NamedEntityGraph Returns All Columns and Objects - spring

I am trying to utilize a NamedEntityGraph to limit the return data for specific queries. Mainly I do not want to return full object details when listing the object. A very simple class example is below.
#Entity
#Table(name="playerreport",schema="dbo")
#NamedEntityGraphs({
#NamedEntityGraph(name = "report.simple",
attributeNodes =
{#NamedAttributeNode(value="intId")
}
)
})
public class PlayerReportEntity {
#Id
#Column(name="intid",columnDefinition="uniqueidentifier")
private String intId;
#Column(name="plyid",columnDefinition="uniqueidentifier")
#Basic(fetch=FetchType.LAZY)
private String plyId;
#ManyToOne(fetch=FetchType.LAZY)
#JoinColumn(name = "plyid", insertable=false,updatable=false)
private PlayerEntity player;
No matter what I do to plyId and player are always returned. Is there any way to only return the requested columns (intId) ?
As for the collection Hibernate does not do the join for the player object but it still returns player as null. So that part is working to an extent.
I am using a JPARepository below to generate Crud Statements for me
public interface PlayerReportRepository extends JpaRepository<PlayerReportEntity, String> {
#EntityGraph(value="report.simple")
List<PlayerIntelEntity> findByPlyId(#Param(value = "playerId") String playerId);
#Override
#EntityGraph(value="report.simple")
public PlayerIntelEntity findOne(String id);
}

A chunk of text from here - "Hence it seems that the #NamedEntityGraph only affects fields that are Collections, but fields that are not a Collection are always loaded." from JIRA
Please use the Example 47 on this page and use repositories accordingly.
In essence, hibernate is right now loading all the feilds in the class and for collections it will work if you follow the example stated above.
Thanks.

Related

Dynamic JPA query

I have two entities Questions and UserAnswers. I need to make an api in spring boot which returns all the columns from both the entities based on some conditions.
Conditions are:
I will be give a comparator eg: >, <, =, >=, <=
A column name eg: last_answered_at, last_seen_at
A value of the above column eg: 28-09-2020 06:00:18
I will need to return an inner join of the two entities and filter based on the above conditions.
Sample sql query based on above conditions will be like:
SELECT q,ua from questions q INNER JOIN
user_answers ua on q.id = ua.question_id
WHERE ua.last_answered_at > 28-09-2020 06:00:18
The problem I am facing is that the column name and the comparator for the query needs to be dynamic.
Is there an efficient way to do this using spring boot and JPA as I do not want to make jpa query methods for all possible combinations of columns and operators as it can be a very large number and there will be extensive use of if else?
I have developed a library called spring-dynamic-jpa to make it easier to implement dynamic queries with JPA.
You can use it to write the query templates. The query template will be built into different query strings before execution depending on your parameters when you invoke the method.
This sounds like a clear custom implementation of a repository method. Firstly, I will make some assumptions about the implementation of your entities. Afterwards, I will present an idea on how to solve your challenge.
I assume that the entities look basically like this (getters, setters, equals, hachCode... ignored).
#Entity
#Table(name = "questions")
public class Question {
#Id
#GeneratedValue
private Long id;
private LocalDateTime lastAnsweredAt;
private LocalDateTime lastSeenAt;
// other attributes you mentioned...
#OneToMany(mappedBy = "question", cascade = CascadeType.ALL, orphanRemoval = true)
private List<UserAnswer> userAnswers = new ArrayList();
// Add and remove methods added to keep bidirectional relationship synchronised
public void addUserAnswer(UserAnswer userAnswer) {
userAnswers.add(userAnswer);
userAnswer.setQuestion(this);
}
public void removeUserAnswer(UserAnswer userAnswer) {
userAnswers.remove(userAnswer);
userAnswer.setQuestion(null);
}
}
#Entity
#Table(name = "user_answers")
public class UserAnswer {
#Id
#GeneratedValue
private Long id;
#ManyToOne(fetch = FetchType.LAZY)
#JoinColumn(name = "task_release_id")
private Question question;
}
I will write the code with the knowledge about the JPA of Hibernate. For other JPAs, it might work similarly or the same.
Hibernate often needs the name of attributes as a String. To circumvent the issue of undetected mistakes (especially when refactoring), I suggest the module hibernate-jpamodelgen (see the class names suffixed with an underscore). You can also use it to pass the names of the attributes as arguments to your repository method.
Repository methods try to communicate with the database. In JPA, there are different ways of implementing database requests: JPQL as a query language and the Criteria API (easier to refactor, less error prone). As I am a fan of the Criteria API, I will use the Criteria API together with the modelgen to tell the ORM Hibernate to talk to the database to retrieve the relevant objects.
public class QuestionRepositoryCustomImpl implements QuestionRepository {
#PersistenceContext
private EntityManager entityManager;
#Override
public List<Question> dynamicFind(String comparator, String attribute, String value) {
CriteriaBuilder cb = entityManager.getCriteriaBuilder();
CriteriaQuery<Question> cq = cb.createQuery(Question.class);
// Root gets constructed for first, main class in the request (see return of method)
Root<Question> root = cq.from(Question.class);
// Join happens based on respective attribute within root
root.join(Question_.USER_ANSWER);
// The following ifs are not the nicest solution.
// The ifs check what comparator String contains and adds respective where clause to query
// This .where() is like WHERE in SQL
if("==".equals(comparator)) {
cq.where(cb.equal(root.get(attribute), value));
}
if(">".equals(comparator)) {
cq.where(cb.gt(root.get(attribute), value));
}
if(">=".equals(comparator)) {
cq.where(cb.ge(root.get(attribute), value));
}
if("<".equals(comparator)) {
cq.where(cb.lt(root.get(attribute), value));
}
if("<=".equals(comparator)) {
cq.where(cb.le(root.get(attribute), value));
}
// Finally, query gets created and result collected and returned as List
// Hint for READ_ONLY is added as lists are often just for read and performance is better.
return entityManager.createQuery(cq).setHint(QueryHints.READ_ONLY, true).getResultList();
}
}

