How to Define Dynamic Model in Spring Framework - spring

I am using Spring Framework as my back end
I have define know as Entity class The Entity class know contain 5 Fields
Below is the class , The code below dose not have setter getter part to make shorter and cleaner
#Entity
#Table(name="TblKnow")
public class Know {
#Id
private Double idKnow;
private String SubjectKnow;
private String BodyKnow;
private String ImgKnow;
private double CountView;
In JpaRepository interface i want to only query two column not all of columns.
public interface KnowRepository extends JpaRepository<Know,Double> {
#Query("SELECT idKnow,SubjectKnow FROM Know")
public Page<Know> findCByOrderByIdKnowDesc(Pageable pageable);
Problem: i try to run but i get below exception
java.lang.IllegalArgumentException: Cannot create TypedQuery for query with more than one return using requested result type [java.lang.Long]
But if i use without below query it is fine
public Page<Know> findAllByOrderByIdKnowDesc(Pageable pageable);

You can create a custom constructor and use that to select only some fields in JPA query.
public Know(Double idKnow, String SubjectKnow) {
this.idKnow = idKnow;
this.SubjectKnow = SubjectKnow;
}
And the use this constructor in JPA query. Make sure you use complete path of class with package.
#Query("SELECT NEW packagePath.Know(idKnow,SubjectKnow) FROM Know")

query :
public Page<Know> findAllByOrderByIdKnowDesc(Pageable pageable);
works dut to you select Know objects with fields that are mapped correct into Know class (and after wrapped into Page).
with query :
#Query("SELECT idKnow,SubjectKnow FROM Know")
public Page<Know> findCByOrderByIdKnowDesc(Pageable pageable);
returns some custome bean/object that spring data can't map in correct way into Know class (as you declared it as expected return class wrapped into Page). add counstructor into Know with idKnow,SubjectKnow fields , or you can wrap it into some DTO with idKnow,SubjectKnow fields.

Related

Can we fetch data using JPA without using primary key

My Table:
Emotion
EID (Primary Key)
user_mood
Latitude
Longitude
uid
#Entity
public class Emotion
{
#Id
#GeneratedValue(strategy=GenerationType.IDENTITY)
private Integer eId;
#Column(name = "user_mood")
private String mood;
private String latitude;
private String longitude;
private String uid;
}
My Interface:
public interface EmotionRepository extends JpaRepository<Emotion, String>{}
When I try to fetch values using uid
emotionRepo.findById(uid)
I am getting below type mismatch error
Error message:
org.hibernate.TypeMismatchException: Provided id of the wrong type for class mood_sensor.mood_sensor.pojos.Emotion. Expected: class java.lang.Integer, got class java.lang.String
Can I retrieve data using uid or I should use only the primaryKey(Integer) to retrieve data?
Yes, it's possible, but the method findById, from CrudRepository<T, ID> always expect to receive the ID as the argument - see here (be aware that JpaRepository extends from CrudRepository).
You can query by another field in a variety of ways, but as I saw that your use case is pretty simple, I'd suggest you to make use of Spring Data's query creation deriving from the method's name - see here and here
It's really simple, just create a method called findByUid in your EmotionRepository and Spring will take care of the rest by deriving the desired query from the method's name:
public interface EmotionRepository extends JpaRepository<Emotion, Integer> {
Optional<Emotion> findByUid(String uid);
}
Spring Data will then generate something like the following query:
SELECT e.* FROM Emotion e WHERE e.uid = <the-value>
Now, you can use findByUid method to query using the uid field.
And another fix: change from
EmotionRepository extends JpaRepository<Emotion, String>
to
EmotionRepository extends JpaRepository<Emotion, Integer>
this is because the id from Emotion is of type Integer, not String.
Inside your JpaRepository (EmotionRepository), you can make a function signature, anmd and the JPA will create a method to fetch whatever data you need.
This doc should help you understand exactly what is happening: http://static.springsource.org/spring-data/data-jpa/docs/1.0.0.M1/reference/html/#jpa.query-methods.query-creation

Spring Data JPA Repository Method Query Cannot Parse Property Name Following OrderBy

