@ManyToOne
@JoinColumn(name = "delivery_agent_id", referencedColumnName = "id")
private DeliveryAgent deliveryAgent;
@Column(name = "is_project_lead")
private boolean isProjectLead;
}
Still need the association PK class. Notice the fields names should correspond exactly to the field names in the association class, but the types should be the type of the id in the associated type.
public class EmployerDeliveryAgentId implements Serializable {
private int employer;
private int deliveryAgent;
// getters/setters and most importantly equals() and hashCode()
}
Good solution, thanks
– Renan
Oct 3 '15 at 15:43
1
This is the cleanest solution you can get, well done.
– FrancescoM
Feb 3 '18 at 14:37
anyone can provide insert or delete example too.
– VK321
Oct 21 '18 at 7:37
Very cool, thank you thank you. Question... ...i get a NPE if i also put a generated @id column in the association class (EmployerDeliveryAgent in your example). But I want a RDBMS long PK... what say you? is there a way to do it? screw my surrogate key?
– tom
Nov 20 '18 at 18:42
add a comment |
OK, I got it working based on the solution available at
http://en.wikibooks.org/wiki/Java_Persistence/ManyToMany#Mapping_a_Join_Table_with_Additional_Columns.
This solution does not generate duplicate attributes on the database, but it does generate duplicate attributes in my JPA entities (which is very acceptable, since you can relay the extra work to a constructor or method - it ends up being transparent). The primary and foreign keys generated in the database are 100% correct.
As stated on the link, I couldn't use @PrimaryKeyJoinColumn and instead used @JoinColumn(name = "projectId", updatable = false, insertable = false, referencedColumnName = "id"). Another thing worth mentioning: I had to use EntityManager.persist(association), which is missing on the example at the link.
So my final solution is:
@Entity
public class Employee {
@Id
private long id;
...
@OneToMany(mappedBy="employee")
private List<ProjectAssociation> projects;
...
}
@Entity
public class Project {
@PersistenceContext
EntityManager em;
@Id
private long id;
...
@OneToMany(mappedBy="project")
private List<ProjectAssociation> employees;
...
// Add an employee to the project.
// Create an association object for the relationship and set its data.
public void addEmployee(Employee employee, boolean teamLead) {
ProjectAssociation association = new ProjectAssociation();
association.setEmployee(employee);
association.setProject(this);
association.setEmployeeId(employee.getId());
association.setProjectId(this.getId());
association.setIsTeamLead(teamLead);
em.persist(association);
this.employees.add(association);
// Also add the association object to the employee.
employee.getProjects().add(association);
}
}
@Entity
@Table(name="PROJ_EMP")
@IdClass(ProjectAssociationId.class)
public class ProjectAssociation {
@Id
private long employeeId;
@Id
private long projectId;
@Column(name="IS_PROJECT_LEAD")
private boolean isProjectLead;
@ManyToOne
@JoinColumn(name = "employeeId", updatable = false, insertable = false,
referencedColumnName = "id")
private Employee employee;
@ManyToOne
@JoinColumn(name = "projectId", updatable = false, insertable = false,
referencedColumnName = "id")
private Project project;
...
}
public class ProjectAssociationId implements Serializable {
private long employeeId;
private long projectId;
...
public int hashCode() {
return (int)(employeeId + projectId);
}
public boolean equals(Object object) {
if (object instanceof ProjectAssociationId) {
ProjectAssociationId otherId = (ProjectAssociationId) object;
return (otherId.employeeId == this.employeeId)
&& (otherId.projectId == this.projectId);
}
return false;
}
}
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
First of all You need to generate a EmployerDeliveryAgentPK
class because It has a multiple PK:
@Embeddable
public class EmployerDeliveryAgentPK implements Serializable {
@Column(name = "EMPLOYER_ID")
private Long employer_id;
@Column(name = "DELIVERY_AGENT_ID")
private Long deliveryAgent_id;
}
Next, You need to create a EmployerDeliveryAgent
class. This class represent many to many table of Employer
and DeliveryAgent
:
@Entity
@Table(name = " EmployerDeliveryAgent")
public class EmployerDeliveryAgent implements Serializable {
@EmbeddedId
private EmployerDeliveryAgentPK id;
@ManyToOne
@MapsId("employer_id") //This is the name of attr in EmployerDeliveryAgentPK class
@JoinColumn(name = "EMPLOYER_ID")
private Employer employer;
@ManyToOne
@MapsId("deliveryAgent_id")
@JoinColumn(name = "DELIVERY_AGENT_ID")
private DeliveryAgent deliveryAgent;
}
After that, in Employer
class You need to add:
@OneToMany(mappedBy = "deliveryAgent")
private Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
And in DeliveryAgent
class You need to add:
@OneToMany(mappedBy = "employer")
private Set<EmployerDeliveryAgent> employer = new HashSet<EmployerDeliveryAgent>();
This is all! Good luck!!