Spring Boot JPA find, filter

As Spring jpa Provides some usefull features to find Items from a repository by defining it in the method name. e .x findByTitle(String title) then Spring is automatically searching the Title Colum for the given String. If i have an int column named numberOfCopies and i want only to find the datasets with >0 greater then null how would define such a method ?
to filter out those books with the numberOfCopies equals 0 = zero
#Entity
public class Book {
#Id
private int id;
private String title;
private int numberOfCopies;
}
can i use the Repomethod
public List findBooksByNumberOfCopies.greater then 0 ?To Use this Spring Feature without some if or for loops
First, you should use Integer, since it is better, in my opinion, to use wrapper classes than to primitives, and enforce not null requirement through annotations, e.g. #Column(nullable = false)
#Entity
public class Book {
#Id
private Integer id;
private String title;
private Integer numberOfCopies;
}
Then you can add the following two methods in your BookRepository;
List<Book> findByNumberOfCopiesGreaterThan(Integer numberOfCopies);
default List<Book> findAllAvailableBooks() {
return findByNumberOfCopiesGreaterThan(0);
}
and use the default findAllAvailableBooks method, with hardcoded 0 value which is your requirement.
you can easily use
List<Book> findByNumberOfCopiesGreaterThanEqual(int numberOfCopies);
Pretty sure this would work:
public interface BookRepo extends JpaRepository<Book, Integer> {
#Query("SELECT b FROM Book b WHERE b.numberOfCopies >= 0")
public Optional<List<Book>> getTheBooksWithMultCopies();
}
// back in your component class:
{
...
Optional<List<Book>> optionalBookList = myBookRepo.getTheBooksWithMultCopies();
if (optionalBookList.isPresent()){
List<Book> bookList = optionalBookList.get();
}
}
Note that the language within the query is called HQL, which is what is used by Hibernate internally (which is used by JPA internally). It's really not very intimidating - just, know that you the objects in your POJO, which map to your database table, rather than your database table directly.
Also, I'd recommend using Integer over int in entity classes, at least if your value is nullable. Otherwise, numberOfCopies will always default to 0, which may not be desirable and may cause exceptions that are difficult to decipher.
GreaterThanEqual takes an Integer not int
List<Book> findByNumberOfCopiesGreaterThanEqual(Integer numberOfCopies);

