saving table with multiple children takes too much time to execute (more than 5 sec)
tried this,
spring.jpa.properties.hibernate.jdbc.batch_size=30
spring.jpa.properties.hibernate.order_inserts=true
Timesheet Model
package com.Erp.Model.TimeSheetModel;
#Data
#Getter
#Setter
#Entity
#Table(name = "timesheet")
#JsonIdentityInfo(generator =
ObjectIdGenerators.PropertyGenerator.class,property="id", scope =
Timesheet.class)
public class Timesheet extends BaseEntity{
#ManyToOne
#JoinColumn(name = "employee_personal_information_id")
private EmployeePersonalInformation employeePersonalInformation;
#OneToMany(cascade = CascadeType.ALL)
#JoinColumn(name = "timesheet_id")
private Collection<TimesheetBudget> timesheetBudget;
#OneToMany(cascade = CascadeType.ALL)
#JoinColumn(name = "timesheet_id")
private Collection<TimesheetDateType> timesheetDateType;
}
TimesheetBudget Model
package com.Erp.Model.TimeSheetModel;
#Setter
#Getter
#Entity
#Table(name = "timesheet_budget")
#JsonIdentityInfo(generator =
ObjectIdGenerators.PropertyGenerator.class,property="id", scope =
TimesheetBudget.class)
public class TimesheetBudget extends BaseEntity {
#OneToMany(cascade = CascadeType.ALL)
#JoinColumn(name = "timesheet_budget_id")
private Collection<TimesheetHours> timesheetHours;
#ManyToOne(fetch= FetchType.LAZY)
#JsonIgnore
private Timesheet timesheet;
}
TimesheetHours Model
package com.Erp.Model.TimeSheetModel;
#Entity
#Getter
#Setter
#Table(name = "timesheet_hours")
public class TimesheetHours extends BaseEntity {
#ManyToOne(fetch=FetchType.LAZY )
#JsonIgnore
private TimesheetBudget timesheetBudget;
}
TimesheetDateType Model
package com.Erp.Model.TimeSheetModel;
#Setter
#Getter
#Entity
#Table(name="timesheet_date_type")
#JsonIdentityInfo(generator =
ObjectIdGenerators.PropertyGenerator.class,property="id", scope =
TimesheetDateType.class)
public class TimesheetDateType extends BaseEntity {
#ManyToOne(fetch = FetchType.LAZY)
#JsonIgnore
private Timesheet timesheet;
}
Timesheet Service
package com.Erp.Service.TimeSheetService;
#Service
public class TimesheetService {
#Autowired
TimesheetRepository timesheetRepository;
public ResponseEntity<String> saveTimesheet(Timesheet timesheet) {
timesheetRepository.save(timesheet);
return ResponseEntity.ok().body("Timesheet Saved Successfully.");
}
}
Timesheet Repository
package com.Erp.Repository.TimesheetRepository;
#Repository
public interface TimesheetRepository extends JpaRepository<Timesheet,
Integer>{
}
Related
i want to implement a many to many association with quantity information in it . like this :
#Entity
#Table(name = "reserves")
#Getter #Setter #NoArgsConstructor
public class Reserve {
#Id
#ManyToOne(cascade = CascadeType.ALL,fetch = FetchType.EAGER)
#JoinColumn(name = "groupe_id")
private GroupeSanguin bloodGroup;
#Id
#ManyToOne(cascade = CascadeType.ALL,fetch = FetchType.EAGER)
private Banque banque;
private int quantity;
}
the GroupSanguin and the Banque are two class stored in the database two . here is the code for the two if you need :
#Entity
#Table(name = "groupe_sanguins")
public class GroupeSanguin {
#Id
private String groupe;
#OneToMany(mappedBy = "groupeSanguin")
private List<Donneur> donneurs;
}
#Entity #Getter #Setter #NoArgsConstructor
public class Banque {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#Column(unique = true,nullable = false)
private String nom;
private String adresse;
#Column(unique = true)
private String telephone;
private String localisation;
}
so my i want to know how to annotate the JpaRepository to take the two as primary key like this and is my annotation good for it to work ?
public interface ReserveRepository extends JpaRepository<
Reserve,
//what to put here ?