My Repository Method Query cannot parse a property name that comes after the OrderBy, but it can if it follows the findBy or findAllBy. The attribute in my entity that is giving me issues is zIndex
Entity Class
#Entity
public class DisplayLayer
{
#Id
#Column(name="DISPLAY_LAYER_ID")
private Long id;
#ManyToOne(cascade = CascadeType.ALL)
#JoinColumn(name="DISPLAY_CONTAINER_ID")
private DisplayContainer displayContainer;
#Column(name="NAME")
private String name;
#Column(name="Z_INDEX")
private Long zIndex;
#Column(name="DESCRIPTION")
private String description;
// Getters & Setters
}
Repository Class
public interface DisplayLayerRepository extends BaseRepository<DisplayLayer, Long>
{
// This one method query works just fine
public List<DisplayLayer> findByZIndex(Long pZIndex);
// These two throw the same error:
// Unable to locate Attribute with the the given name [ZIndex]
//
// public List<DisplayLayer> findAllByOrderByZIndexAsc();
// public List<DisplayLayer> findByDisplayContainerIdOrderByZIndexAsc(Long pDisplayLayerId);
}
So in my first method query, when OrderBy is not used, it correctly parses it as [zIndex], but when it follows the OrderBy for some reason it capitalizes the z: [ZIndex].
Is this a known issue, or am I doing something wrong?
EDIT
What makes me think there might be a bug with the parser is that if I update zIndex to IndexZ in my entity and then change my query methods from OrderByZIndex to OrderByIndexZ everything works.
This solution is not ideal though as the database table is Z_INDEX and that has a specific meaning when talking about layering of graphics.
Yes there is a bug but there is a simple workaround, please try :
displayLayerRepository.findAll(Sort.by(Sort.Order.desc("zIndex").ignoreCase()));
I suppose DisplayLayerRepository extends JpaRepository.
It's similar when you use parameters in your findBy method. Define in your repository empty method :
public List<DisplayLayer> findByDisplayContainerId(Long pDisplayLayerId, Sort sort);
Then you can call it by :
displayLayerRepository.findByDisplayContainerId(1L, Sort.by(Sort.Order.desc("zIndex").ignoreCase()));
You can use JpaSort in query. As in your case, it looks something like this:
Sort sort = JpaSort.unsafe(Sort.Direction.ASC, "zIndex");
List<DisplayLayer> zIndexes = displayLayerRepository.findByZIndex(pZIndex, sort);
and your query in Repository will look like this:
public List<DisplayLayer> findByZIndex(Long pZIndex, Sort sort);

How to get the specific property value from .properties file in Spring Data Repository interface method #Query

I am able to get the property value in Spring classes like below:
#Value("${database.name}")
private String databaseName;
I have to execute a native query by joining different tables which are in different databases.
#Query(value="select t1.* FROM db1.table1 t1 INNER JOIN db2.table2 t2 ON t2.t1_id1 = t1.id1")
Instead of hard coding database names i.e., db1 and db2 here, I have to get them from properties file.
how to get the property value inside the #Query annotation in Spring Data JPA Repository ?
I don't know if it is possible, but if not, you can consider this approach:
Instead of using properties in Repository's #Query directly, you can use params in the query but when you call the actual method - you can provide values from .properties.
Imagine you have simple repository:
public interface UserRepository extends JpaRepository<User, Long> {
// query with param
#Query("select u from User u where u.lastname = :lastname")
User findByLastname(#Param("lastname") String lastname);
}
Then, let's say you have some Service or Controller where you need to use your Repository - you can inject properties there and pass them to your method:
#Service
public class UserService {
// this comes from .properties
#Value("${user.lastName}")
private String userLastName;
#Autowired
private UserRepository userRepository;
public User getUser() {
// you pass it as param to the repo method which
// injects it into query
return userRepository.findByLastname(userLastName);
}
}
This is just an example. But I believe it may be useful.
Happy hacking :)

How to paginate a list of object (not mapped domain object) using custom pagination with spring data

I have a repository called BananaRepositoryImpl that contains a function that return a list of BananaDTO ( the legacy code can't return the mapped entity ( Banana.java ),it's a constraint and I can't change this behavior :( )
public class BananaRepositoryImpl implements BananaRepository{
#Autowired
EntityManager em;
public List<BananaDTO> findAllBananes(){
//logic to get list of bananasDTO object types using Query query = em.createQuery(JPQL_QUERY_HERE);
}
}
Knowing that the BananaDTO object is a DTO for Banana.java class which looks like this :
#Data
#Entity
public class Banana{
private Long id;
private Double price;
private Double weight;
}
What I should do is to implement pagination over the findAllBananes() method so that I can return a Page using spring Data ( or another approach ).
Assuming the attributes of the BananaDTO is a subset of the Banana entities attributes you can use class-based projection support of Spring Data JPA, i.e. you just add a Pageable as a parameter to your method and return a Page<BananaDTO>:
interface BananaRepository extends CrudRepository<Banana, Long> {
Page<BananaDTO> findAllBananes(Pageable page)
}

How can I use Spring's pagination (using Pageable) while writing a dynamic query using QueryDSL?