Thank you very much, I wasted two days tryng to figure out this problem.
– Paolo
Mar 28 '17 at 9:58
I think in theDeliveryAgent
class we should mapEmployerDeliveryAgent
class instead ofEmployer
class. So, the code in theDeliveryAgent
class should look likeprivate Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
, same as in theEmployer
class.
– tylerdurden
Oct 4 '17 at 7:10
@tylerdurden You are rigth! I have already modified it!
– Erik Lucio
Oct 4 '17 at 7:21
Is it required to overrideequals()
andhashcode()
? (assumimng I will notintend to put instances of persistent classes in a Set AND intend to use reattachment of detached instances
)
– user3529850
Oct 29 '18 at 21:13
add a comment |
First of all You need to generate a EmployerDeliveryAgentPK
class because It has a multiple PK:
@Embeddable
public class EmployerDeliveryAgentPK implements Serializable {
@Column(name = "EMPLOYER_ID")
private Long employer_id;
@Column(name = "DELIVERY_AGENT_ID")
private Long deliveryAgent_id;
}
Next, You need to create a EmployerDeliveryAgent
class. This class represent many to many table of Employer
and DeliveryAgent
:
@Entity
@Table(name = " EmployerDeliveryAgent")
public class EmployerDeliveryAgent implements Serializable {
@EmbeddedId
private EmployerDeliveryAgentPK id;
@ManyToOne
@MapsId("employer_id") //This is the name of attr in EmployerDeliveryAgentPK class
@JoinColumn(name = "EMPLOYER_ID")
private Employer employer;
@ManyToOne
@MapsId("deliveryAgent_id")
@JoinColumn(name = "DELIVERY_AGENT_ID")
private DeliveryAgent deliveryAgent;
}
After that, in Employer
class You need to add:
@OneToMany(mappedBy = "deliveryAgent")
private Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
And in DeliveryAgent
class You need to add:
@OneToMany(mappedBy = "employer")
private Set<EmployerDeliveryAgent> employer = new HashSet<EmployerDeliveryAgent>();
This is all! Good luck!!
Thank you very much, I wasted two days tryng to figure out this problem.
– Paolo
Mar 28 '17 at 9:58
I think in theDeliveryAgent
class we should mapEmployerDeliveryAgent
class instead ofEmployer
class. So, the code in theDeliveryAgent
class should look likeprivate Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
, same as in theEmployer
class.
– tylerdurden
Oct 4 '17 at 7:10
@tylerdurden You are rigth! I have already modified it!
