How do I query based on fields of EmbeddedId using JPA Criteria? - java

I've spent the last couple of hours trying to figure this out, mostly by searching since I figure someone has already done it, but I'm not finding an answer that works for me.
Here's an on-the-fly translation of my code to something I can put in public.
#Entity #Table(name="result")
public class Result implements Serializable {
#Embeddable
public static class ResultPK implements Serializable {
#Column(name="result_date", nullable=false)
#Type(type="com.example.HibernateUTC$LocalDateType") // <- UserType
public LocalDate resultDate;
#Column(name="name", nullable=false)
public String name;
#Column(name="category", nullable=false)
public String category;
public ResultPK() {}
public ResultPK(Date resultDate, String name, String category) {
this.resultDate = resultDate;
this.name = name;
this.category = category;
}
// ...more code for hashCode/equals, no setters or getters...
}
#EmbeddedId
#AttributeOverrides({
#AttributeOverride(name="resultDate", column=#Column(name="result_date", nullable = false)),
#AttributeOverride(name="name", column=#Column(name="name", nullable=false)),
#AttributeOverride(name="category", column=#Column(name="category", nullable = false)),
})
private ResultPK resultId;
#Column(name="r_square")
private Double rSq;
#Column(name="p_value")
private pValue;
// ... more code for other fields, setters, getters, but nothing else; vanilla pojo...
}
I have a DAO where queries are hiding; the method that I'm calling is this
#Repository("resultDAO")
public class ResultDAOImpl extends AbstractBaseDAO<Result> implements ResultDAO {
// boilerplate for intializing base class and other queries
#Override
public List<Result> findDateRange(String category, String name, LocalDate begDate, LocalDate endDate) {
EntityManager em = entityManagerFactory.createEntityManager();
CriteriaBuilder cb = em.getCriteriaBuilder();
CriteriaQuery<Result> q = cb.createQuery(Result.class);
Root<Result> root = q.from(Result.class);
Predicate catMatch = cb.equal(root.get("resultId.category"), category);
Predicate nameMatch = cb.equal(root.get("resultId.name"), name);
Predicate dateRange = cb.between(root.get("resultId.resultDate"), begDate, endDate);
q.select(root).where(cb.and(catMatch, nameMatch, dateRange));
return em.createQuery(q).getResultList();
}
}
When I attempt to run the code that executes that query, I end up with an error
Exception in thread "main" java.lang.IllegalArgumentException: Unable to locate Attribute with the the given name [resultId.category] on this ManagedType [com.example.Result]
Some of the similar questions I've found made it look like I needed to use the resultPK or ResultPK in the query. I've tried those, no joy. I have no idea how to specify the fields in the key for the query, or if I need something totaly different to this. I really need a clue...
I'm using Spring 4.3.8.RELEASE and Hibernate 4.3.11.Final, Java 8 (hence the UserType to handle LocalDate).
Edited to correct some inconsistencies in my transcription of the actual code.

You should modify your predicates this way:
Predicate catMatch = cb.equal(root.get("resultId").get("category"), category);
Predicate nameMatch = cb.equal(root.get("resultId").get("name"), name);
Predicate dateRange = cb.between(root.get("resultId").get("resultDate"), begDate, endDate);
By the way no need to use cb.and inside of where statement. You can make the code a bit shorter using
q.select(root).where(catMatch, nameMatch, dateRange);

