I've got two entities that I want to join together using a field they have in common, called shared_id. The field is not the primary key of either entity. The shared_id is unique - each Hipster will have a unique shared_id.
The tables look like:
Hipster Fixie
========= ========
id id
shared_id shared_id
There is a OneToMany relationship between Hipsters and their Fixies. I've tried something like this:
#Entity
public class Hipster {
#Id
#Column(name = "id")
private Integer id;
#Column(name = "shared_id")
private Integer sharedId;
#OneToMany(mappedBy = "hipster")
private List<Fixie> fixies;
}
#Entity
public class Fixie {
#Id
#Column(name = "id")
private Integer id;
#ManyToOne
#JoinColumn(name = "shared_id", referencedColumnName = "shared_id")
private Hipster hipster;
}
#Repository
public class HipsterDAO {
#PersistenceContext
private EntityManager entityManager;
public Hipster getHipsterBySharedId(Integer sharedId) {
String queryString = "SELECT h FROM Hipster h WHERE h.sharedId = :sharedId";
TypedQuery<Hipster> query = entityManager.createQuery(queryString, Hipster.class);
query.setParameter("sharedId", sharedId);
try {
return query.getSingleResult();
} catch (PersistenceException e) {
return null;
}
}
}
Now, my DAO gives me this error:
java.lang.IllegalArgumentException: Can not set java.lang.Integer field Hipster.sharedId to java.lang.Integer
I think it's upset because the sharedId field is used in a relation, rather than just being a basic field. I haven't included the sharedId field in the Fixie entity, but I get the same result if I do. How do I persuade it to run this query for me? Do I need to change the query or the entities?
Related
I created the following models:
"Vendor"
"PickupStation"
And both of them have a OneToMany Relationship to a composite Key
"PickupStationVendorDetails" which has an embedded Id "PickupStationVendorKey"
It works to save the "PickupStationVendorDetails" with the corresponding Vendor and PickupStation but when I want to fetch them from e.g the Vendor nothing is found.
Vendor.java
#Entity
public class Vendor {
...
#OneToMany(mappedBy = "vendor")
private Set<PickupStationVendorDetails> pickupStations;
}
PickupStation.java
#Entity
public class PickupStation {
#OneToMany(mappedBy = "pickupStation")
private Set<PickupStationVendorDetails> vendors;
}
PickupStationVendorDetails.java
#Entity
public class PickupStationVendorDetails {
#EmbeddedId
private PickupStationVendorKey id;
#ManyToOne
#MapsId("vendorId")
#JoinColumn(name = "vendor_id")
private Vendor vendor;
#ManyToOne
#MapsId("pickupStationId")
#JoinColumn(name = "pickup_station_id")
private PickupStation pickupStation;
}
PickupStationVendorKey.java
#Embeddable
public class PickupStationVendorKey implements Serializable {
#Column(name = "vendor_id", columnDefinition = "BINARY(16)")
private UUID vendorId;
#Column(name = "pickup_station_id")
private Long pickupStationId;
public PickupStationVendorKey() {
}
public PickupStationVendorKey(UUID vendorId, Long pickupStationId) {
this.vendorId = vendorId;
this.pickupStationId = pickupStationId;
}
....
}
How I persist the entities:
At first I create the embeddedID and save the details via repository:
PickupStationVendorDetails pickupStationVendorDetails = new PickupStationVendorDetails();
pickupStationVendorDetails.setVendor(vendor);
pickupStationVendorDetails.setPickupStation(pickupStation);
pickupStationVendorDetails.setDeliveryDays(relationship.getDeliveryDays());
PickupStationVendorKey embeddedId = new PickupStationVendorKey(vendor.getId(),pickupStation.getId());
pickupStationVendorDetails.setId(embeddedId);
PickupStationVendorDetails d = pickupStationVendorDetailsRepository.save(pickupStationVendorDetails);
Afterwards I add them to the Set<> of the corresponding Entities and save them too.
vendor.getPickupStations().add(d);
pickupStation.getVendors().add(d);
vendorService.save(vendor);
pickupStationRepository.save(pickupStation);
And when I try to call vendor.getPickupStations() there seems to be no relationship.
Except I call pickupStationVendorDetailsRepository.findAll() the composite Key is correctly persisted and saved, and from there on I would be able to get the PickupStation and the Vendor. But that's not how it should work I guess.
