How di I turn my class into an Entity in Room Persistence? - android-room

This is my first time using Room Persistence Library and I am having difficulties understanding the concept of entity.
So I have these two classes:
public class CapturedTime
{
#SerializedName("startTime")
#Expose
private Long startTime;
#SerializedName("endTime")
#Expose
private Long endTime;}
and
public class CapturedItem implements Parcelable {
private String serialNumber;
private CapturedTime firstCapturedTime;
private CapturedTime secondCapturedTime;}
I believe it's pretty straightforward for the CapturedTime class but I have no idea what I should do for the CapturedItem class. Can I just make the CapturedTime variables into columns or are there steps that I should do first?

First add dependency in buid.gradle file
dependencies {
def room_version = "2.1.0-rc01"
implementation "androidx.room:room-runtime:$room_version"
annotationProcessor "androidx.room:room-compiler:$room_version" // For Kotlin use kapt instead of annotationProcessor
}
After that use #Entity annotation to declare normal POJO class as entity in Room persistence library.
#Entity(tableName = "your_table_name")
public class CapturedTime{
#SerializedName("startTime")
#Expose
#ColumnInfo(name = "start_time")
private long startTime;
#SerializedName("endTime")
#Expose
#ColumnInfo(name = "end_time")
private long endTime;
}
#ColumnInfo used to declare column name in table, by default column name is variable name if you do't provide #ColumnInfo annotation
If you want to store custom object in room you use
#TypeConverters() annotation
In your case,
#TypeConverters(CapturedTime.class)
private CapturedTime firstCapturedTime;
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Related

Spring property binding with multiple separation

I have an application.property like this:
somevalue.api.test=something
somevalue.anotherproperty=stuff
I have made a configuration bean like this:
#Configuration
#ConfigurationProperties("somevalue")
public class SomeProperties {
#NotNull
private String apiTest;
#NotNull
private String anotherproperty;
}
Is it possible to refer to api.test like apiTest?
Mainly my issue is that I want to use the somevalue starting point for both property. I know if I don't separate with a dot the apiTest and I use it in this way somevalue.api-test I can refer to that with apiTest in my bean, but in my case it's not possible the renaming. So with dot separation can I achieve the same result or I should create two separate config bean, one refering to somevalue.api and the another only to somevalue?
If you can't rename the property then no, you can't reference it using String apiTest. You need an additional class as follows:
#Configuration
#ConfigurationProperties("somevalue")
public class GcssProperties {
#NotNull
private GcssApiProperties api;
#NotNull
private String anotherproperty;
}
public class GcssApiProperties {
#NotNull
private String test;
}
This should work.

Service cannot auto-wire in Entity class

I needed a RoleMappingService class(which is annotated by #Service) object into a Employee class (which is annotated by #Entity)
below are my classes
********************* RoleMappingsService class **********************
#Service
public class RoleMappingsService {
#Autowired
RolesMappingDao rolesMappingDao;
public List<RolesMappings> getRolesMappingByauthSystemRole(String authSystemRole) {
return rolesMappingDao.getRolesMappingByauthSystemRole(authSystemRole);
}
}
############### Employee class
#Configurable
#Component
#Entity
#NamedQuery(name = "Employee.findAll", query = "SELECT e FROM Employee e")
public class Employee implements Serializable, UserDetails {
#Autowired
#Transient
RoleMappingsService roleMappingsService;
public static final String STATUS_ACTIVE = "ACTIVE";
public static final String STATUS_INACTIVE = "INACTIVE";
public static final String STATUS_LOCKED = "LOCKED";
public static final String STATUS_ONLEAVE = "ONLEAVE";
public static final String STATUS_EXPIRED = "EXPIRED";
private static final long serialVersionUID = 1L;
#Id
#Column(name = "emp_id")
private String empId;
#Column(name = "emp_password")
private String empPassword;
#Column(name = "emp_email")
private String empEmail;
#Column(name = "emp_address")
private String empAddress;
#Column(name = "emp_age")
private int empAge;
#Column(name = "emp_firstname")
private String empFirstname;
}
Here Autowire is not working for roleMappingsService and the object is always found null. However I tried to autowire same object in some other service and there Autowire is perfectly working.
( I know Entity class is only used for representing database table but in my case I need to set some field values which depend on another table so need to fetch data using service)
JB Nizet is totally right
I'll try to provide more explanations here.
Spring can Autowire only beans, objects that it manages, and not arbitrary objects.
Entities are usually created from within a JPA (Hibernate) and are not something that you want to manage by Spring.
There is a related discussion available here but bottom line you should never do something like this.
Why not?
Here are a couple of questions/reasons:
Maybe these entities will go outside spring context at all (serialization), what should that reference contain? Should we also serialize the service? How?
What will happen if the method that turns to the service will be called "outside" the spring driven application (maybe even in different JVM)?
If there are, say 1000 objects returned by that query, do you really want all of them to reside in application context? Or maybe should they be of "prototype" scope?
As you see, it doesn't play nice with spring concepts. I think the reason for it is that Hibernate and JPA do not "support" an idea of methods inside entities, it's just a different framework. I know there are other frameworks that do allow such a concept, but Hibernate/JPA just doesn't, period
So instead of trying to inject the service into the entity bean, probably you should redesign the application so that the service method will be called from outside, maybe via some facade, and entities will be just populated by query, and then "enriched" with additional information if we're talking about SELECT queries, or, alternatively, some information should be pre-set on entity objects, generated by the Business Logic Layer and only then the entity object should be stored in DB