– Erik Lucio
Oct 4 '17 at 7:21
Is it required to overrideequals()
andhashcode()
? (assumimng I will notintend to put instances of persistent classes in a Set AND intend to use reattachment of detached instances
)
– user3529850
Oct 29 '18 at 21:13
add a comment |
First of all You need to generate a EmployerDeliveryAgentPK
class because It has a multiple PK:
@Embeddable
public class EmployerDeliveryAgentPK implements Serializable {
@Column(name = "EMPLOYER_ID")
private Long employer_id;
@Column(name = "DELIVERY_AGENT_ID")
private Long deliveryAgent_id;
}
Next, You need to create a EmployerDeliveryAgent
class. This class represent many to many table of Employer
and DeliveryAgent
:
@Entity
@Table(name = " EmployerDeliveryAgent")
public class EmployerDeliveryAgent implements Serializable {
@EmbeddedId
private EmployerDeliveryAgentPK id;
@ManyToOne
@MapsId("employer_id") //This is the name of attr in EmployerDeliveryAgentPK class
@JoinColumn(name = "EMPLOYER_ID")
private Employer employer;
@ManyToOne
@MapsId("deliveryAgent_id")
@JoinColumn(name = "DELIVERY_AGENT_ID")
private DeliveryAgent deliveryAgent;
}
After that, in Employer
class You need to add:
@OneToMany(mappedBy = "deliveryAgent")
private Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
And in DeliveryAgent
class You need to add:
@OneToMany(mappedBy = "employer")
private Set<EmployerDeliveryAgent> employer = new HashSet<EmployerDeliveryAgent>();
This is all! Good luck!!
First of all You need to generate a EmployerDeliveryAgentPK
class because It has a multiple PK:
@Embeddable
public class EmployerDeliveryAgentPK implements Serializable {
@Column(name = "EMPLOYER_ID")
private Long employer_id;
@Column(name = "DELIVERY_AGENT_ID")
private Long deliveryAgent_id;
}
Next, You need to create a EmployerDeliveryAgent
class. This class represent many to many table of Employer
and DeliveryAgent
:
@Entity
@Table(name = " EmployerDeliveryAgent")
public class EmployerDeliveryAgent implements Serializable {
@EmbeddedId
private EmployerDeliveryAgentPK id;
@ManyToOne
@MapsId("employer_id") //This is the name of attr in EmployerDeliveryAgentPK class
@JoinColumn(name = "EMPLOYER_ID")
private Employer employer;
@ManyToOne
@MapsId("deliveryAgent_id")
@JoinColumn(name = "DELIVERY_AGENT_ID")
private DeliveryAgent deliveryAgent;
}
After that, in Employer
class You need to add:
@OneToMany(mappedBy = "deliveryAgent")
private Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
And in DeliveryAgent
class You need to add:
@OneToMany(mappedBy = "employer")
private Set<EmployerDeliveryAgent> employer = new HashSet<EmployerDeliveryAgent>();
This is all! Good luck!!
edited Oct 4 '17 at 7:19
answered Mar 18 '15 at 7:53
Erik LucioErik Lucio
63358
63358
Thank you very much, I wasted two days tryng to figure out this problem.
– Paolo
Mar 28 '17 at 9:58
I think in theDeliveryAgent
class we should mapEmployerDeliveryAgent
class instead ofEmployer
class. So, the code in theDeliveryAgent
class should look likeprivate Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
, same as in theEmployer
class.
– tylerdurden
Oct 4 '17 at 7:10
@tylerdurden You are rigth! I have already modified it!
– Erik Lucio
Oct 4 '17 at 7:21
Is it required to overrideequals()
andhashcode()
? (assumimng I will notintend to put instances of persistent classes in a Set AND intend to use reattachment of detached instances
)
– user3529850
Oct 29 '18 at 21:13
add a comment |
Thank you very much, I wasted two days tryng to figure out this problem.
– Paolo
Mar 28 '17 at 9:58
I think in theDeliveryAgent
class we should mapEmployerDeliveryAgent
class instead ofEmployer
class. So, the code in theDeliveryAgent
class should look likeprivate Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
, same as in theEmployer
class.
– tylerdurden
Oct 4 '17 at 7:10
@tylerdurden You are rigth! I have already modified it!
– Erik Lucio
Oct 4 '17 at 7:21
Is it required to overrideequals()
andhashcode()
? (assumimng I will notintend to put instances of persistent classes in a Set AND intend to use reattachment of detached instances
)
– user3529850
Oct 29 '18 at 21:13
Thank you very much, I wasted two days tryng to figure out this problem.
– Paolo
Mar 28 '17 at 9:58
Thank you very much, I wasted two days tryng to figure out this problem.