Am I missing something?
Customer said no Primary Key Required in Child Table. So Child table has two column "ID" and "Value" where ID can be duplicated.
When i remove #Id then hibernate says "No identifier specified for entity"
When i keep #Id in code then hibernate says "javax.persistence.EntityExistsException: a different object with the same identifier value was already associated with the session" ; while persisting
So crux is that i need to keep #Id but how to persist two same ID in one session with #Id annotation.
Following is code:
Main Entity:
public class CustomerAgreement implements Serializable {
#OneToMany(mappedBy = "customerAgreement", orphanRemoval = true, fetch = FetchType.LAZY, cascade = {CascadeType.PERSIST})
private List<CustomerAgreementComputerAttachments> autoAttachComputersFromOrganizations;
Composed Entity:
public class CustomerAgreementComputerAttachments implements Serializable{
private static final long serialVersionUID = 1L;
#Id
#ManyToOne
#JoinColumn(name = "ID")
private CustomerAgreement customerAgreement;
Main Program:
public static List<CustomerAgreement> create() {
List<CustomerAgreement> li = new ArrayList<CustomerAgreement>();
CustomerAgreement cAgreement = new CustomerAgreement();
cAgreement.setId(2222l);
cAgreement.setName("Tillu");;
cAgreement.setCustomerId("140");
List<CustomerAgreementComputerAttachments> catl = new ArrayList<>();
CustomerAgreementComputerAttachments catt = new CustomerAgreementComputerAttachments();
catt.setAttachmentValue("TEST");
catt.setCustomerAgreement(cAgreement);
CustomerAgreementComputerAttachments tatt = new CustomerAgreementComputerAttachments();
tatt.setAttachmentValue("TESTy");
tatt.setCustomerAgreement(cAgreement);
catl.add(catt);
catl.add(tatt);
cAgreement.setAutoAttachComputersFromOrganizations(catl);
li.add(cAgreement);
return li;
}
public static void main(String[] args) {
EntityManagerFactory emf = Persistence.createEntityManagerFactory("IntegratorMasterdataPU");
em = emf.createEntityManager();
em.getTransaction().begin();
for(CustomerAgreement ca: create()) {
em.persist(ca);
}
em.getTransaction().commit();
}
An Entity must be identifiable by a unique key. This doesn't need to correspond to any database primary key but there must a unique column or columns that can be used to identity an entity.
If this is not possible, then you need to make CustomerAgreementComputerAttachment an #Embeddable.
An #Embeddable unlike an entity has no independent identity (no #ID). See further here:
What is difference between #Entity and #embeddable
#Entity
public class CustomerAgreement {
#ElementCollection
#JoinTable(name="...", joinColumn = "id")
private List<CustomerAgreementComputerAttachment> attachments;
}
and
#Embeddable
public class CustomerAgreementComputerAttachments {
//No back reference to CustomerAgreement
//Other fields as required.
}
Have a "full Entity" class:
#Entity(name = "vacancy_dec_to_words")
public class VacancyDescriptionToWords {
#Id
#Column(name = "id")
#GeneratedValue(strategy = GenerationType.IDENTITY)
private long id;
#JoinColumn(name = "vacancy_description_id")
#ManyToOne(cascade = CascadeType.ALL)
private VacancyDescription vacancyDescription;
#JoinColumn(name = "words_id")
#ManyToOne
private Words words;
#Column(name = "qty")
private int qty;
#Column(name = "create_date")
private Date date;
//...getters and setters
In some methods I need use only 2 column from this database: word_id and qty
I try the follow way:
Projections
https://docs.spring.io/spring-data/jpa/docs/2.1.2.RELEASE/reference/html/#projections
public interface QtyWords {
Long getWords();
Integer getQty();
}
JpaReposytory:
*Query, that I use tested and it workable, I use him in JpaRepository:
#Repository
public interface SmallVDTWRepository extends JpaRepository<VacancyDescriptionToWords, Long> {
#Query(nativeQuery = true,
value = "SELECT sum(qty), words_id FROM vacancy_desc_to_words WHERE vacancy_description_id IN (" +
"SELECT id FROM vacancy_description WHERE vacancy_id IN (" +
"SELECT id FROM vacancy WHERE explorer_id = :exp))" +
"GROUP BY words_id")
List<QtyWords> getDistinctWordsByExplorer(#Param("exp") long exp);
}
But I get some interesting result when I get list of entities:
List<QtyWords> list = vdtwService.getByExplorerId(72);
I am not get any exceptions, but I have the list with are unknowns objects. This objects contains my data, which I need(qty and words_id), but I cannot get them from him.