Related

returning a list of instances by the foreign key

I have two models, Owner and Contract. A contract has an instance of an owner, owner does not have a list of contracts. I'm trying to query my list of contracts, to return a list filtered by owner, ie, a list of contracts by owner.
I had tried to follow previous examples and use Criteria to write a custom query, but, following suggestions I've checked the docks and tried to use named queries instead, however, I'm still really struggling.
There was an unexpected error (type=Internal Server Error, status=500).
Named parameter not bound : ownerId; nested exception is org.hibernate.QueryException: Named parameter not bound : ownerId
My models look like this:
#Entity
#Table(name="Contracts")
#NamedQueries({
#NamedQuery(
name = "Contract.allContractsByOwner",
query = "SELECT c FROM Contract c WHERE c.owner.id LIKE :ownerId"
)
})
public class Contract {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column
private Long id;
#ManyToOne
private Owner owner;
#Column
private double price;
#Column
private String deliverDate;
public Contract(Owner owner, double price, String deliverDate) {
this.id = id;
this.owner = owner;
this.price = price;
this.deliverDate = deliverDate;
}
and
#Entity
#Table(name="Owners")
public class Owner {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column
private Long id;
#Column
private String name;
public Owner(String name){
this.name = name;
}
my contractRepoImpl
#Service
public class ContractRepositoryImpl implements ContractRepositoryCustom {
ContractRepository contractRepository;
#Autowired
EntityManager entityManager;
public List allContractsByOwner(Long ownerId) {
List contracts = entityManager.createQuery(
"SELECT c FROM Contract c WHERE c.owner.id LIKE :ownerId", Contract.class)
.getResultList();
return contracts;
}
}
which I name in my ContractRepo and ContractRepoCustom files, and then in my controller I map to it like so. But, when I query it in my browser I get the error in my terminal.
#GetMapping(value="/owners/{ownerId}/contracts")
public List allContractsByOwner(#PathVariable("ownerId") Long ownerId){
return contractRepository.allContractsByOwner(ownerId);
}
I appreciate this is probably beginners mistakes, I am trying to follow docs but get a bit stuck with syntax & where annotations need to go.
Thanks JB Nizet, got there in the end
I added parameters to my contractRepoImpl
#Service
public class ContractRepositoryImpl implements ContractRepositoryCustom {
ContractRepository contractRepository;
#Autowired
EntityManager entityManager;
public List allContractsByOwner(Long id) {
List contracts = entityManager.createQuery(
"SELECT c FROM Contract c WHERE c.owner.id = :ownerId", Contract.class)
.setParameter("ownerId", id)
.getResultList();
return contracts;
}
}
that then produced a SQL error, which I fixed by changing my #NamedQuery from 'LIKE' to '=' in my Contract class...
#NamedQueries({
#NamedQuery(
name = "Contract.allContractsByOwner",
query = "SELECT c FROM Contract c WHERE c.owner.id = :ownerId"
)
})

JPA - How to query using Specification and #EmbeddedId?

I'm using a JPA query that uses a specification to retrieve entities. When I execute the query, I'm getting the error:
org.springframework.data.mapping.PropertyReferenceException: No property name found for type Task!
I've looked at the answers to similar questions that have been asked on this site previously & tried to model my code to follow the patterns that were recommended but the code is still failing.
When I step through the code with a debugger, the expanded path in the criteria builder is returning the embedded ID class, but when the specification is actually used in the query it looks like the attribute is being applied to the base entity class.
Am I missing something obvious?
Here is the entity class:
#Entity
#Table(name = "TASKS")
public class Task implements Serializable {
#EmbeddedId
private TaskId id;
...more attributes, getters and setters
}
Here is the embedded ID entity class:
#Embeddable
public class TaskId implements Serializable {
#Column(name = "NAME", length = 100)
private String name;
...more attributes, getters and setters
}
Here is the specification builder that matches on the embedded id 'name' attribute:
public class HasTaskNameSpec {
private HasTaskNameSpec() {
}
public static Specification<Task> equals(String name) {
return (root, query, criteriaBuilder) -> {
return criteriaBuilder.equal(root.get("id").get("name"), taskName);
};
}
}
The query is executed on the repository as follows:
List<Task> results = taskRepository.findAll(HasTaskNameSpec.equals("foo"));
The repository itself is very simple:
public interface TaskRepository extends JpaRepository<Task, TaskId>, JpaSpecificationExecutor<Task> {
List<Task> findByIdName(String name);
Page<Task> findByIdName(String name, Pageable page);
}
** EDIT added methods to repository as was suggested below **
Ahh, the root cause was totally in our codebase. There was a sort order being specified on the page that didn't include the embedded "id" attribute. The above code works.
'root.get({embeddedIdName}).get({subPropertyName})' is used to query on embeddedId using specification.
#Embeddable
public class ProjectId implements Serializable{
private static final long serialVersionUID = 1L;
#Column(name = "PROJECT_NAME")
private String projectName;
#Column(name = "ORGANIZATION")
private String organization;
......
......
}
#Entity
#Table(name = "projects")
public class Project {
#EmbeddedId
private ProjectId projectId;
#Column(name = "STARTED_TIME")
private Timestamp startedTime;
#Column(name = "ACTIVE")
private String active;
#Column(name = "DESCRIPTION")
private String description;
......
......
}
In the above snippet, ProjectId is an embedded id. To query on projectName, we should use below snippet.
expression = root.get("projectId").get("projectName");
Demo application link.
Take a look at this link which has a similar query.
EmbbededId Lookup
The final answer suggests that you can add a method to your TaskRepository thus.
public interface TaskRepository extends JpaRepository<Task, TaskId>, JpaSpecificationExecutor<Task> {
public List<Task> findByIdName(String name);
}