– Paolo
Mar 28 '17 at 9:58
I think in the
DeliveryAgent
class we should map EmployerDeliveryAgent
class instead of Employer
class. So, the code in the DeliveryAgent
class should look like private Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
, same as in the Employer
class.– tylerdurden
Oct 4 '17 at 7:10
I think in the
DeliveryAgent
class we should map EmployerDeliveryAgent
class instead of Employer
class. So, the code in the DeliveryAgent
class should look like private Set<EmployerDeliveryAgent> employerDeliveryAgent = new HashSet<EmployerDeliveryAgent>();
, same as in the Employer
class.– tylerdurden
Oct 4 '17 at 7:10
@tylerdurden You are rigth! I have already modified it!
– Erik Lucio
Oct 4 '17 at 7:21
@tylerdurden You are rigth! I have already modified it!
– Erik Lucio
Oct 4 '17 at 7:21
Is it required to override
equals()
and hashcode()
? (assumimng I will not intend to put instances of persistent classes in a Set AND intend to use reattachment of detached instances
)– user3529850
Oct 29 '18 at 21:13
Is it required to override
equals()
and hashcode()
? (assumimng I will not intend to put instances of persistent classes in a Set AND intend to use reattachment of detached instances
)– user3529850
Oct 29 '18 at 21:13
add a comment |
Both answers from Eric Lucio and Renan helped, but there use of the ids in the association table is redundant. You have both the associated entities and their ids in the class. This is not required. You can simple map the associated entity in the association class with the @Id
on the associated entity field.
@Entity
public class Employer {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@OneToMany(mappedBy = "employer")
private List<EmployerDeliveryAgent> deliveryAgentAssoc;
// other properties and getters and setters
}
@Entity
public class DeliveryAgent {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@OneToMany(mappedBy = "deliveryAgent")
private List<EmployerDeliveryAgent> employerAssoc;
// other properties and getters and setters
}
The association class
@Entity
@Table(name = "employer_delivery_agent")
@IdClass(EmployerDeliveryAgentId.class)
public class EmployerDeliveryAgent {
@Id
@ManyToOne
@JoinColumn(name = "employer_id", referencedColumnName = "id")
private Employer employer;
@Id
@ManyToOne
@JoinColumn(name = "delivery_agent_id", referencedColumnName = "id")
private DeliveryAgent deliveryAgent;
@Column(name = "is_project_lead")
private boolean isProjectLead;
}
Still need the association PK class. Notice the fields names should correspond exactly to the field names in the association class, but the types should be the type of the id in the associated type.
public class EmployerDeliveryAgentId implements Serializable {
private int employer;
private int deliveryAgent;
// getters/setters and most importantly equals() and hashCode()
}
Good solution, thanks
– Renan
Oct 3 '15 at 15:43
1
This is the cleanest solution you can get, well done.
– FrancescoM
Feb 3 '18 at 14:37
anyone can provide insert or delete example too.
– VK321
Oct 21 '18 at 7:37
Very cool, thank you thank you. Question... ...i get a NPE if i also put a generated @id column in the association class (EmployerDeliveryAgent in your example). But I want a RDBMS long PK... what say you? is there a way to do it? screw my surrogate key?
– tom
Nov 20 '18 at 18:42
add a comment |
Both answers from Eric Lucio and Renan helped, but there use of the ids in the association table is redundant. You have both the associated entities and their ids in the class. This is not required. You can simple map the associated entity in the association class with the @Id
on the associated entity field.
@Entity
public class Employer {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@OneToMany(mappedBy = "employer")
private List<EmployerDeliveryAgent> deliveryAgentAssoc;
// other properties and getters and setters
}
@Entity
public class DeliveryAgent {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@OneToMany(mappedBy = "deliveryAgent")
private List<EmployerDeliveryAgent> employerAssoc;
// other properties and getters and setters
}
The association class
@Entity
@Table(name = "employer_delivery_agent")
@IdClass(EmployerDeliveryAgentId.class)
public class EmployerDeliveryAgent {
@Id
@ManyToOne
@JoinColumn(name = "employer_id", referencedColumnName = "id")
private Employer employer;
@Id
@ManyToOne
@JoinColumn(name = "delivery_agent_id", referencedColumnName = "id")
private DeliveryAgent deliveryAgent;
@Column(name = "is_project_lead")
private boolean isProjectLead;
}
Still need the association PK class. Notice the fields names should correspond exactly to the field names in the association class, but the types should be the type of the id in the associated type.
public class EmployerDeliveryAgentId implements Serializable {
private int employer;
private int deliveryAgent;
// getters/setters and most importantly equals() and hashCode()
}
Good solution, thanks
– Renan
Oct 3 '15 at 15:43
1
This is the cleanest solution you can get, well done.