Get entity property with Spring JPA

I'm using Spring JPA in my DAO layer. I have an entity Projet having inside an entity property Client:
Project.java
#Entity
public class Project {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private int projetId;
private String libelle;
#OneToOne(fetch = FetchType.LAZY, cascade = CascadeType.ALL)
#JoinColumn(name="client_id")
private Client client;
// ... constructors, getters & setters
}
Client.java
#Entity
public class Client {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private int clientId;
private String denomination;
// ... constructors, getters & setters
}
in my DAO interface I have the following specifications:
ProjetDao.java
#Repository
#Transactional
public interface ProjetDao extends CrudRepository<Projet, Integer> {
#Transactional
public Projet findByLibelle(String libelle);
#Transactional
public Projet findByProjetId(int projetId);
}
My question is: How can I specify in my DAO interface a method that will return all clients distinct in List<Client>?
From the documentation and JIRA:
List<Project> findAllDistinctBy();
The query builder mechanism built into Spring Data repository infrastructure is useful for building constraining queries over entities of the repository. The mechanism strips the prefixes find…By, read…By, query…By, count…By, and get…By from the method and starts parsing the rest of it. The introducing clause can contain further expressions such as a Distinct to set a distinct flag on the query to be created. However, the first By acts as delimiter to indicate the start of the actual criteria. At a very basic level you can define conditions on entity properties and concatenate them with And and Or.
You are dealing with a one-to-one relationship, in this case I guess the list that you need is not really related to specific project, you just want a distinct list of clients.
You will need to create another repository (ClientRepository) for the Client entity and add a findAllDistinct method in this repository.

spring jpa projection nested bean

is it possible to have a projection with nested collection with Spring JPA?
I have the following 2 simple entity (to explain the problem)
#Entity
#Table(name = "person")
public class Person implements Serializable {
private Integer id;
private String name;
#OneToMany
private List<Address> addressList = new ArrayList<>();
}
#Entity
#Table(name = "address")
public class Address implements Serializable {
private Integer id;
private String city;
private String street;
}
Is it possible to have a projection of Person with following attributes filled in ? {person.name, address.city}
I might be wrong in semantics of word Projection. but the problem is what i need to achieve. Maybe it is not possible with Projection, but is there another way to achieve the end goal? Named Entity graph perhaps ?
P.S. please suggest a solution for Spring JPA not Spring Jpa REST
thanks in advance
You're right, Entity Graphs serve this exact purpose - control field loading.
Create entity graphs dynamically from the code or annotate target entities with Named Entity Graphs and then just use their name.
Here is how to modify your Person class to use Named Entity Graphs:
#Entity
#Table(name = "person")
#NamedEntityGraph(name = "persion.name.with.city",
attributeNodes = #NamedAttributeNode(value = "addressList", subgraph = "addresses.city"),
subgraphs = #NamedSubgraph(name = "addresses.city", attributeNodes = #NamedAttributeNode("city")))
public class Person implements Serializable {
private Integer id;
private String name;
#OneToMany
private List<Address> addressList;
}
And then when loading your person:
EntityGraph graph = em.getEntityGraph("person.name.with.city");
Map hints = new HashMap();
hints.put("javax.persistence.fetchgraph", graph);
return em.find(Person.class, personId, hints);
The same applies for queries, not only em.find method.
Look this tutorial for more details.
I think that that's not usual scenario of Data JPA usage. But you can achieve your goal with pure JPQL:
SELECT a.street, a.person.name FROM Address a WHERE …
This solution has 2 drawbacks:
It forces you to have bidirectional relationship Address ←→ Person
It returns List
Another solution (and that's preferred JPA way) is to create DTO like this:
class MyPersonDTO {
private String personName;
private List<String> cities;
public MyPersonDTO(String personName, List<Address> adresses) {
this.personName = personName;
cities = adresses
.stream()
.map(Address::getCity)
.collect(Collectors.toList());
}
}
And the execute JPQL query like this:
SELECT NEW package.MyPersonDTO(p.name, p.addressList) FROM Person p WHERE …
Return type will be List<MyPersonDTO> in that case.
Of course you can use any of this solutions inside #Query annotation and it should work.