PersistenceException, Column 'id' specified twice

I have the following files in Play Framework 2.2.3
Controller:
public class Comment extends Controller
{
public Result create(UUID id)
{
models.blog.Blog blog = models.blog.Blog.finder.byId(id);
Result result;
if(blog == null)
{
result = notFound(main.render("404", error404.render()));
}
else
{
Form<models.blog.Comment> commentForm = Form.form(models.blog.Comment.class);
commentForm = commentForm.bindFromRequest();
if(commentForm.hasErrors())
{
result = badRequest(Json.toJson(commentForm));
}
else
{
models.blog.Comment comment = commentForm.get();
comment.setId(UUID.randomUUID());
comment.setTimeCreated(new Date());
comment.setBlogId(blog.getId());
comment.save();
result = ok(Json.toJson(comment));
}
}
return result;
}
}
And two models
#Entity
#Table(name="blog")
public class Blog extends Model
{
private static final SimpleDateFormat MONTH_LITERAL = new SimpleDateFormat("MMMMM"),
DAY_NUMBER = new SimpleDateFormat("d"),
YEAR_NUMBER = new SimpleDateFormat("yyyy");
public static Finder<UUID, Blog> finder = new Finder<UUID, Blog>(UUID.class, Blog.class);
#Id
#Column(name="id",length=36, nullable=false)
public UUID id;
#OneToOne
#JoinColumn(name="author_id")
public User author;
#Column(name="title",length=255)
public String title;
#Column(name="summary",length=255)
public String summary;
#Column(name="url",length=255)
public String url;
#Column(name="content")
public String content;
#Column(name="time_updated")
public Date time_created;
#Column(name="time_created", nullable=false)
public Date time_updated;
#OneToMany(cascade = CascadeType.ALL)
#JoinColumn(name="blog_id")
public List<Comment> comments;
#ManyToMany(cascade = CascadeType.ALL)
#JoinTable(
name="blog_tag_map",
joinColumns={ #JoinColumn(name="blog_id", referencedColumnName="id") },
inverseJoinColumns={ #JoinColumn(name="tag_id", referencedColumnName="id") }
)
public List<Tag> tags;
public List<Comment> getComments()
{
return this.comments;
}
}
#Entity
#Table(name="blog_comment")
public class Comment extends Model
{
private static final SimpleDateFormat MONTH_LITERAL = new SimpleDateFormat("MMMMM"),
DAY_NUMBER = new SimpleDateFormat("d"),
YEAR_NUMBER = new SimpleDateFormat("yyyy");
#Id
#Column(name="id",length=36, nullable=false)
public UUID id;
#Column(name="blog_id", length=36)
public UUID blog_id;
#ManyToOne
public Blog blog;
#Column(name="content", length=500)
public String content;
#Column(name="website", length=255)
public String website;
#Column(name="name", length=255)
public String name;
#Column(name="time_created", updatable=false)
public Date time_created;
}
I have excluded some setters and getters from these models for brevity, so it doesn't clog up this post.
When I attempt to make a POST request to the aforementioned controller, everything goes fine until I get to the "comment.save()" statement in the controller file, then I get the following error.
I'm unsure why this save isn't going through, and why there is a column conflict.
Help much appreciated
The issue lies in the fact that you have defined basically two foreign key columns for Blog in your Comment's entity:
#Column(name = "blog_id", length = 36)
public UUID blog_id;
#ManyToOne
public Blog blog;
The default column name for your 'blog' field is: blog_id
However, you've already named your 'blog_id' column that.
Interestingly, no error/warning is thrown when creating this table...
So when you call comment.save(), the following insert statement is generated:
insert into blog_comment (id, blog_id, content, website, name, time_created, blog_id) values (?,?,?,?,?,?,?)
Notice a reference to 'blog_id' column twice, which is invalid.
And this is because of the above double mapping.
To fix, just give your 'blog' property a different name to use for the foreign key column:
#Column(name = "blog_id", length = 36)
public UUID blog_id;
#ManyToOne
#JoinColumn(name = "blogId")
public Blog blog;
I'm not sure why you're mapping your entities like this (perhaps legacy schema?) but the 'blog_id' fields seem to be redundant (and confusing) as you already have an entity mapping in the form of your 'blog' property.
This question is pretty old, but for any future reference i have found this answer that solved my problem.
After numerous searchers around the web I found this answer here - thanks to jtal!
Just to summaries the problem:
Using Ebean i have made a #ManyToOne entity that is not implemented in the database in anyway,
even more the join field, in your case
blogId
is a valid field that has values of its own.
when trying to join the column on that field, it will always fail because it creates this sql query:
SELECT
*
FROM
blog_comment;
select
t0.id c0,
t0.blog_id c1,
t0.content c2,
t0.website c3,
t0.time_created c4,
t0.blog_id c5 <---- notice this duplicate
from
blog_comment t0
in order to solve this, i tell ebean not to use the second set of properties.
your new ebean element should look something like this:
#ManyToOne
#JoinColumn(name = "blogId", insertable = false, updatable = false)
public Blog blog;
hope this helps! =)