– FrancescoM
Feb 3 '18 at 14:37
anyone can provide insert or delete example too.
– VK321
Oct 21 '18 at 7:37
Very cool, thank you thank you. Question... ...i get a NPE if i also put a generated @id column in the association class (EmployerDeliveryAgent in your example). But I want a RDBMS long PK... what say you? is there a way to do it? screw my surrogate key?
– tom
Nov 20 '18 at 18:42
add a comment |
Both answers from Eric Lucio and Renan helped, but there use of the ids in the association table is redundant. You have both the associated entities and their ids in the class. This is not required. You can simple map the associated entity in the association class with the @Id
on the associated entity field.
@Entity
public class Employer {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@OneToMany(mappedBy = "employer")
private List<EmployerDeliveryAgent> deliveryAgentAssoc;
// other properties and getters and setters
}
@Entity
public class DeliveryAgent {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@OneToMany(mappedBy = "deliveryAgent")
private List<EmployerDeliveryAgent> employerAssoc;
// other properties and getters and setters
}
The association class
@Entity
@Table(name = "employer_delivery_agent")
@IdClass(EmployerDeliveryAgentId.class)
public class EmployerDeliveryAgent {
@Id
@ManyToOne
@JoinColumn(name = "employer_id", referencedColumnName = "id")
private Employer employer;
@Id
@ManyToOne
@JoinColumn(name = "delivery_agent_id", referencedColumnName = "id")
private DeliveryAgent deliveryAgent;
@Column(name = "is_project_lead")
private boolean isProjectLead;
}
Still need the association PK class. Notice the fields names should correspond exactly to the field names in the association class, but the types should be the type of the id in the associated type.
public class EmployerDeliveryAgentId implements Serializable {
private int employer;
private int deliveryAgent;
// getters/setters and most importantly equals() and hashCode()
}
Both answers from Eric Lucio and Renan helped, but there use of the ids in the association table is redundant. You have both the associated entities and their ids in the class. This is not required. You can simple map the associated entity in the association class with the @Id
on the associated entity field.
@Entity
public class Employer {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@OneToMany(mappedBy = "employer")
private List<EmployerDeliveryAgent> deliveryAgentAssoc;
// other properties and getters and setters
}
@Entity
public class DeliveryAgent {
@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
@OneToMany(mappedBy = "deliveryAgent")
private List<EmployerDeliveryAgent> employerAssoc;
// other properties and getters and setters
}
The association class
@Entity
@Table(name = "employer_delivery_agent")
@IdClass(EmployerDeliveryAgentId.class)
public class EmployerDeliveryAgent {
@Id
@ManyToOne
@JoinColumn(name = "employer_id", referencedColumnName = "id")
private Employer employer;
@Id
@ManyToOne
@JoinColumn(name = "delivery_agent_id", referencedColumnName = "id")
private DeliveryAgent deliveryAgent;
@Column(name = "is_project_lead")
private boolean isProjectLead;
}
Still need the association PK class. Notice the fields names should correspond exactly to the field names in the association class, but the types should be the type of the id in the associated type.
public class EmployerDeliveryAgentId implements Serializable {
private int employer;
private int deliveryAgent;
// getters/setters and most importantly equals() and hashCode()
}
edited May 23 '17 at 12:34
Community♦
11
11
answered Oct 3 '15 at 8:37
Paul SamsothaPaul Samsotha
155k21304499
155k21304499
Good solution, thanks
– Renan
Oct 3 '15 at 15:43
1
This is the cleanest solution you can get, well done.
– FrancescoM
Feb 3 '18 at 14:37
anyone can provide insert or delete example too.