Spring Data Rest - sort by nested property

I have a database service using Spring Boot 1.5.1 and Spring Data Rest. I am storing my entities in a MySQL database, and accessing them over REST using Spring's PagingAndSortingRepository. I found this which states that sorting by nested parameters is supported, but I cannot find a way to sort by nested fields.
I have these classes:
#Entity(name = "Person")
#Table(name = "PERSON")
public class Person {
#ManyToOne
protected Address address;
#ManyToOne(targetEntity = Name.class, cascade = {
CascadeType.ALL
})
#JoinColumn(name = "NAME_PERSON_ID")
protected Name name;
#Id
protected Long id;
// Setter, getters, etc.
}
#Entity(name = "Name")
#Table(name = "NAME")
public class Name{
protected String firstName;
protected String lastName;
#Id
protected Long id;
// Setter, getters, etc.
}
For example, when using the method:
Page<Person> findByAddress_Id(#Param("id") String id, Pageable pageable);
And calling the URI http://localhost:8080/people/search/findByAddress_Id?id=1&sort=name_lastName,desc, the sort parameter is completely ignored by Spring.
The parameters sort=name.lastName and sort=nameLastName did not work either.
Am I forming the Rest request wrong, or missing some configuration?
Thank you!
The workaround I found is to create an extra read-only property for sorting purposes only. Building on the example above:
#Entity(name = "Person")
#Table(name = "PERSON")
public class Person {
// read only, for sorting purposes only
// #JsonIgnore // we can hide it from the clients, if needed
#RestResource(exported=false) // read only so we can map 2 fields to the same database column
#ManyToOne
#JoinColumn(name = "address_id", insertable = false, updatable = false)
private Address address;
// We still want the linkable association created to work as before so we manually override the relation and path
#RestResource(exported=true, rel="address", path="address")
#ManyToOne
private Address addressLink;
...
}
The drawback for the proposed workaround is that we now have to explicitly duplicate all the properties for which we want to support nested sorting.
LATER EDIT: another drawback is that we cannot hide the embedded property from the clients. In my original answer, I was suggesting we can add #JsonIgnore, but apparently that breaks the sort.
I debugged through that and it looks like the issue that Alan mentioned.
I found workaround that could help:
Create own controller, inject your repo and optionally projection factory (if you need projections). Implement get method to delegate call to your repository
#RestController
#RequestMapping("/people")
public class PeopleController {
#Autowired
PersonRepository repository;
//#Autowired
//PagedResourcesAssembler<MyDTO> resourceAssembler;
#GetMapping("/by-address/{addressId}")
public Page<Person> getByAddress(#PathVariable("addressId") Long addressId, Pageable page) {
// spring doesn't spoil your sort here ...
Page<Person> page = repository.findByAddress_Id(addressId, page)
// optionally, apply projection
// to return DTO/specifically loaded Entity objects ...
// return type would be then PagedResources<Resource<MyDTO>>
// return resourceAssembler.toResource(page.map(...))
return page;
}
}
This works for me with 2.6.8.RELEASE; the issue seems to be in all versions.
From Spring Data REST documentation:
Sorting by linkable associations (that is, links to top-level resources) is not supported.
https://docs.spring.io/spring-data/rest/docs/current/reference/html/#paging-and-sorting.sorting
An alternative that I found was use #ResResource(exported=false).
This is not valid (expecially for legacy Spring Data REST projects) because avoid that the resource/entity will be loaded HTTP links:
JacksonBinder
BeanDeserializerBuilder updateBuilder throws
com.fasterxml.jackson.databind.exc.MismatchedInputException: Cannot construct instance of ' com...' no String-argument constructor/factory method to deserialize from String value
I tried activate sort by linkable associations with help of annotations but without success because we need always need override the mappPropertyPath method of JacksonMappingAwareSortTranslator.SortTranslator detect the annotation:
if (associations.isLinkableAssociation(persistentProperty)) {
if(!persistentProperty.isAnnotationPresent(SortByLinkableAssociation.class)) {
return Collections.emptyList();
}
}
Annotation
#Retention(RetentionPolicy.RUNTIME)
#Target(ElementType.FIELD)
public #interface SortByLinkableAssociation {
}
At project mark association as #SortByLinkableAssociation:
#ManyToOne
#SortByLinkableAssociation
private Name name;
Really I didn't find a clear and success solution to this issue but decide to expose it to let think about it or even Spring team take in consideration to include at nexts releases.
Please see https://stackoverflow.com/a/66135148/6673169 for possible workaround/hack, when we wanted sorting by linked entity.

