I have the following NodeEntity:
#NodeEntity(label = "Book")
public class Book{
private Long id;
private String content;
#Relationship(direction = Relationship.OUTGOING, type="WRITTEN_BY")
private User author;
}
where User is
#NodeEntity(label = "User)
public class User{
private Long id;
private String username;
}
and the BookRepository
#Repository
public interface BookRepository extends GraphRepository<Book> {
}
I've build a simple Rest-Controller to store a Book in the DB.
#Controller
#ResponseBody
#RequestMapping(path = "/books", consumes = MediaType.APPLICATION_JSON_VALUE, produces = MediaType.APPLICATION_JSON_VALUE)
#Transactional
public class BookController {
#Autowired
private BookRepository bookRepo;
#RequestMapping(method = RequestMethod.POST)
public Book createBook(#RequestBody Book book) {
return bookRepo.save(book);
}
When I now POST a Book the JSON
{
"content":"Test content",
"author":{
"username":"Test-Username"
}
}
to the controller, two things happen that confuse me:
First, the author in the book-object is null, although both the Book and the User have their default constructor.
Secoundly: The Book doesn't get persisted. There is no error, I just get the same book-object returned (with the author still null) but still with a null id.
Querying MATCH n RETURN n on the neo4j client also yields nothing.
I tried removing the User object from the `Book, thinking the fault was there, but I still get the same behavior.
What am I doing wrong?
Sometimes talking about a problem alone solves it... I've forgotten to put the package that contained the Book into the
#Bean
public SessionFactory getSessionFactory() {
return new SessionFactory(/*Package of Book should have been here*/);
}
in the Application
Related
This is Repository interface
#Repository
public interface MenuStructureTblService extends JpaRepository<MenuStructureTbl, Integer> {
}
MenuStructureTbl Table
#Data
#Entity(name = "menu_structure_tbl")
public class MenuStructureTbl {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
Integer menuId;
public String menuTitle;
public String menuText;
public Integer menuMotherId;
public String createDate;
public String updateDate;
}
I am banging my head on the wall, Returned list is empty.
This one returns null
#Repository
public interface MenuStructureTblService extends JpaRepository<MenuStructureTbl, Integer> {
#Query(value="SELECT * FROM menu_structure_tbl", nativeQuery = true)
Object justGetIT();
}
Returns null; I have checked the query, generates result. But, I do not know what is going on.
Let me know if you have any question.
My mistake, Conflict of Environment, i was checking against wrong environment.
i'm facing this issue while using Spring JPA and trying to retrieve a List of objects.
This is the class i'm trying to retrieve
#Entity
#Table(name="OBJECTSTERMIC")
public class TermicObject {
#Id
#Column(name="TERMICID")
private long termicId;
#MapsId
#OneToOne(fetch = FetchType.LAZY)
#JoinColumn(name="OBJECTID",columnDefinition="INTEGER")
private Object object;
#Column(name="CONTECA_RIF")
private int contecaRif;
#Column(name="CONTECA_VAL")
private int contecaVal;
#Column(name="TYPE")
private String type;
//getters and setters
The Object class has the primary key on MySQL stored as an Integer, indeed this is Object
#Entity
public class Object {
#Column(name="OBJECTID")
#Id
#JsonProperty("OBJECTID")
private int objectId;
....
So, nowhere is set a Long...
Now, i simply call in a service class
#Override
public List<TermicObject> findAll() {
return repository.findAll();
}
and got this exception
org.springframework.dao.InvalidDataAccessApiUsageException: org.hibernate.TypeMismatchException: Provided id of the wrong type for class it.besmart.db_eipo.persistence.model.Object. Expected: class java.lang.Integer, got class java.lang.Long; nested exception is java.lang.IllegalArgumentException: org.hibernate.TypeMismatchException: Provided id of the wrong type for class it.besmart.db_eipo.persistence.model.Object. Expected: class java.lang.Integer, got class java.lang.Long
Where is set that Object Id should be Long?