>
This isn't a JPA question in fact, it's a relationnal database conception.
If Reserve has is own data and links with other entity it has it own Id
You can add unicity constraint
#Entity
#Table(name = "reserves", uniqueConstraints={
#UniqueConstraint(columnNames = {"banque_id", "groupe_id"})
#Getter #Setter #NoArgsConstructor
public class Reserve {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
#ManyToOne(cascade = CascadeType.ALL,fetch = FetchType.EAGER)
#JoinColumn(name = "groupe_id")
private GroupeSanguin bloodGroup;
#ManyToOne(cascade = CascadeType.ALL,fetch = FetchType.EAGER)
#JoinColumn(name = "banque_id")
private Banque banque;
private int quantity;
}
i've found this solutions too.
#Entity
#Table(name = "reserves")
#Getter #Setter #NoArgsConstructor
#IdClass(ReserveId.class) //this annotation will tell that id that the
// the id will be represented by a class
public class Reserve {
#Id
#ManyToOne(cascade = CascadeType.ALL,fetch = FetchType.EAGER)
#JoinColumn(name = "groupe_id")
private GroupeSanguin groupeSanguin;
#Id
#ManyToOne(cascade = CascadeType.ALL,fetch = FetchType.EAGER)
#JoinColumn(name = "banque_id")
private Banque banque;
private int quantity;
}
and the id class should implements Serializable like this :
#Getter #Setter
public class ReserveId implements Serializable {
private Banque banque;
private GroupeSanguin groupeSanguin;
}
and finally the repository will be like that :
#Repository
public interface ReserveRepo extends JpaRepository<Reserve, ReserveId>{}
See your Reserve class has nowhere mentioned composite primary key. First you need to fix the model, You can refer to the solution here How to create and handle composite primary key in JPA
I have the following entity classes:
#Embeddable
#Getter
#Setter
public class OrganizationCyclePlageKey implements Serializable {
#Column(name = "organization_id")
Long organizationId;
#Column(name = "cycle_plages_id")
Long cyclePlagesId;
...
equals() and hashCode() methods come here
#Entity
#Table(name = "organization_cycle_plages")
#Getter
#Setter
public class OrganizationCyclePlage {
#EmbeddedId
private OrganizationCyclePlageKey id;
#ManyToOne
#MapsId("organizationId")
#JoinColumn(name = "organization_id")
Organization organization;
#ManyToOne
#MapsId("cyclePlagesId")
#JoinColumn(name = "cycle_plages_id")
CyclePlage cyclePlage;
...
other attributes
}
#Entity
#Getter
#Setter
public class CyclePlage extends AbstractEntity {
#OneToMany(mappedBy = "cyclePlage")
private Set<OrganizationCyclePlage> organizationCyclePlages;
...
}
#Entity
#DynamicUpdate
#Getter
#Setter
public class Organization extends AbstractEntity {
#OneToMany(mappedBy = "organization")
private Set<OrganizationCyclePlage> organizationCyclePlages = new HashSet<>();
...
}
SpringBoot app starts up normally without errors.
But when I try to save an instance of OrganizationCyclePlage:
OrganizationCyclePlage ocp = new OrganizationCyclePlage();
ocp.setOrganization(organization);
ocp.setCyclePlage(cyclePlage);
organizationCyclePlageRepository.save(ocp);
it raises the error when calling organizationCyclePlageRepository.save(ocp):
org.hibernate.PropertyAccessException: Could not set field value [361] value by reflection : [class com.XXXX.OrganizationCyclePlageKey.cyclePlagesId] setter of com.XXXX.OrganizationCyclePlageKey.cyclePlagesId
What's wrong with these relations?