Can I use this method (Projection) to implement this task and, in general, how to correctly implement the 'Light Entity' in this case?
Spring provides two mechanisms that can be used to limit data to be fetched.
Projections
Projections can help you to reduce data, retrieved from database, by setting what exactly attributes you want to fetch.
Example:
#Entity
class Person {
#Id UUID id;
String firstname, lastname;
#OneToOne
Address address;
}
#Entity
static class Address {
#Id UUID id;
String zipCode, city, street;
}
interface NamesOnly {
String getFirstname();
String getLastname();
}
#Repository
interface PersonRepository extends Repository<Person, UUID> {
Collection<NamesOnly> findByLastname(String lastname);
}
Entity graph
Annotation EntityGraph can help you to reduce amount of queries to database, by setting what exactly related entities you need to fetch.
Example:
#Entity
#NamedEntityGraph(name = "GroupInfo.detail", attributeNodes = #NamedAttributeNode("members"))
public class GroupInfo {
#Id UUID id;
#ManyToMany //default fetch mode is lazy.
List<GroupMember> members = new ArrayList<GroupMember>();
}
#Repository
public interface GroupRepository extends CrudRepository<GroupInfo, String> {
#EntityGraph(value = "GroupInfo.detail", type = EntityGraphType.LOAD)
GroupInfo getByGroupName(String name); //Despite of GroupInfo.members has FetchType = LAZY, it will be fetched because of using EntityGraph
}
There are two types of EntityGraph:
EntityGraphType.LOAD - is used to specify an entity graph, attributes that are specified by attribute nodes of the entity graph are treated as FetchType.EAGER and attributes that are not specified are treated according to their specified or default FetchType.
EntityGraphType.FETCH - is used to specify an entity graph, attributes that are specified by attribute nodes of the entity graph are treated as FetchType.EAGER and attributes that are not specified are treated as FetchType.LAZY.
PS: Also remember that you can set lazy fetch type: #ManyToOne(fetch = FetchType.LAZY) and JPA will not fetching child entities when parent is being fetched.
Lets say I have two objects, say one is a User object and the other is a State Object. The state object is basically the 50 states of America so it doesn't ever have to change. The user object however has a Collection of States where the user has been. So like this:
#Entity
#Table(name="tbl_users")
class User {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name="id", unique=true, nullable = false)
private int id;
#Column(name="user_name")
private String name;
#OneToMany(targetEntity=State.class, orphanRemoval = false)
#Column(name="states")
private Collection<State> states;
//getters and setters
}
and the States entity looks like this:
#Entity
#Table(name="tbl_states")
class State {
#Id
#Column(name="id", unique=true, nullable=false)
private int id;
#Column(name="state")
private String state;
// getters and setters
}
Code for adding user (using hibernate):
public int addUser(User user) {
em.persist(user);
em.flush();
return user.getId();
}
Code for getting state by id:
public State getStateById(int id) {
return em.createQuery("SELECT s FROM State s WHERE s.id =:id, State.class)
.setParameter("id", id)
.getSingleResult();
}
but when I try to create a User and pick several states, I get a PSQLException:
org.postgresql.util.PSQLException: ERROR: duplicate key value violates unique constraint "uk_g6pr701i2pcq7400xrlb0hns"
2017-06-21T22:54:35.959991+00:00 app[web.1]: Detail: Key (states_id)=(5) already exists.
I tried looking up the Cascade methods to see if I could use any, but Cascade.MERGE and Cascade.PERSIST seem to do the same thing, and the rest I don't think I need (REMOVE, DETACH, etc). My question is:
How do I add states to the User object without having that error?