hibernate inheritance : How to protect base class entry on child class deletion

I have some trouble with Hibernate 4 and inheritance:
I use a ChildData class which inherit from BaseData by a JOIN inheritance strategy. My mapping is done by annotation in classes.
Everything is working fine except that when I delete a ChildData instance (with session.delete() or with a Hql query) the BaseData entry is also deleted.
I understand that in most case this is the awaited behavior, but for my particular case, I would like to preserve the BaseData entry no matter what for history purpose.
In other words I want all actions on the child class to be cascaded to base class except deletion.
I have already tried #OnCascade on the child class, with no success.
Is it a way to achieve this by code or do I have to use a SQL Trigger ON DELETE ?
EDIT :
Base Class
#Entity
#Table(name = "dbBenchHistory", uniqueConstraints = #UniqueConstraint(columnNames = "Name"))
#Inheritance(strategy = InheritanceType.JOINED )
public class DbBenchHistory implements java.io.Serializable {
private int id;
private String name;
private String computer;
private String eap;
private Date lastConnexion;
private Set<DbPlugin> dbPlugins = new HashSet<DbPlugin>(0);
private Set<DbSequenceResult> dbSequenceResults = new HashSet<DbSequenceResult>(
0);
public DbBenchHistory() {
}
public DbBenchHistory(int id, String name) {
this.id = id;
this.name = name;
}
public DbBenchHistory(int id, String name, String computer, String eap,
Date lastConnexion, Set<DbPlugin> dbPlugins,
Set<DbSequenceResult> dbSequenceResults) {
this.id = id;
this.name = name;
this.computer = computer;
this.eap = eap;
this.lastConnexion = lastConnexion;
this.dbPlugins = dbPlugins;
this.dbSequenceResults = dbSequenceResults;
}
#Id
#Column(name = "Id", unique = true, nullable = false)
#GeneratedValue(strategy=GenerationType.IDENTITY)
public int getId() {
return this.id;
}
public void setId(int id) {
this.id = id;
}
//Getters/Setters
Child Class :
#Entity
#Table(name = "dbBench")
#OnDelete(action=OnDeleteAction.NO_ACTION)
public class DbBench extends DbBenchHistory {
private Set<DbProgram> dbPrograms = new HashSet<DbProgram>(0);
private Set<DbUser> dbUsers = new HashSet<DbUser>(0);
public DbBench() {
}
public DbBench(Set<DbProgram> dbPrograms,
Set<DbUser> dbUsers) {
this.dbPrograms = dbPrograms;
this.dbUsers = dbUsers;
}
//Getters/Setters
But I'm starting to think that I was wrong from the beginning and that inheritance was not the good way to handle this. If nothing shows up I will just go for BenchHistory - Bench being a simple one-to-one relationship
EDIT2 :
I edit while I can't answer my own question for insuficient reputation
I feel completly stupid now that I found the solution, that was so simple :
As I said, I was using hibernate managed methods : session.delete() or hql query. Hibernate was doing what he was supposed to do by deletintg the parent class, like it would have been in object inheritance.
So I just bypass hibernate by doing the deletion of the child class with one of the simplest SqlQuery on earth. And the base class entry remain untouched.
I understand that I somehow violate the object inheritance laws, but in my case it is really handy.
Thanks to everyone for your time, and believ me when I say I'm sorry.
I don't think Hibernate/JPA supports this. What you basically want is conversion from a subclass to a superclass, and not a cascading delete. When you have an object of the subclass, the members from the superclass are treated no different than the members of the subclass.
This can be solved through writing some logic for it though:
public void deleteKeepSuperclassObject(final ChildData childData) {
final BaseData baseDataToKeep = new BaseData();
//populate baseDataToKeep with data from the childData to remove
em.persist(baseDataToKeep);
em.remove(childData);
}

