I have to query on timestamp field for current date and get only 1 record.
I can write the query like:
#Query("Select i From Log i Where i.createdBy = :userId And DATE(i.createdDate) = CURRENT_DATE")
JPA Entity:
#Entity
#Table(name = "log")
public class Log {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "log_id")
private Long logId;
#Column(name = "address")
private String address;
#Column(name = "created_date", updatable = false)
#Temporal(TemporalType.TIMESTAMP)
#DateTimeFormat(style = "yyyy-MM-dd HH:mm:ss")
private Calendar createdDate;
//SETTERS AND GETTERS
}
TABLE:
CREATE TABLE `log` (
`log_id` BIGINT(20) NOT NULL AUTO_INCREMENT,
`address` VARCHAR(30) NOT NULL,
`created_date` TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP,
PRIMARY KEY (`log_id`)
)
ENGINE=InnoDB
;
here i am not able to limit the records,
I know we can limit the records by sending Pageable, but again i have to get the record from list.
Is there any way to do this ?
how can we do it in spring data jpa method name ?
You can use a Spring Data repository to do this like the following:
public interface LogRepository extends JpaRepository<Log, Long> {
// just get one or null
Log findOneByCreatedByAndCreatedDate(Instant createdBy, Instant createdDate);
// get all, but pagable
List<Log> findAllByCreatedByAndCreatedDate(Instant createdBy, Instant createdDate, Pageable pageable);
}
I assume you use Instant as timestamp, but this should also work for the other Java 8 date types or the old Date class.
Within your business logic you can now call:
Log log = logRepository.findOneByCreatedByAndCreatedDate(YOUR_TIMESTAMP, Instant.now());
// or
Log allLogs = logRepository.findOneByCreatedByAndCreatedDate(YOUR_TIMESTAMP, Instant.now(), PageRequest.of(0, 50));
Related
The user can search for products if any product shown in the result exists in the user_favorites table so the show flag tells the front-end this product was added for this user by user_id and product_id. with spring boot and spring data.
My Entity :
#Id
#Column(name = "catId")
private Integer catId;
#Column(name = "cat_no")
private String catNo;
#Column(name = "cat_sn")
private String catSn;
#Column(name = "doc_ref")
private String docRef;
#Column(name = "user_id")
private Integer userId;
#Column(name = "updated_at")
private String updatedAt;
#Column(name = "created_at")
private String createdAt;
I tried that using #Formula but nothing happing always returns null. and if it's done by #Formula how can i add parameters to #Formula
#Formula(value = "SELECT count(*) as checker FROM fb_user_favorites WHERE cat_id = 34699 AND user_id = '52') ")
#Transient
private String checker;
#Transient is part of JPA spec. In Hibernate fields marked with this annotation just simply ignored/excluded from any JPA engine/runtime logic.
#Formula is part of Hibernate. Fields, marked with it, don't persisted by Hibernate (first argument do not use #Transient as redundant), values are calculated by provided SQL when executing query for entity.
So for Hibernate to see this fields, they should not be excluded by #Transient
TL;DR remove #Transient annotation
Complicated but fast working way.
Adding isFavorite field to the entity:
#Transient
private boolean isFavorite;
Create an entity linking Product and User:
public class ProductFavorite {
#Id
#GeneratedValue(strategy = IDENTITY)
private Long id;
#ManyToOne(optional = false, fetch = LAZY)
private Product product;
#ManyToOne(optional = false, fetch = LAZY)
private User user;
}
Then create a repository with a method to find the user's favorite products:
#Repository
public interface ProductLikeRepository extends JpaRepository<ProductFavorite, Long> {
#Query("select f.product.id from ProductFavorite f where f.product in ?1 and f.user = ?2")
Set<Integer> findProductIdsByIdsAndUser(List<Product> products, User user);
}
And at the end, write a method that will fill in the isFavorite field:
public void fillFavorite(List<Product> products, User user) {
if (products.isEmpty()) {
return;
}
var likedIds = favoriteRepository.findProductIdsByIdsAndUser(products, user);