I am trying to use pagination with QueryDSL - using the com.mysema.querydsl package.
All my Querydsl query types look like this -
#Generated("com.mysema.query.codegen.EntitySerializer")
public class QCountry extends EntityPathBase<Country> {...}
Currently, my repository implementation class looks something like this -
#Override
public Page<Country> findPaginatedCountries(String country, Optional<String> status, Pageable pageable) {
QCountry qCountry= QCountry.someObject;
QActiveCountry qActiveCountry = QActiveCountry.activeCountry;
JPAQuery jpaQuery = new JPAQuery(entityManager);
QueryBase queryBase = jpaQuery.from(qCountry).innerJoin(qActiveCountry).fetch()
.where(qCountry.codeLeft.country.upper().eq(country.toUpperCase()))
.where(qCountry.codeRight.country.upper().eq(country.toUpperCase()));
if(status.isPresent()){
queryBase = queryBase.where(qActiveCountry.id(qCountry.active.id))
.where(qActiveCountry.status.upper().eq(status.get().toUpperCase()));
}
.......}
Now, I want this dynamic query to return a paginated response. I want to use Spring's pagination to do that and not manually set offset, size etc.
I know I can use QueryDslRepositorySupport class - as implemented here - https://github.com/keke77/spring-data-jpa-sample/blob/master/spring-data-jpa/src/main/java/com/gmind7/bakery/employee/EmployeeRepositoryImpl.java
Sample code from the above link -
#Override
public Page<Employees> QFindByOfficeCode(long officeCode, Pageable pageable) {
//JPAQuery query = new JPAQuery(em);
JPQLQuery query = from(QEmployees.employees).where(QEmployees.employees.officeCode.eq(officeCode));
query = super.getQuerydsl().applyPagination(pageable, query);
SearchResults<Employees> entitys = query.listResults(QEmployees.employees);
return new PageImpl<Employees>(entitys.getResults(), pageable, entitys.getTotal());
}
However, to do that -
I need to pass JPQLQuery object to the applyPagination method. How can I do that without changing my code (Ofcourse, the repository class will extend QueryDslRepositorySupport class). Currently, I am using JPAQuery as you can see.
OR
I probably need to change my QueryDSL types by having them extend EntityPath instead of EntityPathBase so that I can use JPQLQuery.from() to generate the query and then use the applyPagination method, which requires a JPQLQuery object. However, my Q classes are extending EntityPathBase class instead. Should I be use com.querydsl package instead of com.mysemsa.querydsl package to generate query types?
OR
Other option is to use the following - http://docs.spring.io/spring-data/commons/docs/current/api/org/springframework/data/querydsl/QueryDslPredicateExecutor.html#findAll-com.querydsl.core.types.Predicate-org.springframework.data.domain.Pageable-
Code snippet below -
Page<T> page = QueryDslPredicateExecutor.findAll(org.springframework.data.querydsl.Predicate predicate, Pageable pageable)
However, I am making joins between two tables and then filtering results with a where clause (as you can see above in my code). How can I pass a predicate object in the findAll method above? Not sure how to include a join in it.
Please let me know if the problem is not clear, I can add more details.
EDIT: There is a many to one relationship between Country and ActiveCountry.
Country class has an ActiveCountry reference. And we have to do a join between both ids. Is is possible that Country can have null ActiveCountry. Therefore, we want an inner join - only non null values for active country
#ManyToOne
#JoinColumn(name="id")
ActiveCountry active;
Step 1: Annotate the entity class with #QueryEntity
#Entity
#QueryEntity
public class Country {}
This seems to have been addressed already since the question shows Q classes.
Step 2: Have the repository interface extend QueryDslPredicateExecutor
public interface CountryRepository
extends PagingAndSortingRepository<Country, Long>
, QueryDslPredicateExecutor<Country> {
}
Step 3: Invoke the Page<T> findAll(Predicate query, Pageable page) method provided by QueryDslPredicateExecutor
public Page<Country> getCountries(String country, Optional<String> status, Pageable page) {
QCountry root = QCountry.country;
BooleanExpression query = root.codeLeft.country.equalsIgnoreCase(country);
query = query.and(root.codeRight.country.equalsIgnoreCase(country));
if (status.isPresent()) {
query = query.and(root.active.status.equalsIgnoreCase(status));
}
return countryRepository.findAll(query, page);
}

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