JPA: How to get entity based on field value other than ID?

In JPA (Hibernate), when we automatically generate the ID field, it is assumed that the user has no knowledge about this key. So, when obtaining the entity, user would query based on some field other than ID. How do we obtain the entity in that case (since em.find() cannot be used).
I understand we can use a query and filter the results later. But, is there a more direct way (because this is a very common problem as I understand).
It is not a "problem" as you stated it.
Hibernate has the built-in find(), but you have to build your own query in order to get a particular object. I recommend using Hibernate's Criteria :
Criteria criteria = session.createCriteria(YourClass.class);
YourObject yourObject = criteria.add(Restrictions.eq("yourField", yourFieldValue))
.uniqueResult();
This will create a criteria on your current class, adding the restriction that the column "yourField" is equal to the value yourFieldValue. uniqueResult() tells it to bring a unique result. If more objects match, you should retrive a list.
List<YourObject> list = criteria.add(Restrictions.eq("yourField", yourFieldValue)).list();
If you have any further questions, please feel free to ask. Hope this helps.
if you have repository for entity Foo and need to select all entries with exact string value boo (also works for other primitive types or entity types). Put this into your repository interface:
List<Foo> findByBoo(String boo);
if you need to order results:
List<Foo> findByBooOrderById(String boo);
See more at reference.
Basically, you should add a specific unique field. I usually use xxxUri fields.
class User {
#Id
// automatically generated
private Long id;
// globally unique id
#Column(name = "SCN", nullable = false, unique = true)
private String scn;
}
And you business method will do like this.
public User findUserByScn(#NotNull final String scn) {
CriteriaBuilder builder = manager.getCriteriaBuilder();
CriteriaQuery<User> criteria = builder.createQuery(User.class);
Root<User> from = criteria.from(User.class);
criteria.select(from);
criteria.where(builder.equal(from.get(User_.scn), scn));
TypedQuery<User> typed = manager.createQuery(criteria);
try {
return typed.getSingleResult();
} catch (final NoResultException nre) {
return null;
}
}
Best practice is using #NaturalId annotation. It can be used as the business key for some cases it is too complicated, so some fields are using as the identifier in the real world.
For example, I have user class with user id as primary key, and email is also unique field. So we can use email as our natural id
#Entity
#Table(name="user")
public class User {
#Id
#Column(name="id")
private int id;
#NaturalId
#Column(name="email")
private String email;
#Column(name="name")
private String name;
}
To get our record, just simply use 'session.byNaturalId()'
Session session = sessionFactory.getCurrentSession();
User user = session.byNaturalId(User.class)
.using("email","huchenhai#qq.com")
.load()
This solution is from Beginning Hibernate book:
Query<User> query = session.createQuery("from User u where u.scn=:scn", User.class);
query.setParameter("scn", scn);
User user = query.uniqueResult();
I solved a similar problem, where I wanted to find a book by its isbnCode not by your id(primary key).
#Entity
public class Book implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue(strategy=GenerationType.IDENTITY)
private Integer id;
private String isbnCode;
...
In the repository the method was created like #kamalveer singh mentioned. Note that the method name is findBy+fieldName (in my case: findByisbnCode):
#Repository
public interface BookRepository extends JpaRepository<Book, Integer> {
Book findByisbnCode(String isbnCode);
}
Then, implemented the method in the service:
#Service
public class BookService {
#Autowired
private BookRepository repo;
public Book findByIsbnCode(String isbnCode) {
Book obj = repo.findByisbnCode(isbnCode);
return obj;
}
}
Write a custom method like this:
public Object findByYourField(Class entityClass, String yourFieldValue)
{
CriteriaBuilder criteriaBuilder = entityManager.getCriteriaBuilder();
CriteriaQuery<Object> criteriaQuery = criteriaBuilder.createQuery(entityClass);
Root<Object> root = criteriaQuery.from(entityClass);
criteriaQuery.select(root);
ParameterExpression<String> params = criteriaBuilder.parameter(String.class);
criteriaQuery.where(criteriaBuilder.equal(root.get("yourField"), params));
TypedQuery<Object> query = entityManager.createQuery(criteriaQuery);
query.setParameter(params, yourFieldValue);
List<Object> queryResult = query.getResultList();
Object returnObject = null;
if (CollectionUtils.isNotEmpty(queryResult)) {
returnObject = queryResult.get(0);
}
return returnObject;
}
Edit: Just realized that #Chinmoy was getting at basically the same thing, but I think I may have done a better job ELI5 :)
If you're using a flavor of Spring Data to help persist / fetch things from whatever kind of Repository you've defined, you can probably have your JPA provider do this for you via some clever tricks with method names in your Repository interface class. Allow me to explain.
(As a disclaimer, I just a few moments ago did/still am figuring this out for myself.)
For example, if I am storing Tokens in my database, I might have an entity class that looks like this:
#Data // << Project Lombok convenience annotation
#Entity
public class Token {
#Id
#Column(name = "TOKEN_ID")
private String tokenId;
#Column(name = "TOKEN")
private String token;
#Column(name = "EXPIRATION")
private String expiration;
#Column(name = "SCOPE")
private String scope;
}
And I probably have a CrudRepository<K,V> interface defined like this, to give me simple CRUD operations on that Repository for free.
#Repository
// CrudRepository<{Entity Type}, {Entity Primary Key Type}>
public interface TokenRepository extends CrudRepository<Token, String> { }
And when I'm looking up one of these tokens, my purpose might be checking the expiration or scope, for example. In either of those cases, I probably don't have the tokenId handy, but rather just the value of a token field itself that I want to look up.
To do that, you can add an additional method to your TokenRepository interface in a clever way to tell your JPA provider that the value you're passing in to the method is not the tokenId, but the value of another field within the Entity class, and it should take that into account when it is generating the actual SQL that it will run against your database.
#Repository
// CrudRepository<{Entity Type}, {Entity Primary Key Type}>
public interface TokenRepository extends CrudRepository<Token, String> {
List<Token> findByToken(String token);
}
I read about this on the Spring Data R2DBC docs page, and it seems to be working so far within a SpringBoot 2.x app storing in an embedded H2 database.
No, you don't need to make criteria query it would be boilerplate code you just do simple thing if you working in Spring-boot:
in your repo declare a method name with findBy[exact field name].
Example-
if your model or document consist a string field myField and you want to find by it then your method name will be:
findBymyField(String myField);
All the answers require you to write some sort of SQL/HQL/whatever. Why? You don't have to - just use CriteriaBuilder:
Person.java:
#Entity
class Person {
#Id #GeneratedValue
private int id;
#Column(name = "name")
private String name;
#Column(name = "age")
private int age;
...
}
Dao.java:
public class Dao {