for (Product product : products) {
product.setFavorite(likedIds.contains(product.getId()));
}
}
You need to call it manually:
List<Product> products = productRepository.findAll();
fillFavorite(products, currentUser());
Hi below is my schema definition
CREATE TABLE LOANS (
LOAN_ID NUMBER(9,0) PRIMARY KEY,
CORR_ID VARCHAR(5) NULL
);
CREATE TABLE DV_LOAN_PARTICIPANTS (
LOAN_ID NUMBER(9,0) ,
DVP_PARTICIPANT_NAME VARCHAR(50) NOT NULL,
DVP_PARTICIPANT_TYPE VARCHAR(50) NOT NULL,
PRIMARY KEY ("LOAN_ID", "DVP_PARTICIPANT_NAME")
);
LOANS Entity
#Table(name = "LOANS")
#Entity
public class Loans {
#Id
#Column(name = "LOAN_ID")
private Long loanId;
#OneToMany(cascade = CascadeType.ALL,fetch = FetchType.EAGER)
#JoinColumn(name = "LOAN_ID")
#MapKey(name = "dvpParticipantName")
private Map<String, DVLoanParticipants> dvLoanParticipantsMap;
// getter and setters
}
DV_LOAN_PARTICIPANTS Entity
#Table(name = "DV_LOAN_PARTICIPANTS")
#Entity
public class DVLoanParticipants implements Serializable {
#Id
#Column(name = "LOAN_ID")
private Long loanId;
#Id
#Column(name = "DVP_PARTICIPANT_NAME")
private String dvpParticipantName;
#Column(name = "DVP_PARTICIPANT_TYPE")
private String dvpParticipantType;
// getters and setters
}
Service Class is
DVLoanParticipants dvLoanParticipants = new DVLoanParticipants();
dvLoanParticipants.setLoanId(Long.valueOf("196801758"));
dvLoanParticipants.setDvpParticipantName("VKP");
dvLoanParticipants.setDvpParticipantType("Developer");
Loans loanInsert = new Loans();
loanInsert.setLoanId(Long.valueOf("196801758"));
Map<String,DVLoanParticipants> partyMap = new HashMap<>();
partyMap.put("VKP",dvLoanParticipants);
loanInsert.setDvLoanParticipantsMap(partyMap);
repository.save(loanInsert);
But when i am executing the save i am getting error as
NULL not allowed for column "LOAN_ID"; SQL statement:
insert into dv_loan_participants (dvp_participant_type, loan_id, dvp_participant_name) values (?, ?,
?)
Git Hub Code
https://github.com/vinoykp/spring-jpa/tree/master/spring-boot-hibernate-crud-demo
I had the similar question
Why Value is not getting assigned in JPA for insert statement
What is the issue in association?
In my database I have a user who can have multiple email addresses. An email address can have only one user. I have following two tables in my database to handle this.
CREATE TABLE IF NOT EXISTS w4a_user (
id INTEGER NOT NULL AUTO_INCREMENT,
login_id VARCHAR(100) NOT NULL UNIQUE,
first_name VARCHAR(100),
last_name VARCHAR(100),
division INTEGER NOT NULL,
created_date TIMESTAMP NOT NULL,
last_active DATE,
PRIMARY KEY (id),
FOREIGN KEY (login_id) REFERENCES w4a_authentication_data (login_id) ON DELETE RESTRICT,
FOREIGN KEY (division) REFERENCES w4a_division (id) ON DELETE RESTRICT
);
CREATE TABLE IF NOT EXISTS w4a_email_address(
email_address VARCHAR(100) NOT NULL,
user_id INTEGER NOT NULL,
is_confirmed BOOLEAN NOT NULL DEFAULT FALSE,
PRIMARY KEY (email_address),
FOREIGN KEY (user_id) REFERENCES w4a_user (id) ON DELETE CASCADE
);
In my Spring boot application, I have following entity classes to handle this.
User.java
#Entity
#Table(name = "w4a_user")
public class User {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "id")
private int id;
#Column(name = "first_name")
#Size(max = 100, message = GlobalConstants.ErrorMessageConstants.ERROR_FIRST_NAME_LENGTH_EXCEEDED)
private String firstName;
#Column(name = "last_name")
#Size(max = 100, message = GlobalConstants.ErrorMessageConstants.ERROR_LAST_NAME_LENGTH_EXCEEDED)
private String lastName;
#Column(name = "created_date")
private Date createdDate;
#Column(name = "last_active")
private Date lastActive;
#ManyToOne
#JoinColumn(name = "division", referencedColumnName = "id")
private Division division;
#OneToMany(mappedBy = "userId", cascade = CascadeType.ALL, orphanRemoval = true)
#Size(min = 1)
private List<ContactNumber> contactNumberList;
#OneToMany(mappedBy = "userId", cascade = CascadeType.ALL, orphanRemoval = true)
#Size(min = 1)
private List<EmailAddress> emailAddresses;
.
.
