How to convert this UML to JPA entities - java

I have this UML diagram.
And I tried to build entities like this (I renamed Entity class to Entidad)
RelationshipType.java
#Entity
#Table(name = "relationship_type")
public class RelationshipType {
#Id
#GeneratedValue
private Long id;
private String type;
#OneToMany(mappedBy = "relationshipType", fetch = FetchType.EAGER)
private Set<Relationship> relationships = new HashSet<Relationship>();
//Getters and Setters
Relationship.java
#Entity
#Table(name = "relationship")
public class Relationship {
#Id
#ManyToOne
private RelationshipType relationshipType;
#Id
#ManyToOne
private Entidad entity;
//Getters and Setters
Entidad.java
#Entity
#Table(name = "entity")
public class Entidad {
#Id
#GeneratedValue
private Long id;
private String image;
private String foundationNotes;
private String alias;
private Boolean excludeNotifications;
private String notes;
//[...]
#ManyToOne
private Relationship related;
#OneToMany(mappedBy = "entity", fetch = FetchType.EAGER)
private Set<Relationship> relationships = new HashSet<Relationship>();
But when I launch app throws this:
Foreign key (FK_9d8afoh1pv9r59iwjkbcpnud1:entity [])) must have same number of columns as the referenced primary key (relationship [relationshipType_id,entity_id])
At now, I don't know where is the problem and need do this well because I'm using this entities to build the DB schema.

Related

How to map master detail on hibernate?

I can not change the db schema and this is what I got so far:
public class User{
#Id
private String userId;
#OneToMany
#JoinTable(
name = "user_invoice",
joinColumns = #JoinColumn(name="user_id"),
inverseJoinColumns = #JoinColumn(name = "invoice_id")
)
private List<InvoiceItem> invoiceItems;
}
public class InvoiceItem{
#Id
private String invoiceId;
private String invoiceItemId;
}
This configuration does not allow invoice_id to be duplicated on invoice_item table(it should since I can have multiple items on a given invoice)
If I make invoice_item_id composite pk I would need to add an extra column on user_invoice table which I can not.
How can I map this?
You could split up the many-to-many association into two one-to-many associations and an entity for the join table. You can map it like this:
public class User{
#Id
private String userId;
#OneToMany(mappedBy = "user")
private List<Invoice> invoices;
}
#Table(name = "user_invoice")
public class Invoice{
#Id
#ManyToOne(fetch = LAZY)
#JoinColumn(name="user_id")
private User user;
#Id
private String invoiceId;
#OneToMany(mappedBy = "invoice")
private List<InvoiceItem> invoiceItems;
}
public class InvoiceItem{
#Id
#ManyToOne(fetch = LAZY)
private Invoice invoice;
#Id
private String invoiceItemId;
}

Jpa, linked class take main class id

I have a Product :
#Data
#Entity
#Table(name = "products", schema = "laboratory", catalog = "laboratory")
#JsonIgnoreProperties({ "hibernateLazyInitializer", "handler" })
public class Product {
#Id
#GeneratedValue
private int id;
#ManyToOne(fetch = FetchType.LAZY, cascade= CascadeType.ALL)
#JoinColumn(name = "project_id")
#Transient
private Project project; // this one is for read only
#Column(name="project_id") // this will save the id in db for the project
private int projectId;
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name="id")
private Inspection inspection;
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name="id")
private Information information;
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name="id")
private Departure departure;
private String un;
private String tc;
}
There is 3 class that this product needs in order to be a Product : Information, Inpection, Departure
All 3 of these classes are similar.
I want to link them by the Product.id witch is a #GeneratedValue AI in sql.
Here is one of the 3 class :
Information
#Data
#Entity
#Table(name = "products_informations", schema = "laboratory", catalog = "laboratory")
#JsonIgnoreProperties({ "hibernateLazyInitializer", "handler" })
public class Information {
#Id
private int productId;
private String description;
private String model;
private int year;
private String serialNumber;
private int odometre;
private int noCrochet;
private int nbKeys;
private String localisation;
private String cemeteryPosition;
#JsonFormat(pattern = "yyyy-MM-dd")
private Date receptionDate;
}
I want, WHEN I save() the product, that the private String productId in this class to automatically take the Id from the Product class without having to do it manually in my controller.
You have the mappings backwards in your model.
By using
public class Product {
#Id
#GeneratedValue
private int id;
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name="id")
private Information information;
You've told JPA to use the PRODUCT.ID primary key as a foreign key to the Information table; foreign keys are controlled by the relationship, so it means your ID value is pulled from the information.productId value. Opposite of what you are asking for and it means you have 4 mappings trying to set the PRODUCT.ID column value (set them different and see for yourself).
Try this instead:
public class Product {
#Id
#GeneratedValue
private int id;
#OneToOne(mappedby="product", cascade = CascadeType.ALL)
private Information information;
..
}
public class Information {
#Id
private int productId;
#MapsId
private Product product;
..
}
With this you will need to set the Information.product reference, but JPA will use that to set your productId value, using the one you set within the product.id property. You just need to set this relationship when you add an Information instance to a product. Do the same for the other relationships