Spring Data JPA and Generics

I have an entity that looks like this
#Entity(name = "encounter_pdf_export")
public class EncounterPDFExport<T extends Encounter> implements Serializable {
public static final long serialVersionUID = 1L;
#Id
#GeneratedValue
private Long pdfExportId;
#Any(metaColumn = #Column(name = "encounter_type"))
#Cascade(CascadeType.ALL)
#AnyMetaDef(
idType = "long",
metaType = "string",
metaValues = {
#MetaValue(value = "FooEncounter", targetEntity = FooEncounter.class)
})
#JoinColumn(name = "encounter_id")
private T encounter;
The abstract type that I'm extending is:
public abstract class Encounter {
public abstract Long getEncounterId();
}
Here is my Spring Data Repository
#Repository
public interface EncounterPDFExportRepository extends PagingAndSortingRepository<EncounterPDFExport, Long> {
EncounterPDFExport findOneByEncounter_encounterId(#Param("encounterId") Long encounterId);
}
I am getting a stack trace when starting up the application related to to the findOneByEncounter_encounterId method:
Caused by: java.lang.IllegalArgumentException: Unable to locate Attribute with the the given name [encounter] on this ManagedType [com.iimassociates.distiller.domain.EncounterPDFExport]
at org.hibernate.jpa.internal.metamodel.AbstractManagedType.checkNotNull(AbstractManagedType.java:144)
at org.hibernate.jpa.internal.metamodel.AbstractManagedType.getAttribute(AbstractManagedType.java:130)
at org.springframework.data.jpa.repository.query.QueryUtils.toExpressionRecursively(QueryUtils.java:468)
at org.springframework.data.jpa.repository.query.JpaQueryCreator$PredicateBuilder.getTypedPath(JpaQueryCreator.java:300)
at org.springframework.data.jpa.repository.query.JpaQueryCreator$PredicateBuilder.build(JpaQueryCreator.java:243)
at org.springframework.data.jpa.repository.query.JpaQueryCreator.toPredicate(JpaQueryCreator.java:148)
at org.springframework.data.jpa.repository.query.JpaQueryCreator.create(JpaQueryCreator.java:88)
at org.springframework.data.jpa.repository.query.JpaQueryCreator.create(JpaQueryCreator.java:46)
at org.springframework.data.repository.query.parser.AbstractQueryCreator.createCriteria(AbstractQueryCreator.java:109)
at org.springframework.data.repository.query.parser.AbstractQueryCreator.createQuery(AbstractQueryCreator.java:88)
at org.springframework.data.repository.query.parser.AbstractQueryCreator.createQuery(AbstractQueryCreator.java:73)
at org.springframework.data.jpa.repository.query.PartTreeJpaQuery$QueryPreparer.<init>(PartTreeJpaQuery.java:116)
at org.springframework.data.jpa.repository.query.PartTreeJpaQuery$CountQueryPreparer.<init>(PartTreeJpaQuery.java:237)
at org.springframework.data.jpa.repository.query.PartTreeJpaQuery.<init>(PartTreeJpaQuery.java:65)
at org.springframework.data.jpa.repository.query.JpaQueryLookupStrategy$CreateQueryLookupStrategy.resolveQuery(JpaQueryLookupStrategy.java:100)
I am assuming that either Spring Data JPA doesn't support abstracted/generic fields? If that's the case, would creating a #Query be a sufficient workaround?
Not sure if this will be helpful to anyone, but I did get this working.
Removed the abstract class and made it an interface with a single public getEncounterId() method
Modified FooEncounter to implement the above interface
Removed generics from the EncounterPDFExport class
Modified the encounter field to utilize the above interface rather than a generic
Apparently, I'm hitting some Hibernate bug/limitation when accessing fields within FooEncounter. Accessing Encounter within EncounterPDFExport works OK, though. I modified my Spring Data JPA Repository to look like the following (note the modification from finding by encounter.encounterId vs. just encounter):
#Repository
public interface EncounterPDFExportRepository extends PagingAndSortingRepository<EncounterPDFExport, Long> {
EncounterPDFExport findOneByEncounter(#Param("encounter") Encounter encounter);
}
The Hibernate bug in question seems to be related to https://jira.spring.io/browse/DATAJPA-836.

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