Why is JPA query so slow?

I am implementing queries in my web application with JPA repositories. The two main tables I am querying from are FmReportTb and SpecimenTb.
Here are the two entity classes (only important attributes are listed).
//FmReportTb.java
#Entity
#Table(name="FM_REPORT_TB")
public class FmReportTb implements Serializable {
#Column(name="ROW_ID")
private long rowId;
#Column(name="FR_BLOCK_ID")
private String frBlockId;
#Column(name="FR_FULL_NAME")
private String frFullName;
#OneToOne
#JoinColumn(name="SPECIMEN_ID")
private SpecimenTb specimenTb;
FmReportTb has OneToOne relationship with SpecimenTb.
#Entity
#Table(name="SPECIMEN_TB")
public class SpecimenTb implements Serializable {
private String mrn;
#OneToOne(mappedBy="specimenTb", cascade=CascadeType.ALL)
private FmReportTb fmReportTb;
The query I am working on is to find all records in FmReportTb and show a few attributes from FmReportTb plus mrn from SpecimenTb.
Here is my JPA repository for FmReportTb:
#Repository
public interface FmReportRepository extends JpaRepository<FmReportTb, Long> {
#Query("select f from FmReportTb f where f.deleteTs is not null")
public List<FmReportTb> findAllFmReports();
Since, I am only showing part of the attributes from FmReportTb and one attribute from SpecimenTb, I decided to create a Value Object for FmReportTb. The constructor of the VO class assigns attributes from FmReportTb and grabs mrn attribute from SpecimenTb based on the OneToOne relationship. Another reason for using VO is because table FmReportTb has a lot of OneToMany children entities. For this particular query, I don't need any of them.
public class FmReportVO {
private String frBlockId;
private Date frCollectionDate;
private String frCopiedPhysician;
private String frDiagnosis;
private String frFacilityName;
private String frFullName;
private String frReportId;
private String filepath;
private String mrn;
public FmReportVO(FmReportTb fmReport) {
this.frBlockId = fmReport.getFrBlockId();
this.frCollectionDate = fmReport.getFrCollectionDate();
this.frCopiedPhysician = fmReport.getFrCopiedPhysician();
this.frDiagnosis = fmReport.getFrDiagnosis();
this.frFacilityName = fmReport.getFrFacilityName();
this.frFullName = fmReport.getFrFullName();
this.frReportId = fmReport.getFrReportId();
this.mrn = fmReport.getSpecimenTb().getMrn();
}
I implemented findall method in servicebean class to return a list of FmReportTb VOs.
//FmReportServiceBean.java
#Override
public List<FmReportVO> findAllFmReports() {
List<FmReportTb> reports = fmReportRepository.findAllFmReports();
if (reports == null) {
return null;
}
List<FmReportVO> fmReports = new ArrayList<FmReportVO>();
for (FmReportTb report : reports) {
FmReportVO reportVo = new FmReportVO(report);
String filepath = fileLoadRepository.findUriByFileLoadId(report.getFileLoadId().longValue());
reportVo.setFilepath(filepath);
fmReports.add(reportVo);
}
return fmReports;
}
Lastly, my controller looks like this:
#RequestMapping(
value = "/ristore/foundation/",
method = RequestMethod.GET,
produces = "application/json")
public ResponseEntity<List<FmReportVO>> getAllFmReports() {
List<FmReportVO> reports = ristoreService.findAllFmReports();
if (reports == null) {
return new ResponseEntity<List<FmReportVO>>(HttpStatus.NOT_FOUND);
}
return new ResponseEntity<List<FmReportVO>>(reports, HttpStatus.OK);
}
There are about 200 records in the database. Surprisingly, it took almost 2 full seconds to retrieve all the records in JSON. Even though I did not index all the tables, this is way too slow. Similar query takes about probably a few ms on the database directly. Is it because I am using Value Objects or JPA query tends to be this slow?
EDIT 1
This may have to do with the fact that FmReportTb has almost 20 OneToMany entities. Although the fetchmode of these child entities are set to LAZY, JPA Data repository tends to ignore the fetchmode. So I ended up using NamedEntityGraph to specify the attributes EAGER. This next section is added to the head of my FmReportTb entity class.
#Entity
#NamedEntityGraph(
name = "FmReportGraph",
attributeNodes = {
#NamedAttributeNode("fileLoadId"),
#NamedAttributeNode("frBlockId"),
#NamedAttributeNode("frCollectionDate"),
#NamedAttributeNode("frDiagnosis"),
#NamedAttributeNode("frFullName"),
#NamedAttributeNode("frReportId"),
#NamedAttributeNode("specimenTb")})
#Table(name="FM_REPORT_TB")
And then #EntityGraph("FmReportGraph") was added before the JPA repository query to find all records. After doing that, the performance is improved a little bit. Now fetching 1500 records only takes about 10 seconds. However, it still seems too slow given each json object is fairly small.
Answering for the benefit of others with slow JPA queries...
As #Ken Bekov hints in the comments, foreign keys can help a lot with JPA.
I had a couple of tables with a many to one relationship - a query of 100,000 records was taking hours to perform. Without any code changes I reduced this to seconds just by adding a foreign key.
In phpMyAdmin you do this by creating a Relationship from the "many" table to the "one" table. For a detailed explanation see this question: Setting up foreign keys in phpMyAdmin?
and the answer by #Devsi Odedra