This code works:
class Example {
#Test
public void workingTest() {
EntityManagerFactory emf = Persistence.createEntityManagerFactory("testPU");
EntityManager em = emf.createEntityManager();
// Creating three states
State alabama = new State(state: 'Alabama');
State louisiana = new State(state: 'Louisiana');
State texas = new State(state: 'Texas');
em.getTransaction().begin();
em.persist(alabama);
em.persist(louisiana);
em.persist(texas);
em.getTransaction().commit();
List<State> states = em.createQuery('FROM State').getResultList();
// Assert only three states on DB
assert states.size() == 3;
User userFromAlabama = new User();
User userFromAlabamaAndTexas = new User();
em.getTransaction().begin();
State alabamaFromDB = em.find(State, alabama.getId());
State texasFromDB = em.find(State, texas.getId());
userFromAlabama.getStates().add(alabamaFromDB);
userFromAlabamaAndTexas.getStates().add(alabamaFromDB);
userFromAlabamaAndTexas.getStates().add(texasFromDB);
em.persist(userFromAlabama);
em.persist(userFromAlabamaAndTexas);
em.getTransaction().commit();
states = em.createQuery('FROM State').getResultList();
// Assert only three states on DB again
assert states.size() == 3;
// Assert one user
User userFromDB = em.find(User, userFromAlabama.getId());
assert userFromDB.getStates().size() == 1;
userFromDB = em.find(User, userFromAlabamaAndTexas.getId());
assert userFromDB.getStates().size() == 2;
}
}
#Entity
#Table(name="tbl_users")
class User {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private int id
#Column(name="user_name")
private String name
#ManyToMany
private Collection<State> states = Lists.newArrayList()
// Getters and setters
}
#Entity
#Table(name="tbl_states")
class State {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private int id;
#Column(name="state")
private String state;
// Getters and setters
}
You should change your mapping to #ManyToMany!
And you must have 3 tables on DB like this:
TBL_USERS, TBL_STATES and TBL_USERS_TBL_STATES
The TBL_USERS_TBL_STATES table is the default table name that Hibernate uses when a property is annotated with #ManyToMany. If you want to change the tablename of TBL_USERS_TBL_STATES, use the #JoinTable annotation too. See the docs here
With this configuration, you should be able to fetch a State from database, add it to a new User and then persist it. I made a unit test and It works!
In your case it might be better to use a manytomany association with manytomany hibernate dont generate unicity constraint.
Hibernate auto generation scheme behavior is a little bit strange with onetoMany but you can use this workaround.
Try this:
#ManyToMany
#JoinTable(name = "user_state")
private List<State> states;
I'm trying to set up the following tables using JPA/Hibernate:
User:
userid - PK
name
Validation:
userid - PK, FK(user)
code
There may be many users and every user may have max one validation code or none.
Here's my classes:
public class User
{
#Id
#Column(name = "userid")
#GeneratedValue(strategy = GenerationType.IDENTITY)
protected Long userId;
#Column(name = "name", length = 50, unique = true, nullable = false)
protected String name;
...
}
public class Validation
{
#Id
#Column(name = "userid")
protected Long userId;
#OneToOne(cascade = CascadeType.ALL)
#PrimaryKeyJoinColumn(name = "userid", referencedColumnName = "userid")
protected User user;
#Column(name = "code", length = 10, unique = true, nullable = false)
protected String code;
...
public void setUser(User user)
{
this.user = user;
this.userId = user.getUserId();
}
...
}
I create a user and then try to add a validation code using the following code:
public void addValidationCode(Long userId)
{
EntityManager em = createEntityManager();
EntityTransaction tx = em.getTransaction();
try
{
tx.begin();
// Fetch the user
User user = retrieveUserByID(userId);
Validation validation = new Validation();
validation.setUser(user);
em.persist(validation);
tx.commit();
}
...
}
When I try to run it I get a org.hibernate.PersistentObjectException: detached entity passed to persist: User
I have also tried to use the following code in my Validation class:
public void setUserId(Long userId)
{
this.userId = userId;
}
and when I create a validation code I simply do:
Validation validation = new Validation();
validation.setUserId(userId);
em.persist(validation);
tx.commit();
But then since User is null I get org.hibernate.PropertyValueException: not-null property references a null or transient value: User.code
Would appreciate any help regarding how to best solve this issue!
I have been able to solve this problem of "OneToOne between two tables with shared primary key" in pure JPA 2.0 way(Thanks to many existing threads on SOF). In fact there are two ways in JPA to handle this. I have used eclipselink as JPA provider and MySql as database. To highlight once again no proprietary eclipselink classes have been used here.
First approach is to use AUTO generation type strategy on the Parent Entity's Identifier field.