public static Person getPersonByName(String name) {
SessionFactory sessionFactory = new Configuration().configure().buildSessionFactory();
Session session = sessionFactory.openSession();
session.beginTransaction();
CriteriaBuilder cb = session.getCriteriaBuilder();
CriteriaQuery<Person> cr = cb.createQuery(Person.class);
Root<Person> root = cr.from(Person.class);
cr.select(root).where(cb.equal(root.get("name"), name)); //here you pass a class field, not a table column (in this example they are called the same)
Query query = session.createQuery(cr);
query.setMaxResults(1);
List<Person> resultList = query.getResultList();
Person result = resultList.get(0);
return result;
}
}
example of use:
public static void main(String[] args) {
Person person = Dao.getPersonByName("John");
System.out.println(person.getAge()); //John's age
}
Have a look at:
JPA query language: The Java Persistence Query Language
JPA Criteria API: Using the Criteria API to Create Queries
I've written a library that helps do precisely this. It allows search by object simply by initializing only the fields you want to filter by: https://github.com/kg6zvp/GenericEntityEJB
Refer - Spring docs for query methods
We can add methods in Spring Jpa by passing diff params in methods like:
List<Person> findByEmailAddressAndLastname(EmailAddress emailAddress, String lastname);
// Enabling static ORDER BY for a query
List<Person> findByLastnameOrderByFirstnameAsc(String lastname);
In my Spring Boot app I resolved a similar type of issue like this:
#Autowired
private EntityManager entityManager;
public User findByEmail(String email) {
User user = null;
Query query = entityManager.createQuery("SELECT u FROM User u WHERE u.email=:email");
query.setParameter("email", email);
try {
user = (User) query.getSingleResult();
} catch (Exception e) {
// Handle exception
}
return user;
}
This is very basic query :
Entity : Student
#Entity
#Data
#NoArgsConstructor
public class Student{
#Id
#GeneratedValue(generator = "uuid2", strategy = GenerationType.IDENTITY)
#GenericGenerator(name = "uuid2", strategy = "uuid2")
private String id;
#Column(nullable = false)
#Version
#JsonIgnore
private Integer version;
private String studentId;
private String studentName;
private OffsetDateTime enrollDate;
}
Repository Interface : StudentRepository
#Repository
public interface StudentRepository extends JpaRepository<Student, String> {
List<Student> findByStudentName(String studentName);
List<Student> findByStudentNameOrderByEnrollDateDesc(String studentName);
#Transactional
#Modifying
void deleteByStudentName(String studentName);
}
Note:
findByColumnName : give results by criteria
List findByStudentName(String studentName)
Internally convert into query : select * from Student where name='studentName'
#Transactional
#Modifying
Is useful when you want to remove persisted data from database.
Using CrudRepository and JPA query works for me:
import org.springframework.data.jpa.repository.Query;
import org.springframework.data.repository.CrudRepository;
import org.springframework.data.repository.query.Param;
public interface TokenCrudRepository extends CrudRepository<Token, Integer> {
/**
* Finds a token by using the user as a search criteria.
* #param user
* #return A token element matching with the given user.
*/
#Query("SELECT t FROM Token t WHERE LOWER(t.user) = LOWER(:user)")
public Token find(#Param("user") String user);
}
and you invoke the find custom method like this:
public void destroyCurrentToken(String user){
AbstractApplicationContext context = getContext();
repository = context.getBean(TokenCrudRepository.class);
Token token = ((TokenCrudRepository) repository).find(user);
int idToken = token.getId();
repository.delete(idToken);
context.close();
}

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