Join 3 tables in Spring Jpa Data

I've been struggling lately to join 3 tables with spring data jpa. I have 3 entities, Series, Dossier and Item. Series has many Dossiers, and Dossier has many Items (Relationships). I do something like Series.join(Dossier_.series).join(Dossier_.items) and I end up with a Join set. I want to make the following query:
Select Items from Series,Dossier,Item
Where Series.Id=Dossier.seriesId
and Dossier.id=Item.dossierId
and series.projectId = :param
I can't express this statement with Spring Specifications and criteria api....Please shed some light
It is more a JPA question.
First, I always emphasize, you are not access "tables". You should view them as domain entities. Lot of misuse of JPA/Hibernate/other ORMs actually comes from direct "translate" of SQL or database concepts.
Back to your question, the answer is simple. First make sure you actually have the "relationships" in your domain entities. Storing IDs is not helping to build a concrete domain model. For example, you have something like :
#Entity
class Series {
#Id
Long id;
#OneToMany(mappedBy="series")
List<Dossier> dossiers;
}
#Entity
class Dossier{
#Id
Long id;
#ManyToOne
Series series;
#OneToMany(mappedBy="dossier"
List<Item> items;
}
#Entity
class Item{
#Id
Long id;
#ManyToOne
Dossier dossier;
}
The query is straight-forward:
select s.dossiers.items from Series s where s.projectId = :param
Or, if it is more reasonable to have only the #ManyToOnes and omit the #OneToManys, the query is still straight-forward:
from Item where i.dossier.series.projectId = :param
[Still rocky here] Maybe i didn't make myself clear.I know how to express the query in HQL.The problem is to use Spring Data's Specifications,with the help of the criteria api to build that query.
//Let's exampine the following piece of code
public class CustomItemSpecs {
public static Specification<Item> createSpecificationFromSearchForm(final SearchForm searchForm) {
return new Specification<Item>() {
#Override
public Predicate toPredicate(Root<Item> root, CriteriaQuery<?> query, CriteriaBuilder cb) {
CriteriaQuery<Item> cq = cb.createQuery(Item.class);
CriteriaQuery<Series> sb = cb.createQuery(Series.class);
Root<Series> series = sb.from(Series.class);
Join<Series,Dossier> join1 = series.join(Series_.dossiers);
Join<Dossier, Item> join2 = join1.join(Dossier_.items);
}
}
}
As you can see i manage to do two seperate joins.The problem is when i want to join Series,Dossier And Items to execute the above query.Notice that join2 is a Dossier-Item set.I can't make criteria like cb.equals(join2.get(Series_.projectId),)
im using spring data interface projection .
example like this
note items must be interface class
Repository Class
#Repository
public interface ItemRepository extends JpaRepository<Items,Long> {
#Query(nativeQuery = true,value = "Select Items from Series,Dossier,Item Where Series.Id=Dossier.seriesId and Dossier.id=Item.dossierId and series.projectId = :param")
public Items findTransaksisByAccountIdOrderById(#Param("param") Long projectId);
}

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