– VK321
Oct 21 '18 at 7:37
Very cool, thank you thank you. Question... ...i get a NPE if i also put a generated @id column in the association class (EmployerDeliveryAgent in your example). But I want a RDBMS long PK... what say you? is there a way to do it? screw my surrogate key?
– tom
Nov 20 '18 at 18:42
add a comment |
Good solution, thanks
– Renan
Oct 3 '15 at 15:43
1
This is the cleanest solution you can get, well done.
– FrancescoM
Feb 3 '18 at 14:37
anyone can provide insert or delete example too.
– VK321
Oct 21 '18 at 7:37
Very cool, thank you thank you. Question... ...i get a NPE if i also put a generated @id column in the association class (EmployerDeliveryAgent in your example). But I want a RDBMS long PK... what say you? is there a way to do it? screw my surrogate key?
– tom
Nov 20 '18 at 18:42
Good solution, thanks
– Renan
Oct 3 '15 at 15:43
Good solution, thanks
– Renan
Oct 3 '15 at 15:43
1
1
This is the cleanest solution you can get, well done.
– FrancescoM
Feb 3 '18 at 14:37
This is the cleanest solution you can get, well done.
– FrancescoM
Feb 3 '18 at 14:37
anyone can provide insert or delete example too.
– VK321
Oct 21 '18 at 7:37
anyone can provide insert or delete example too.
– VK321
Oct 21 '18 at 7:37
Very cool, thank you thank you. Question... ...i get a NPE if i also put a generated @id column in the association class (EmployerDeliveryAgent in your example). But I want a RDBMS long PK... what say you? is there a way to do it? screw my surrogate key?
– tom
Nov 20 '18 at 18:42
Very cool, thank you thank you. Question... ...i get a NPE if i also put a generated @id column in the association class (EmployerDeliveryAgent in your example). But I want a RDBMS long PK... what say you? is there a way to do it? screw my surrogate key?
– tom
Nov 20 '18 at 18:42
add a comment |
OK, I got it working based on the solution available at
http://en.wikibooks.org/wiki/Java_Persistence/ManyToMany#Mapping_a_Join_Table_with_Additional_Columns.
This solution does not generate duplicate attributes on the database, but it does generate duplicate attributes in my JPA entities (which is very acceptable, since you can relay the extra work to a constructor or method - it ends up being transparent). The primary and foreign keys generated in the database are 100% correct.
As stated on the link, I couldn't use @PrimaryKeyJoinColumn and instead used @JoinColumn(name = "projectId", updatable = false, insertable = false, referencedColumnName = "id"). Another thing worth mentioning: I had to use EntityManager.persist(association), which is missing on the example at the link.
So my final solution is:
@Entity
public class Employee {
@Id
private long id;
...
@OneToMany(mappedBy="employee")
private List<ProjectAssociation> projects;
...
}
@Entity
public class Project {
@PersistenceContext
EntityManager em;
@Id
private long id;
...
@OneToMany(mappedBy="project")
private List<ProjectAssociation> employees;
...
// Add an employee to the project.
// Create an association object for the relationship and set its data.
public void addEmployee(Employee employee, boolean teamLead) {
ProjectAssociation association = new ProjectAssociation();
association.setEmployee(employee);
association.setProject(this);
association.setEmployeeId(employee.getId());
association.setProjectId(this.getId());
association.setIsTeamLead(teamLead);
em.persist(association);
this.employees.add(association);
// Also add the association object to the employee.
employee.getProjects().add(association);
}
}
@Entity
@Table(name="PROJ_EMP")
@IdClass(ProjectAssociationId.class)
public class ProjectAssociation {
@Id
private long employeeId;
@Id
private long projectId;
@Column(name="IS_PROJECT_LEAD")
private boolean isProjectLead;
@ManyToOne
@JoinColumn(name = "employeeId", updatable = false, insertable = false,
referencedColumnName = "id")
private Employee employee;
@ManyToOne
@JoinColumn(name = "projectId", updatable = false, insertable = false,
referencedColumnName = "id")
private Project project;
...