}
EmailAddress.java
#Entity
#Table(name = "w4a_email_address")
public class EmailAddress {
#Id
#Column(name = "email_address")
#Email(message = GlobalConstants.ErrorMessageConstants.ERROR_EMAIL_INCORRECT_FORMAT,
regexp = GlobalConstants.RegexList.EMAIL_REGEX)
#Size(max = 100, message = GlobalConstants.ErrorMessageConstants.ERROR_EMAIL_LENGTH_EXCEEDED)
private String emailAddress;
#ManyToOne
#JoinColumn(name = "user_id", referencedColumnName = "id")
private User userId;
#Column(name = "is_confirmed")
private Boolean isConfirmed;
.
.
}
I use following method to persist entitites to my database.
#PersistenceContext
private EntityManager em;
#Override
public T createEntity(T entity) {
this.em.unwrap(Session.class).save(entity);
return entity;
}
I set email address list in the user entity and perform above method to create a new user.
The issue I have is when adding a user with an email address already used by an existing user. In this case, the database entry for the email address gets updated with the id of the new user. Instead I want to give an error saying the email address is already in use. What is the best way of handling this?
getting
com.microsoft.sqlserver.jdbc.SQLServerException: Invalid column name
'partnerIdPartner'.
application.properties:
spring.jpa.hibernate.naming.implicit-strategy=org.hibernate.boot.model.naming.ImplicitNamingStrategyLegacyJpaImpl
spring.jpa.hibernate.naming.physical-strategy=org.hibernate.boot.model.naming.PhysicalNamingStrategyStandardImpl
Table creation:
CREATE TABLE [partnersystem] (
[idPartnerSystem] INT IDENTITY(1,1) ,
[partner_idPartner] INT NOT NULL DEFAULT NULL ,
[Name] NVARCHAR(45) NULL DEFAULT NULL ,
[Domain] NVARCHAR(45) NULL DEFAULT NULL ,
[Code] NVARCHAR(45) NULL DEFAULT NULL ,
[PartnerSystem_idSystem] INT NOT NULL DEFAULT NULL ,
[UpdateUser] NVARCHAR(45) NULL DEFAULT NULL ,
[UpdateDT] DATETIME NULL DEFAULT NULL ,
CONSTRAINT [partnersystem_PRIMARY] PRIMARY KEY CLUSTERED ([idPartnerSystem]), CONSTRAINT [partnersystem_fk_PartnerSystem_partner] FOREIGN KEY ("partner_idPartner") REFERENCES "partner" ( "idPartner" ) ON UPDATE NO ACTION ON DELETE NO ACTION, CONSTRAINT [partnersystem_fk_PartnerSystem_System] FOREIGN KEY ("PartnerSystem_idSystem") REFERENCES "system" ( "idSystem" ) ON UPDATE NO ACTION ON DELETE NO ACTION);
CREATE INDEX [partnersystem_fk_PartnerSystem_partner] ON [partnersystem] ([partner_idPartner]);
CREATE INDEX [partnersystem_fk_PartnerSystem_System] ON [partnersystem] ([PartnerSystem_idSystem]);
JPA Entity:
#Entity
#Table(name = "partnersystem")
public class PartnerSystem {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "idPartnerSystem")
private int idPartnerSystem;
#Column(name = "partner_idPartner" )
private int partnerIdPartner;
#Column(name = "Name")
private String name;
#Column(name = "Domain" )
private String domain;
#Column(name = "Code" )
private String code;
#Column(name = "PartnerSystem_idSystem" )
private int partnerSystemIdSystem;
#Column(name = "UpdateUser" )
private String updateUser;
my repository:
#Repository
public interface PartnerSystemRepository extends JpaRepository<PartnerSystem,
Integer>{
public PartnerSystem findByPartnerIdPartner(int partnerIdPartner);
}
executing simple query throws an error.
public List<Object[]> findAllPartnerSystem(int id) {
String test =
"SELECT idPartnerSystem, partnerIdPartner, name, domain, code, partnerSystemId" +
" FROM PartnerSystem " +
"WHERE partnerIdPartner = ?"
;
Query query = em.createNativeQuery(test);
query.setParameter(1, id);
List<Object[]> results = query.getResultList();
for (Object[] row : results) {
}
return results;
}
In native queries you have to use the column name not the property name:
"SELECT idPartnerSystem, partner_idPartner, name, domain, code, PartnerSystem_idSystem" +
" FROM partnersystem " +
"WHERE partner_idPartner=
But I suggest using JPQL queries and not native queries.