Parent Entity must contain the Child Entity Type member in OneToOne relationship(cascade type PERSIST and mappedBy = Parent Entity Type member of Child Entity)
#Entity
#Table(name = "USER_LOGIN")
public class UserLogin implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name="USER_ID")
private Integer userId;
#OneToOne(cascade = CascadeType.PERSIST, mappedBy = "userLogin")
private UserDetail userDetail;
// getters & setters
}
Child Entity must not contain an identifier field. It must contain a member of Parent Entity Type with Id, OneToOne and JoinColumn annotations. JoinColumn must specify the ID field name of the DB table.
#Entity
#Table(name = "USER_DETAIL")
public class UserDetail implements Serializable {
#Id
#OneToOne
#JoinColumn(name="USER_ID")
private UserLogin userLogin;
// getters & setters
}
Above approach internally uses a default DB table named SEQUENCE for assigning the values to the identifier field. If not already present, This table needs to be created as below.
DROP TABLE TEST.SEQUENCE ;
CREATE TABLE TEST.SEQUENCE (SEQ_NAME VARCHAR(50), SEQ_COUNT DECIMAL(15));
INSERT INTO TEST.SEQUENCE(SEQ_NAME, SEQ_COUNT) values ('SEQ_GEN', 0);
Second approach is to use customized TABLE generation type strategy and TableGenerator annotation on the Parent Entity's Identifier field.
Except above change in identifier field everything else remains unchanged in Parent Entity.
#Entity
#Table(name = "USER_LOGIN")
public class UserLogin implements Serializable {
#Id
#TableGenerator(name="tablegenerator", table = "APP_SEQ_STORE", pkColumnName = "APP_SEQ_NAME", pkColumnValue = "USER_LOGIN.USER_ID", valueColumnName = "APP_SEQ_VALUE", initialValue = 1, allocationSize = 1 )
#GeneratedValue(strategy = GenerationType.TABLE, generator = "tablegenerator")
#Column(name="USER_ID")
private Integer userId;
#OneToOne(cascade = CascadeType.PERSIST, mappedBy = "userLogin")
private UserDetail userDetail;
// getters & setters
}
There is no change in Child Entity. It remains same as in the first approach.
This table generator approach internally uses a DB table APP_SEQ_STORE for assigning the values to the identifier field. This table needs to be created as below.
DROP TABLE TEST.APP_SEQ_STORE;
CREATE TABLE TEST.APP_SEQ_STORE
(
APP_SEQ_NAME VARCHAR(255) NOT NULL,
APP_SEQ_VALUE BIGINT NOT NULL,
PRIMARY KEY(APP_SEQ_NAME)
);
INSERT INTO TEST.APP_SEQ_STORE VALUES ('USER_LOGIN.USER_ID', 0);
If you use Hibernate you can also use
public class Validation {
private Long validationId;
private User user;
#Id
#GeneratedValue(generator="SharedPrimaryKeyGenerator")
#GenericGenerator(name="SharedPrimaryKeyGenerator",strategy="foreign",parameters = #Parameter(name="property", value="user"))
#Column(name = "VALIDATION_ID", unique = true, nullable = false)
public Long getValidationId(){
return validationId;
}
#OneToOne
#PrimaryKeyJoinColumn
public User getUser() {
return user;
}
}
Hibernate will make sure that the ID of Validation will be the same as the ID of the User entity set.
Are you using JPA or JPA 2.0 ?
If Validation PK is a FK to User, then you do not need the Long userId attribute in validation class, but instead do the #Id annotation alone. It would be:
Public class Validation
{
#Id
#OneToOne(cascade = CascadeType.ALL)
#PrimaryKeyJoinColumn(name = "userid", referencedColumnName = "userid")
protected User user;
#Column(name = "code", length = 10, unique = true, nullable = false)
protected String code;
...
public void setUser(User user)
{
this.user = user;
this.userId = user.getUserId();
}
...
}
Try with it and tell us your results.
You need to set both userId and user.
If you set just the user, then the id for Validation is 0 and is deemed detached. If you set just the userId, then you need to make the user property nullable, which doesn't make sense here.
To be safe, you can probably set them both in one method call:
#Transient
public void setUserAndId(User user){
this.userId = user.getId();
this.user = user;
}
I marked the method #Transient so that Hibernate will ignore it. Also, so you can still have setUser and setUserId work as expected with out any "side effects."