many-to-one

How to establish a many to one relationship between 2 tables for a field

断了今生、忘了曾经 提交于 2019-12-23 05:06:52
问题 I am building an database schema for a project and I am bit stuck on this issue. I have 2 tables: USER table: Id Name Contact_ID 1 Arun 2 2 Barath 3 3 Charan 2 4 Dinesh 1 CONTACT table: ID Name Phone Mail 1 Mahesh 1234 Mahesh@Yahoo.com 2 Suresh 54321 Sureh@Google.com 3 Jayesh 9876 Jayesh@Bing.com 4 Ganesh 98754 Gahesh@Safari.com Each of the users in USER will have a contact in CONTACT. If a user has a single contact then I can use a foreign key relationship on Contact_ID in USER and build the

In JPA, having a many-to-one as primary key throws referential integrity constraint violation

你说的曾经没有我的故事 提交于 2019-12-23 03:18:10
问题 I have defined the following entities: @Entity public class Child implements Serializable { @Id @ManyToOne(cascade = CascadeType.ALL) public Parent parent; @Id public int id; } @Entity public class Parent { @Id public int id; } When I try to persist a Child with the following code: Parent p = new Parent(); p.id = 1; Child c1 = new Child(); c1.id = 1; c1.parent = p; em.persist(c1); Hibernate throws a 'Referential integrity constraint violation' error: Caused by: org.h2.jdbc.JdbcSQLException:

Hibernate @Filter on @ManyToOne

别来无恙 提交于 2019-12-22 08:55:23
问题 Could some please explain me why Having a @Filter on a @ManyToOne relation does not work? Here i have a really simple example showing this: I have created two simple table in my database (foo and bar) Foo: id/id_bar bar: id/name/state (state can be active or inactive) and my Entities: @Entity @FilterDef(name = "foo_active") @Table(name = "foo") public class Foo extends AbstractTimestampEntity implements Serializable { private static final long serialVersionUID = 1L; @Id @Column(name = "id")

Symfony ManyToOne relationship getter returns empty object

僤鯓⒐⒋嵵緔 提交于 2019-12-22 06:47:03
问题 I'll simplifly my code, I have te next: Doctor entity: use ...\...\Entity\Paciente; class Doctor extends Usuario { public function __construct() { ... $this->pacientes = new ArrayCollection(); ... } /** * Número de colegiado - numColegiado * * @var string * * @ORM\Column(name="numColegiado", type="string", length=255, unique=true) */ protected $numColegiado; /** * @ORM\OneToMany(targetEntity="Paciente", mappedBy="doctor") * @var \Doctrine\Common\Collections\ArrayCollection */ private

JPA @OneToMany and composite PK

给你一囗甜甜゛ 提交于 2019-12-22 03:58:10
问题 I am working on a JPA project. I need to use a @OneToMany mapping on a class that has three primary keys. You can find the errors and the classes after this. javax.persistence.PersistenceException: No Persistence provider for EntityManager named JTA_pacePersistence: Provider named oracle.toplink.essentials.PersistenceProvider threw unexpected exception at create EntityManagerFactory: javax.persistence.PersistenceException javax.persistence.PersistenceException: Exception [TOPLINK-28018]

How to lazy load a one-to-one composition via hql

≯℡__Kan透↙ 提交于 2019-12-22 02:06:06
问题 If have an entity A with a bidirectional one-or-zero-to-one mapping with entity B. The mapping is as follows: <class name="EntityA" table="TABLE_A" mutable="true" lazy="true"> <id name="idA" type="long" column="pk_a" unsaved-value="null"> <generator class="sequence"> <param name="sequence">pk_a_seq</param> </generator> </id> <one-to-one name="propertyB" class="EntityB" property-ref="propertyA" constrained="true" outer-join="false"/> </class> and <class name="EntityB" table="TABLE_B" mutable=

How to lazy load a one-to-one composition via hql

被刻印的时光 ゝ 提交于 2019-12-22 02:05:22
问题 If have an entity A with a bidirectional one-or-zero-to-one mapping with entity B. The mapping is as follows: <class name="EntityA" table="TABLE_A" mutable="true" lazy="true"> <id name="idA" type="long" column="pk_a" unsaved-value="null"> <generator class="sequence"> <param name="sequence">pk_a_seq</param> </generator> </id> <one-to-one name="propertyB" class="EntityB" property-ref="propertyA" constrained="true" outer-join="false"/> </class> and <class name="EntityB" table="TABLE_B" mutable=

Python many-to-one mapping (creating equivalence classes)

 ̄綄美尐妖づ 提交于 2019-12-18 13:04:39
问题 I have a project of converting one database to another. One of the original database columns defines the row's category. This column should be mapped to a new category in the new database. For example, let's assume the original categories are: parrot, spam, cheese_shop, Cleese, Gilliam, Palin Now that's a little verbose for me, And I want to have these rows categorized as sketch, actor - That is, define all the sketches and all the actors as two equivalence classes. >>> monty={'parrot':

JPA many-to-one relation - need to save only Id

一曲冷凌霜 提交于 2019-12-18 10:46:41
问题 I have 2 classes: Driver and Car. Cars table updated in separate process. What I need is to have property in Driver that allows me to read full car description and write only Id pointing to existing Car. Here is example: @Entity(name = "DRIVER") public class Driver { ... ID and other properties for Driver goes here ..... @ManyToOne(fetch=FetchType.LAZY) @JoinColumn(name = "CAR_ID") private Car car; @JsonView({Views.Full.class}) public Car getCar() { return car; } @JsonView({Views.Short.class}

JPA Composite key with ManyToOne getting org.hibernate.PropertyAccessException: could not set a field value by reflection setter of

☆樱花仙子☆ 提交于 2019-12-18 05:45:21
问题 I have a composite key ContractServiceLocationPK made out of three id's ( contractId , locationId , serviceId ) of type long in an embeddable class. The class which uses this composite key, ContractServiceLocation , maps these ids, using @MapsId annotation, to their objects. Here's how it looks like (removed setters/getters and irrelevant properties): Contract @Entity @Table(name = "Contract") public class Contract implements Serializable { public Contract() { } @Id @GeneratedValue private