hibernate-mapping

org.hibernate.MappingException: Could not determine type for: java.util.Set, at table: USERS, for columns: [org.hibernate.mapping.Column(invoices)]

依然范特西╮ 提交于 2019-12-28 10:51:12
问题 I have a problem that Hibernate is unable to determine the type for Set at the table USERS. I am trying to create a foreign key of table INVOICES through one-to-many relationship. One User can generate many Invoices. My User.java is given below. @Entity @Table(name="USERS") public class User { @Id @Column(name="User_Id",nullable=false) @GeneratedValue(strategy=GenerationType.AUTO) private Integer user_id; @Column(name="NAME") private String name; @Column(name="Address") private String address

How to create hibernate composite key using annotations

雨燕双飞 提交于 2019-12-28 05:07:09
问题 I am trying to use hibernate annotations to insert data to a MySQL database table which doesn't have a primary key defined. However the fact is 2 fields of that table together are unique in the table.how can i achieve the same using hibernate annotation?. here is my code.. @Entity @Table(name = "RolesMenuItems") public class RolesMenuItems { @Column(name = "RoleID") private String roleID; @Column(name = "MenuItemID") private String menuItemID; /*setter getter methods */ } 回答1: You can use

Hibernate - A collection with cascade=”all-delete-orphan” was no longer referenced by the owning entity instance

家住魔仙堡 提交于 2019-12-27 11:39:51
问题 I'm having the following issue when trying to update my entity: "A collection with cascade=”all-delete-orphan” was no longer referenced by the owning entity instance". I have a parent entity and it has a Set<...> of some children entities. When I try to update it, I get all the references to be set to this collections and set it. The following code represents my mapping: @OneToMany(mappedBy = "parentEntity", fetch = FetchType.EAGER) @Cascade({ CascadeType.ALL, CascadeType.DELETE_ORPHAN })

Hibernate - A collection with cascade=”all-delete-orphan” was no longer referenced by the owning entity instance

丶灬走出姿态 提交于 2019-12-27 11:38:12
问题 I'm having the following issue when trying to update my entity: "A collection with cascade=”all-delete-orphan” was no longer referenced by the owning entity instance". I have a parent entity and it has a Set<...> of some children entities. When I try to update it, I get all the references to be set to this collections and set it. The following code represents my mapping: @OneToMany(mappedBy = "parentEntity", fetch = FetchType.EAGER) @Cascade({ CascadeType.ALL, CascadeType.DELETE_ORPHAN })

Hibernate - A collection with cascade=”all-delete-orphan” was no longer referenced by the owning entity instance

你离开我真会死。 提交于 2019-12-27 11:37:03
问题 I'm having the following issue when trying to update my entity: "A collection with cascade=”all-delete-orphan” was no longer referenced by the owning entity instance". I have a parent entity and it has a Set<...> of some children entities. When I try to update it, I get all the references to be set to this collections and set it. The following code represents my mapping: @OneToMany(mappedBy = "parentEntity", fetch = FetchType.EAGER) @Cascade({ CascadeType.ALL, CascadeType.DELETE_ORPHAN })

How to map calculated properties with JPA and Hibernate

本小妞迷上赌 提交于 2019-12-27 09:56:14
问题 My Java bean has a childCount property. This property is not mapped to a database column . Instead, it should be calculated by the database with a COUNT() function operating on the join of my Java bean and its children. It would be even better if this property could be calculated on demand / "lazily", but this is not mandatory. In the worst case scenario, I can set this bean's property with HQL or the Criteria API, but I would prefer not to. The Hibernate @Formula annotation may help, but I

How to map calculated properties with JPA and Hibernate

柔情痞子 提交于 2019-12-27 09:56:01
问题 My Java bean has a childCount property. This property is not mapped to a database column . Instead, it should be calculated by the database with a COUNT() function operating on the join of my Java bean and its children. It would be even better if this property could be calculated on demand / "lazily", but this is not mandatory. In the worst case scenario, I can set this bean's property with HQL or the Criteria API, but I would prefer not to. The Hibernate @Formula annotation may help, but I

Confusion in giving hibernate association annotation

风格不统一 提交于 2019-12-25 09:16:06
问题 I have a parent object lets say Campaign and child object Ad Segments. Campaign is holding one to many relation with Ad Segments. following is the annotations I defined. class CampaignModel { OneToMany(mappedBy="campaign", cascade=CascadeType.PERSIST, fetch = FetchType.EAGER) @Setter @Column(updatable=false) private List<AdSegmentModel> segments; ..... } class AdSegmentModel { @ManyToOne(fetch = FetchType.LAZY) @JoinColumn(name="campaign_id") @Getter @Setter private CampaignModel campaign;

Bidirectional @OneToMany relation with JPA annotations doesn't seem to work

霸气de小男生 提交于 2019-12-25 08:18:09
问题 This is in reference to this answer. Entities- // Many to One @Entity @Table public class Address { @Id @GeneratedValue @Column private int addressIdentity; @Column private int houseNo; @Column private char streetNo; @Column private int pincode; @Column private String city; @Column private String state; @Column private String country; @ManyToOne @JoinTable(name="PersonAddress", joinColumns=@JoinColumn(name="addressId", insertable = false, updatable = false), inverseJoinColumns=@JoinColumn

Hibernate SQLQuery with transient properties

怎甘沉沦 提交于 2019-12-25 02:39:19
问题 I need to make an hibernate SQLQuery with db2 and this query is returning me some fields which are calculated and have no relation with any columns in database. The goal is setting the values of these sum() calculations from SQLQuery on three new transient fields in a Java Object which already existed. The SQLQuery uses the syntax: SELECT id as {entityObject.id}, name as {entityObject.name}, order as {entityObject.order}, SUM(CASE WHEN pv.value_id = 1 AND pv.value=1 THEN 1 ELSE 0 END) AS