}
public class ProjectAssociationId implements Serializable {
private long employeeId;
private long projectId;
...
public int hashCode() {
return (int)(employeeId + projectId);
}
public boolean equals(Object object) {
if (object instanceof ProjectAssociationId) {
ProjectAssociationId otherId = (ProjectAssociationId) object;
return (otherId.employeeId == this.employeeId)
&& (otherId.projectId == this.projectId);
}
return false;
}
}
add a comment |
OK, I got it working based on the solution available at
http://en.wikibooks.org/wiki/Java_Persistence/ManyToMany#Mapping_a_Join_Table_with_Additional_Columns.
This solution does not generate duplicate attributes on the database, but it does generate duplicate attributes in my JPA entities (which is very acceptable, since you can relay the extra work to a constructor or method - it ends up being transparent). The primary and foreign keys generated in the database are 100% correct.
As stated on the link, I couldn't use @PrimaryKeyJoinColumn and instead used @JoinColumn(name = "projectId", updatable = false, insertable = false, referencedColumnName = "id"). Another thing worth mentioning: I had to use EntityManager.persist(association), which is missing on the example at the link.
So my final solution is:
@Entity
public class Employee {
@Id
private long id;
...
@OneToMany(mappedBy="employee")
private List<ProjectAssociation> projects;
...
}
@Entity
public class Project {
@PersistenceContext
EntityManager em;
@Id
private long id;
...
@OneToMany(mappedBy="project")
private List<ProjectAssociation> employees;
...
// Add an employee to the project.
// Create an association object for the relationship and set its data.
public void addEmployee(Employee employee, boolean teamLead) {
ProjectAssociation association = new ProjectAssociation();
association.setEmployee(employee);
association.setProject(this);
association.setEmployeeId(employee.getId());
association.setProjectId(this.getId());
association.setIsTeamLead(teamLead);
em.persist(association);
this.employees.add(association);
// Also add the association object to the employee.
employee.getProjects().add(association);
}
}
@Entity
@Table(name="PROJ_EMP")
@IdClass(ProjectAssociationId.class)
public class ProjectAssociation {
@Id
private long employeeId;
@Id
private long projectId;
@Column(name="IS_PROJECT_LEAD")
private boolean isProjectLead;
@ManyToOne
@JoinColumn(name = "employeeId", updatable = false, insertable = false,
referencedColumnName = "id")
private Employee employee;
@ManyToOne
@JoinColumn(name = "projectId", updatable = false, insertable = false,
referencedColumnName = "id")
private Project project;
...
}
public class ProjectAssociationId implements Serializable {
private long employeeId;
private long projectId;
...
public int hashCode() {
return (int)(employeeId + projectId);
}
public boolean equals(Object object) {
if (object instanceof ProjectAssociationId) {
ProjectAssociationId otherId = (ProjectAssociationId) object;
return (otherId.employeeId == this.employeeId)
&& (otherId.projectId == this.projectId);
}
return false;
}
}
add a comment |
OK, I got it working based on the solution available at
http://en.wikibooks.org/wiki/Java_Persistence/ManyToMany#Mapping_a_Join_Table_with_Additional_Columns.
This solution does not generate duplicate attributes on the database, but it does generate duplicate attributes in my JPA entities (which is very acceptable, since you can relay the extra work to a constructor or method - it ends up being transparent). The primary and foreign keys generated in the database are 100% correct.
As stated on the link, I couldn't use @PrimaryKeyJoinColumn and instead used @JoinColumn(name = "projectId", updatable = false, insertable = false, referencedColumnName = "id"). Another thing worth mentioning: I had to use EntityManager.persist(association), which is missing on the example at the link.
So my final solution is:
@Entity
public class Employee {
@Id
private long id;
...
@OneToMany(mappedBy="employee")
private List<ProjectAssociation> projects;
...
}
@Entity
public class Project {
@PersistenceContext
EntityManager em;
@Id
private long id;
...
@OneToMany(mappedBy="project")
private List<ProjectAssociation> employees;
...
// Add an employee to the project.