I have a postgres database and I am trying to make a simple REST service with Spring Boot. I have a problem with jpa ManytoMany relationship.
Person Entity:
#Entity
#Table(name = "person", schema = "persons")
public class Person implements Serializable {
#Id
#GeneratedValue(strategy= GenerationType.IDENTITY)
private Integer id;
#Column(nullable = false)
private String name;
#Column(nullable = false)
private String email;
#Column
private Integer age;
#ManyToOne
#JoinColumn(name = "country_id", referencedColumnName = "id")
private Country countryOfBirth;
#ManyToMany
#JoinTable(
name="persons_countries_residence",
joinColumns=#JoinColumn(name="person_id", referencedColumnName="id"),
inverseJoinColumns=#JoinColumn(name="country_id", referencedColumnName="id"))
private List<Country> countriesOfResidence;
// getters and setters and to String method overriden
}
Country Entity:
#Entity
#Table(name = "country", schema = "persons")
public class Country implements Serializable {
#Id
#GeneratedValue(strategy= GenerationType.IDENTITY)
private Integer id;
#Column(name = "country_name")
private String name;
#Column(name = "country_code")
private String code;
// getters and setters and to String method overriden
}
The postgres schema is the following:
Person Table:
CREATE TABLE persons.person
(
id serial NOT NULL,
name character varying(50) NOT NULL,
email character varying(40) NOT NULL,
age integer,
country_id serial NOT NULL,
CONSTRAINT id PRIMARY KEY (id),
CONSTRAINT country_id FOREIGN KEY (id)
REFERENCES persons.country (id) MATCH SIMPLE
ON UPDATE NO ACTION ON DELETE NO ACTION
)
Country table:
CREATE TABLE persons.country
(
id serial NOT NULL,
country_name character varying(45) NOT NULL,
country_code character varying(10) NOT NULL,
CONSTRAINT country_id PRIMARY KEY (id)
)
Join table:
CREATE TABLE persons.persons_countries_residence
(
person_id integer NOT NULL,
country_id integer NOT NULL,
CONSTRAINT person_country_id PRIMARY KEY (person_id, country_id),
CONSTRAINT persons_countries_residence_country_id_fkey FOREIGN KEY (country_id)
REFERENCES persons.country (id) MATCH SIMPLE
ON UPDATE CASCADE ON DELETE NO ACTION,
CONSTRAINT persons_countries_residence_person_id_fkey FOREIGN KEY (person_id)
REFERENCES persons.person (id) MATCH SIMPLE
ON UPDATE CASCADE ON DELETE CASCADE
)
When i make an HTTP method call, I don't get the Countries of residence.
Service code:
#RequestMapping(method = RequestMethod.GET, produces = {MediaType.APPLICATION_JSON_VALUE, MediaType.APPLICATION_XML_VALUE})
public List<Person> getAllPersons() {
retutn jpaPersonRepository.findAll();
}
Any help appreciated.
Maybe, you need to specify a schema name in the join table name:
#JoinTable(
name="persons_countries_residence", schema="persons",
joinColumns=#JoinColumn(name="person_id", referencedColumnName="id"),
inverseJoinColumns=#JoinColumn(name="country_id", referencedColumnName="id"))
Update your Country class code like :
#Entity
#Table(name = "country", schema = "persons")
public class Country implements Serializable {
#Id
#GeneratedValue(strategy= GenerationType.IDENTITY)
private Integer id;
#Column(name = "country_name")
private String name;
#Column(name = "country_code")
private String code;
#ManyToMany(mappedBy = "countriesOfResidence")
private List<Person> persons;
// getters and setters and to String method overriden
}
Although a ManyToMany relationship is always bi-directional on the
database, the object model can choose if it will be mapped in both
directions, and in which direction it will be mapped in. If you choose
to map the relationship in both directions, then one direction must be
defined as the owner and the other must use the mappedBy attribute to
define its mapping. This also avoids having to duplicate the JoinTable
information in both places.
Do you mean that the country list is null? #ManyToMany associations are lazily loaded by default, you need to enable eager-fetching for it to work straight away.
#ManyToMany(fetch = FetchType.EAGER)
The solution is this:
#ManyToMany
#JoinTable(
name="persons_countries_residence", schema = "persons",
joinColumns=#JoinColumn(name="person_id", referencedColumnName="id"),
inverseJoinColumns=#JoinColumn(name="country_id", referencedColumnName="id"))
private List<Country> countriesOfResidence;
The schema had to be specified at the #JoinTable annotation as well.