I have to entities exposed by spring boot application powered by Spring data REST.
#Entity
#Table(name = "joke")
#Data
public class Joke {
#Id
#Column(name = "joke_id")
private Long id;
#Column(name = "content")
private String content;
#JsonProperty("category")
#JoinColumn(name = "category_fk")
#ManyToOne(fetch = FetchType.EAGER)
private Category category;
}
and category
#Entity
#Table(name = "category")
#Data
public class Category {
#Id
#Column(name = "category_id")
private int id;
#Column(name = "name")
private String name;
}
It is working fine and exposing the HAL+Json format. I'm using Traverson client which is working fine:
Traverson client = new Traverson(URI.create("http://localhost:8080/api/"),
MediaTypes.HAL_JSON);
HashMap<String, Object> parameters = Maps.newHashMap();
parameters.put("size", "2");
PagedModel<JokesDTO> jokes = client
.follow("jokes")
.withTemplateParameters(parameters)
.toObject(new PagedModelType<JokesDTO>() {
});
return jokes;
where JokesDTO is:
#Builder(toBuilder = true)
#Value
#JsonDeserialize(builder = JokesDTO.JokesDTOBuilder.class)
#JsonInclude(Include.NON_NULL)
public class JokesDTO {
private String content;
#JsonPOJOBuilder(withPrefix = "")
#JsonIgnoreProperties(ignoreUnknown = true)
public static class JokesDTOBuilder {
}
}
I'm new in HAL and HateOS and I would like to achieve 2 things (and question is - is it possible, and how):
Base on Traverson client call, how to retrieve category (or link to category) in one call? How to extend what I wrote. And I'm not talking about adding additional #JsonProperty annotation to my class definition.
Is it possible to expose the inner query from Spring data REST, so I would be able to get all data with one call, is it possible with #RepositoryRestResource?
Related
Repost from here
Given entities and repository:
#Entity
public final class Partner {
#Id
private String id;
#OneToMany(mappedBy = "partner", fetch = FetchType.LAZY)
private List<Merchant> merchants;
...
}
#Entity
public final class Merchant {
#Id
private String id;
#Column
private String name;
#ManyToOne(fetch = FetchType.LAZY)
private Partner partner;
...
}
public interface PartnerRepository
extends JpaRepository<Partner, String>, QueryDslPredicateExecutor<Partner> {
}
If there is only one partner having two merchants in the DB then the following code incorrectly returns list with two instances of the same parnter.
partnerRepository.findAll(new Sort("merchants.name"));
This is caused internally by the DB join. By creating custom implementation that adds the distinct to the selection the result is correctly the single partner.
Wouldn't it be correct to do distinct selection per default?
Try
#OneToMany(mappedBy = "partner", fetch = FetchType.LAZY)
#OrderBy("name")
private List<Merchant> merchants;
I want to do select like this in my jpa spring repository
SELECT sicknes_id, count(symptomp_id) as ilosc FROM symptomp_sicknes where symptomp_id IN (1,2) group by sicknes_id Order by ilosc DESC;
My enitity
#Entity
#Table(name = "symptomp")
public class Symptomp {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "symptomp_id")
private Long symptomp_id;
#Column(name = "name")
private String name;
#Column(name = "description")
private String description;
#ManyToMany(cascade = {CascadeType.DETACH,CascadeType.MERGE,CascadeType.PERSIST,CascadeType.REFRESH}, fetch = FetchType.LAZY)
#JoinTable(name = "symptomp_sicknes",joinColumns = #JoinColumn(name = "symptomp_id"),inverseJoinColumns = #JoinColumn(name = "sicknes_id"))
private Set<Sicknes> sicknes = new HashSet<>();
#Entity
#Table(name = "sicknes")
public class Sicknes {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "sicknes_id")
private Long sicknes_id;
#Column(name = "name")
private String name;
#Column(name = "description")
private String description;
#ManyToOne(cascade = {CascadeType.DETACH,CascadeType.MERGE,CascadeType.PERSIST,CascadeType.REFRESH}, fetch = FetchType.LAZY)
#JoinColumn(name = "speciesId")
private Species species;
My Symptomp repository:
public interface SymptompRepository extends JpaRepository<Symptomp, Long> {
#Query("select p from Symptomp p where name like ?1%")
public List<Symptomp> findAllBySymptompName(String symptomp);
public Symptomp findByName(String symptomp);
public List<Symptomp> findByNameIn(List<String> list);
Integer countDistinctSymptompByName(String id);
}
How I can create this select in my JPA repository?
I try get value like in select but i got error mapping bean.
You can get query result as List<Object[]> using nativeQuery=true parameter
#Query("SELECT sicknes_id, count(symptomp_id) as ilosc FROM symptomp_sicknes where symptomp_id IN (1,2) group by sicknes_id Order by ilosc DESC", nativeQuery=true)
List<Object[]> getQueryResult();
Other option is to create dto class with appropriate constructor
public class QueryResultDto {
Long sicknesId;
Long count;
public QueryResultDto(Long sicknesId, Long count) {
this.sicknesId = sicknesId;
this.count = count;
}
}
Then using JPQL
#Query("select new yourproject.dtopath.QueryResultDto(...")
List<QueryResultDto> getQueryResult(#Param("symptompIds") List<Long> symptompIds);
If you want to avoid a native Query the best way is to create an Entity for that JoinTable. Then you can query it easily. Additional benefit if this is that if in future a requirement will pop up that you have to store additional attributes in that relation you will have the Entity already there to do that easily.