// Create an association object for the relationship and set its data.
public void addEmployee(Employee employee, boolean teamLead) {
ProjectAssociation association = new ProjectAssociation();
association.setEmployee(employee);
association.setProject(this);
association.setEmployeeId(employee.getId());
association.setProjectId(this.getId());
association.setIsTeamLead(teamLead);
em.persist(association);
this.employees.add(association);
// Also add the association object to the employee.
employee.getProjects().add(association);
}
}
@Entity
@Table(name="PROJ_EMP")
@IdClass(ProjectAssociationId.class)
public class ProjectAssociation {
@Id
private long employeeId;
@Id
private long projectId;
@Column(name="IS_PROJECT_LEAD")
private boolean isProjectLead;
@ManyToOne
@JoinColumn(name = "employeeId", updatable = false, insertable = false,
referencedColumnName = "id")
private Employee employee;
@ManyToOne
@JoinColumn(name = "projectId", updatable = false, insertable = false,
referencedColumnName = "id")
private Project project;
...
}
public class ProjectAssociationId implements Serializable {
private long employeeId;
private long projectId;
...
public int hashCode() {
return (int)(employeeId + projectId);
}
public boolean equals(Object object) {
if (object instanceof ProjectAssociationId) {
ProjectAssociationId otherId = (ProjectAssociationId) object;
return (otherId.employeeId == this.employeeId)
&& (otherId.projectId == this.projectId);
}
return false;
}
}
OK, I got it working based on the solution available at
http://en.wikibooks.org/wiki/Java_Persistence/ManyToMany#Mapping_a_Join_Table_with_Additional_Columns.
This solution does not generate duplicate attributes on the database, but it does generate duplicate attributes in my JPA entities (which is very acceptable, since you can relay the extra work to a constructor or method - it ends up being transparent). The primary and foreign keys generated in the database are 100% correct.
As stated on the link, I couldn't use @PrimaryKeyJoinColumn and instead used @JoinColumn(name = "projectId", updatable = false, insertable = false, referencedColumnName = "id"). Another thing worth mentioning: I had to use EntityManager.persist(association), which is missing on the example at the link.
So my final solution is:
@Entity
public class Employee {
@Id
private long id;
...
@OneToMany(mappedBy="employee")
private List<ProjectAssociation> projects;
...
}
@Entity
public class Project {
@PersistenceContext
EntityManager em;
@Id
private long id;
...
@OneToMany(mappedBy="project")
private List<ProjectAssociation> employees;
...
// Add an employee to the project.
// Create an association object for the relationship and set its data.
public void addEmployee(Employee employee, boolean teamLead) {
ProjectAssociation association = new ProjectAssociation();
association.setEmployee(employee);
association.setProject(this);
association.setEmployeeId(employee.getId());
association.setProjectId(this.getId());
association.setIsTeamLead(teamLead);
em.persist(association);
this.employees.add(association);
// Also add the association object to the employee.
employee.getProjects().add(association);
}
}
@Entity
@Table(name="PROJ_EMP")
@IdClass(ProjectAssociationId.class)
public class ProjectAssociation {
@Id
private long employeeId;
@Id
private long projectId;
@Column(name="IS_PROJECT_LEAD")
private boolean isProjectLead;
@ManyToOne
@JoinColumn(name = "employeeId", updatable = false, insertable = false,
referencedColumnName = "id")
private Employee employee;
@ManyToOne
@JoinColumn(name = "projectId", updatable = false, insertable = false,
referencedColumnName = "id")
private Project project;
...
}
public class ProjectAssociationId implements Serializable {
private long employeeId;
private long projectId;
...
public int hashCode() {
return (int)(employeeId + projectId);
}
public boolean equals(Object object) {
if (object instanceof ProjectAssociationId) {
ProjectAssociationId otherId = (ProjectAssociationId) object;
return (otherId.employeeId == this.employeeId)
&& (otherId.projectId == this.projectId);
}
return false;
}
}
edited Nov 22 '18 at 12:20
Sofia Paixão
119113
119113
answered May 26 '14 at 20:25
RenanRenan
5841527
5841527
add a comment |
add a comment |
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