how to use #OrderBy with Embeddable in spring boot jpa

I have 3 entity Unit, Off, Position. in my Unit have an constrain one to many with list offs, off have position. I want to Order by code in position.code, how can I make it
#Entity
#Table(name = "unit")
public class Unit{
#Id
#Column(name = "id")
private Long id;
#OneToMany(fetch = FetchType.LAZY)
#JoinClolumn(name = "unit")
#ElementCollection
#OrderBy("position.code")
private List<Off> offs;
}
this is my Entity Off
#Entity
#Table(name = "off")
#Embeddable
public class Off{
#Id
#Column(name = "id")
private Long id;
#Column(name = "cate_position")
private Long catePosition;
#ManyToOne
#NotFound(action = NotFoundAction.IGNORE)
#JoinClolumn(name = "cate_position")
#Embedded
private Position position;
}
this is my Entity Position
#Entity
#Table(name = "position")
#Embeddable
public class Position{
#Id
#Column(name = "id")
private Long id;
#Column(name = "code")
private String code;
}
How can I sort List offs in Unit entity by 'position.code', it always throws invalid column name 'position'. Many thanks!
Annotate the entity class Off with #Embeddable
and now include the following code above the List
#ElementCollection
#OrderBy("position.code DESC")
private List<Off> offs;

How to use #OnetoMany with Embeddedid

I am working on hibernate and tying to associate mapping with #OneToMany relationship with composite key.
Following are the entities that currently my using .
#Embeddable
#Getter
#Setter
public class AddressKey implements Serializable {
private static final long serialVersionUID = -307823488229761699L;
#Column(name = "id")
private Long id;
#Column(name = "city")
private Long city;
#Column(name = "locale")
private String locale;
#Column(name = "type")
private String type;
#ManyToOne
#JoinColumn(name="id")
private Person person;
}
#Entity
#Table(name = "address", schema = "test")
#Setter
#Getter
public class AddressHistory {
#EmbeddedId
private AddressKey key;
#Column(name = "active")
private boolean active;
#Column(name = "current")
private boolean current;
}
#Entity
#Table(name = "person", schema="test")
#ToString
public class Person {
#Id
#Column(name = "id")
private Long id;
#OneToMany(mappedBy="key.person", fetch=FetchType.EAGER, cascade=CascadeType.ALL)
private Set<AddressHistory> addressHistory;
}
But when I am trying to run this program it gives me following error.
repeated column in mapping for entity AddressHistory.
Someone help me to fix this what's wrong in this mapping.
Thanks in advance
You repeated columns. Remove #JoinColumn(name="id") in AddressKey since you already have one column with the same name or rename it to something else and more maintainable like person_id.

#OneToOne relationship with additional constraint

Suppose, we have two entities, first one:
#Entity
#Table(name = "entitya")
public class EntityA {
#Id
#Column(name = "id")
private Long id;
#Column(name = "name")
private Long name;
#OneToMany(mappedBy = "parent", fetch = FetchType.LAZY)
private Set<EntityB> childEntities;
}
and the second:
#Entity
#Table(name = "entityb")
public class EntityB {
#Id
#Column(name = "id")
private Long id;
#Column(name = "name")
private String name;
#Column(name = "master")
private Boolean master;
#ManyToOne
#JoinColumn(name = "parent")
private EntityA parent;
}
So far, so good. However underlying database tables and constrains enforce that for any entityA there can be only one EntityB with boolean field master set to true. I can extract it by adding following method to entityA:
public entityB getMasterChild() {
for(entityB ent : childEntities) {
if(ent.isMaster()) {
return ent;
}
}
}
The question is, can I create #OneToOne relationship in EntityA that can express that rule, so that entityA can have additional masterChild member of type entityB?
If I understood you correctly you want to create/define a relationship between two entities based on a value of some entity's property. The think is that relationship between entities is defined on entities count (how many entities can has the other entity) and not on some entity's property value.
However
If you really want to use #OneToOne mapping for masterChild I would recommend creating a separate table/entity for it. Once this is done, you can include this new MasterChild entity into EntityA and annotate it with #OneToOne.
Here is new MasterChild entity
#Entity
public class MasterChild extends EntityB{
#Id
#Column(name = "id")
private Long id;
}
Note that I have removed 'master' from EntityB as it is no longer needed
#Entity
#Table(name = "entityb")
public class EntityB {
#Id
#Column(name = "id")
private Long id;
#Column(name = "name")
private String name;
#ManyToOne
#JoinColumn(name = "parent")
private EntityA parent;
}
And here is modified EntityA
#Entity
#Table(name = "entitya")
public class EntityA {
#Id
#Column(name = "id")
private Long id;
#Column(name = "name")
private Long name;
#OneToOne
private MasterChild master;
#OneToMany(mappedBy = "parent", fetch = FetchType.LAZY)
private Set<EntityB> childEntities;
}

Categories

Resources