I am trying to sort a result by nested collection element value. I have a very simple model:
#Entity
public class User {
#Id
#NotNull
#Column(name = "userid")
private Long id;
#OneToMany(mappedBy = "user")
private Collection<Setting> settings = new HashSet<>();
// getters and setters
}
#Entity
public class Setting {
#Id
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "userid")
private User user;
private String key;
private String value;
// getters and setters
}
public interface UserRepository extends JpaRepository<User, Long>, QuerydslPredicateExecutor<User> {
}
I want to have a result returned sorted by the value of one setting.
Is it possible to order by user.settings.value where settings.name = 'SampleName' using Spring Data JPA with QueryDSL?
I've used JpaSpecificationExecutor. let's see findAll for example.
Page<T> findAll(#Nullable Specification<T> spec, Pageable pageable);
Before call this method you can create your specification dynamically (where condition) and Pageable object with dynamic Sort information.
For example
...
Specification<T> whereSpecifications = Specification.where(yourWhereSpeficiation);
Sort sortByProperty = Sort.by(Sort.Order.asc("property"));
PageRequest orderedPageRequest = PageRequest.of(1, 100, sortByProperty);
userRepository.findAll(whereSpecifications, PageRequest.of(page, limit, orderedPageRequest));
I followed by tutorial : http://www.codejava.net/frameworks/hibernate/hibernate-one-to-one-mapping-with-foreign-key-annotations-example
I have following code:
#Entity
#Table(name = DomainConstant.TABLE_USER)
public class User{
#Id
#Column(name = DomainConstant.DOMAIN_USER_ID)
#GeneratedValue
private Long userId;
private UserActivationCode userActivationCode;
///////////////////// CONSTRUCTOR....
/// STANDARD GET AND SET....
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name = DomainConstant.DOMAIN_ACTIVATION_LINK_ID)
public UserActivationCode getUserActivationCode() {
return userActivationCode;
}
}
#Entity
#Table(name = DomainConstant.TABLE_USER_ACTIVATON_LINK)
public class UserActivationCode {
#Id
#Column(name = DomainConstant.DOMAIN_ACTIVATION_LINK_ID)
#GeneratedValue
private Long userActivationCodeId;
#Column(name = DomainConstant.DOMAIN_ACTIVATION_DATE)
#Temporal(javax.persistence.TemporalType.DATE)
private Date date;
#Column(name = DomainConstant.DOMAIN_ACTIVATION_CODE)
private String code;
///////////////////// CONSTRUCTOR....
/// STANDARD GET AND SET....
}
When I save the User object it does not make record in UserActivationCode, why?
Like this:
User newUser = new User();
newUser.setUserActivationCode(new UserActivationCode("this is example"));
userDao.save(newUser);
I have record only in user table.
Can you tell me why?
Your problem is that you are mixing access types. In the User entity you have specified #Id on a field (private Long userId) whereas you have defined the join mapping on a property (the getter to UserActivationCode). If you specify the join mapping on the field, it should work as is.
#Entity
#Table(name = DomainConstant.TABLE_USER)
public class User{
#Id
#Column(name = DomainConstant.DOMAIN_USER_ID)
#GeneratedValue
private Long userId;
#OneToOne(cascade = CascadeType.ALL)
#JoinColumn(name = DomainConstant.DOMAIN_ACTIVATION_LINK_ID)
private UserActivationCode userActivationCode;
///////////////////// CONSTRUCTOR....
/// STANDARD GET AND SET....
public UserActivationCode getUserActivationCode() {
return userActivationCode;
}
}
For more information on access and access types, see Access, Java EE 7
I need 3 entities: User, Contract (which are a many to many relation) and a middle entity: UserContract (this is needed to store some fields).
What I want to know is the correct way to define the relationships between these entities in JPA/EJB 3.0 so that the operations (persist, delete, etc) are OK.
For example, I want to create a User and its contracts and persist them in a easy way.
Currently what I have is this:
In User.java:
#OneToMany(mappedBy = "user", fetch = FetchType.LAZY)
private List<UserContract> userContract;
In Contract.java:
#OneToMany(mappedBy = "contract", fetch = FetchType.LAZY)
private Collection<UserContract> userContract;
And my UserContract.java:
#Entity
public class UserContract {
#EmbeddedId
private UserContractPK userContractPK;
#ManyToOne(optional = false)
private User user;
#ManyToOne(optional = false)
private Contract contract;
And my UserContractPK:
#Embeddable
public class UserContractPK implements Serializable {
#Column(nullable = false)
private long idContract;
#Column(nullable = false)
private String email;
Is this the best way to achieve my goals?
Everything looks right. My advice is to use #MappedSuperclass on top of #EmbeddedId:
#MappedSuperclass
public abstract class ModelBaseRelationship implements Serializable {
#Embeddable
public static class Id implements Serializable {
public Long entityId1;
public Long entityId2;
#Column(name = "ENTITY1_ID")
public Long getEntityId1() {
return entityId1;
}
#Column(name = "ENTITY2_ID")
public Long getEntityId2() {
return entityId2;
}
public Id() {
}
public Id(Long entityId1, Long entityId2) {
this.entityId1 = entityId1;
this.entityId2 = entityId2;
}
}
protected Id id = new Id();
#EmbeddedId
public Id getId() {
return id;
}
protected void setId(Id theId) {
id = theId;
}
}
I omitted obvious constructors/setters for readability. Then you can define UserContract as
#Entity
#AttributeOverrides( {
#AttributeOverride(name = "entityId1", column = #Column(name = "user_id")),
#AttributeOverride(name = "entityId2", column = #Column(name = "contract_id"))
})
public class UserContract extends ModelBaseRelationship {
That way you can share primary key implementation for other many-to-many join entities like UserContract.