I had to add the constructor into the OrganizationCyclePlageKey class to init the foreign keys values as well a default constructor via #NoArgsConstructor annotation:
public OrganizationCyclePlageKey(Long organizationId, Long cyclePlagesId) {
this.organizationId = organizationId;
this.cyclePlagesId = cyclePlagesId;
}
and init the OrganizationCyclePlageKey instance in the OrganizationCyclePlage class:
public class OrganizationCyclePlage {
private OrganizationCyclePlageKey id = new OrganizationCyclePlageKey();
...
}
I have two entities SuperAlbumEntity and AlbumEntity reflecting the same table "albums".
SuperAlbumEntity:
#Entity
#Table(name = "albums")
#Inheritance(strategy = InheritanceType.SINGLE_TABLE)
public class SuperAlbumEntity {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Integer id;
//other fields
}
AlbumEntity:
#EqualsAndHashCode(callSuper = true)
#Entity
#Table(name = "albums")
public class AlbumEntity extends SuperEntity{
//some fields
#Column(name = "country")
private String country;
#OneToMany(fetch = FetchType.EAGER)
#JoinColumn(name = "country_name", referencedColumnName = "country")
private Set<CountryEntity> countrySet = new HashSet<>();
}
AlbumEntity has #OneToMany mapping to CountryEntity:
#Entity
#Table(name = "countries")
public class CountryEntity implements Serializable {
#Id
String id;
String country_name;
//other fields
}
Running my application I get the folowing error:
...
Caused by: org.hibernate.AnnotationException: referencedColumnNames(country) of CountryEntity.countrySet referencing AlbumEntity not mapped to a single property
...
What's interesting is that if I move country field from SuperAlbumEntity to AlbumEntity everything just works fine...
Can someone explain me why I get this error?
I'm not sure but I think is connected with the type of inherence that you used it. Try to modify your superclass to something like this:
SuperAlbumEntity:
#MappedSuperclass
public abstract class SuperAlbumEntity {
}
AlbumEntity:
#Entity
#Inheritance(strategy=InheritanceType.JOINED)
#Table(name = "albums")
public class AlbumEntity extends SuperEntity {
#OneToMany(fetch = FetchType.EAGER)
#JoinColumn(name = "country_name", referencedColumnName = "country")
private Set<CountryEntity> countrySet = new HashSet<>();
}
I'm writing 3 tables in the following relation:
Club class:
#Setter
#Getter
#Entity
#Table(name = "Club")
public class Club {
#Id
#GeneratedValue
private Long id;
private String name;
private String type;
private String mainPage;
private String logo;
#OneToMany(mappedBy="clubProductKey.club", cascade = CascadeType.ALL)
#JsonIgnoreProperties(value = "clubProductKey.club", allowSetters=true)
private Set<ClubProduct> clubProducts;
...
Product class:
#Setter
#Getter
#Entity
#Table(name = "Product")
public class Product {
#Id
#GeneratedValue
private Long id;
#OneToMany(mappedBy="clubProductKey.product", cascade = CascadeType.ALL)
#JsonIgnoreProperties(value = "clubProductKey.product", allowSetters=true)
private Set<ClubProduct> clubProducts;
...
ClubProduct class:
#Setter
#Getter
#Entity
#Table(name = "ClubProduct")
public class ClubProduct {
#EmbeddedId
private ClubProductKey clubProductKey;
...
ClubProductKey class:
#Setter
#Getter
#Embeddable
public class ClubProductKey implements Serializable {
#ManyToOne(cascade = {CascadeType.MERGE,CascadeType.REFRESH })
#JoinColumn(name = "club_id", referencedColumnName = "id")
#JsonIgnoreProperties(value = "clubProducts", allowSetters=true)
private Club club;
#ManyToOne(cascade = {CascadeType.MERGE,CascadeType.REFRESH })
#JoinColumn(name = "product_id", referencedColumnName = "id")
#JsonIgnoreProperties(value = "clubProducts", allowSetters=true)
private Product product;
...
ClubProductRepository class:
public interface ClubProductRepository extends JpaRepository<ClubProduct, ClubProductKey> {
public List<ClubProduct> findByClubProductKeyClub(Club club);
public List<ClubProduct> findByClubProductKeyProduct(Product product);
}
I try to save clubProduct like this:
#Service
public class ClubProductServiceImp implements ClubProductService {
#Autowired
private ClubProductRepository clubProductRepository;
...
ClubProduct savedClubProduct = clubProductRepository.save(clubProduct);
return savedClubProduct;
}
However I find that the clubProduct is not saved in the clubProducts list in the club or product entity, the list is null. Must I add lines like club.getClubProducts.add(clubProduct) or is there any other way to make it added automatically?
Thank you.
The #OnetoMany mapping in your Club class uses the attribute mappedby which means that it represents the owning side of the relation responsible for handling the mapping. However, we still need to have both sides in sync as otherwise, we break the Domain Model relationship consistency, and the entity state transitions are not guaranteed to work unless both sides are properly synchronized.
The answer is yes, you have to manage the java relations yourself so that the clubProducts gets persisted. You are using an instance of the repository class club to persist the data so , you should add a setter method like :
public void addClubProduct(ClubProduct clubProduct) {
if (clubProduct!= null) {
if (clubProduct== null) {
clubProduct= new ArrayList<ClubProduct>();
}
clubProducts.add(clubProduct);
clubProduct.setClubProduct(this);
}
}
also a method to remove it from the list and use these method in your code to set the values to the list properly before initiating save . Read related article
I have a simple problem - but I think "I am standing on the tube".
I have a spring boot rest api with JPA, Modelmapper, Entities and DTOs.
But the mapping doesn't work.
Entities:
#Getter
#Setter
#MappedSuperclass
public class AbstractEntity {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
}
#Getter
#Setter
#Entity(name = "contacts")
public class Contact extends AbstractEntity {
#NotBlank
private String firstName;
#NotBlank
private String lastName;
#Valid
#OneToMany(mappedBy = "contact", cascade = CascadeType.ALL, orphanRemoval = true)
private List<PhoneNumber> phoneNumberList;
}
#Getter
#Setter
#Entity(name = "phone_numbers")
public class PhoneNumber extends AbstractEntity {
#NotBlank
private String label;
#NotBlank
private String number;
#ManyToOne
#JoinColumn(name = "contact_id", referencedColumnName = "id")
#Setter(value = AccessLevel.NONE)
private Contact contact;
}
The DTOs:
#Getter
#Setter
#Builder
#NoArgsConstructor
#AllArgsConstructor
public class ContactDTO {
private Long id;
private String firstName;
private String lastName;
List<PhoneNumberDTO> phoneNumberDTOList = new ArrayList<>();
}
#Data
#Builder
#NoArgsConstructor
#AllArgsConstructor
public class PhoneNumberDTO {
private Long id;
private String label;
private String number;
}
My ModelMapperConfig:
#Bean
public ModelMapper modelMapper() {
ModelMapper modelMapper = new ModelMapper();
modelMapper.getConfiguration()
.setFieldMatchingEnabled(true)
.setFieldAccessLevel(AccessLevel.PRIVATE);
return modelMapper;
}
Repo:
public interface ContactRepository extends JpaRepository<Contact, Long{
}
Service (only the create method):
#Override
public ContactDTO createOne(ContactDTO contactDTO) {
Contact contact = modelMapper.map(contactDTO, Contact.class);
contactRepository.save(contact);
return contactDTO;
}
Is this the correct way to persist the Contact with its multiple phonenumbers?
And how can I create a simple mapping?
If i want to persist it, there comes an error:
Column 'contact_id' cannot be null