Have a look at definition of your repository. Does it have right generic type? do you have Integer as second parameter? IMHO this can be root cause. See proposed correct version:
#RepositoryRestResource
public interface TermicObjectRepository extends JpaRepository<TermicObject, Integer> {
public Optional<TermicObject> findById(Integer id);
public List<TermicObject> findAll()
}
As per #Lubo's answer, in my case I was having compatibility issues between String and Long types and as my model required a Long autogenerated id I had to change the repository from
public interface ProductRepository extends JpaRepository<Product, String> {
}
to
public interface ProductRepository extends JpaRepository<Product, Long> {
}
And my controller from
#RequestMapping(path = "/products/delete/{id}", method = RequestMethod.DELETE)
public void deleteProduct(#PathVariable(name = "id") String id) {
productRepository.deleteById(id);
}
to
#RequestMapping(path = "/products/delete/{id}", method = RequestMethod.DELETE)
public void deleteProduct(#PathVariable(name = "id") Long id) {
productRepository.deleteById(id);
}
You have to define your id as a Long datatype.
#Id
#Column(name="TERMICID")
private Long termicId;
also make a change in your repository interface:
public interface ProductRepository extends JpaRepository<Product, Long> {
}
Got this because
public class MyEntity {
#Id()
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id", nullable = false)
private int id; // <-------- int
...
public long getId() { return id; } // <-------- long
}
Not completely sure, but I think this mapping
#Id
#Column(name="TERMICID")
private long termicId;
#MapsId
#OneToOne(fetch = FetchType.LAZY)
#JoinColumn(name="OBJECTID",columnDefinition="INTEGER")
private Object object;
Makes the id of the Object match the value of termicId which is a long.
use
Long.valueOf(intValue)
to cast int to Long type because you define type Long to #Id
I'm working on a project with Spring Data JPA. I have a table in the database as my_query.
I want to create a method which takes a string as a parameter, and then execute it as a query in the database.
Method:
executeMyQuery(queryString)
As example, when I pass
queryString= "SELECT * FROM my_query"
then it should run that query in DB level.
The repository class is as follows.
public interface MyQueryRepository extends JpaRepository<MyQuery, Long>{
public MyQuery findById(long id);
#Modifying(clearAutomatically = true)
#Transactional
#Query(value = "?1", nativeQuery = true)
public void executeMyQuery(String query);
}
However, it didn't work as I expected. It gives the following error.
Caused by: com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''select * from my_query;'' at line 1
Is there any other way, that I could achieve this goal?
The only part of it you can parameterise are values used in WHERE clause. Consider this sample from official doc:
public interface UserRepository extends JpaRepository<User, Long> {
#Query(value = "SELECT * FROM USERS WHERE EMAIL_ADDRESS = ?1", nativeQuery = true)
User findByEmailAddress(String emailAddress);
}
Using EntityManager you can achieve this .
Suppose your entity class is like bellow:
import javax.persistence.*;
import java.math.BigDecimal;
#Entity
#Table(name = "USER_INFO_TEST")
public class UserInfoTest {
private int id;
private String name;
private String rollNo;
public UserInfoTest() {
}
public UserInfoTest(int id, String name) {
this.id = id;
this.name = name;
}
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "ID", nullable = false, precision = 0)
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
#Basic
#Column(name = "name", nullable = true)
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
#Basic
#Column(name = "roll_no", nullable = true)
public String getRollNo() {
return rollNo;
}
public void setRollNo(String rollNo) {
this.rollNo = rollNo;
}
}
And your query is "select id, name from users where roll_no = 1001".
Here query will return an object with id and a name column. Your Response class is like below:
Your Response class is like:
public class UserObject{
int id;
String name;
String rollNo;
public UserObject(Object[] columns) {
this.id = (columns[0] != null)?((BigDecimal)columns[0]).intValue():0;
this.name = (String) columns[1];
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public String getRollNo() {
return rollNo;
}
public void setRollNo(String rollNo) {
this.rollNo = rollNo;
}
}
here UserObject constructor will get an Object Array and set data with the object.
public UserObject(Object[] columns) {
this.id = (columns[0] != null)?((BigDecimal)columns[0]).intValue():0;
this.name = (String) columns[1];
}
Your query executing function is like bellow :
public UserObject getUserByRoll(EntityManager entityManager,String rollNo) {
String queryStr = "select id,name from users where roll_no = ?1";
try {
Query query = entityManager.createNativeQuery(queryStr);
query.setParameter(1, rollNo);
return new UserObject((Object[]) query.getSingleResult());
} catch (Exception e) {
e.printStackTrace();
throw e;
}
}
Here you have to import bellow packages:
import javax.persistence.Query;
import javax.persistence.EntityManager;
Now your main class, you have to call this function. First get EntityManager and call this getUserByRoll(EntityManager entityManager,String rollNo) function. Calling procedure is given below:
Here is the Imports
import javax.persistence.EntityManager;
import javax.persistence.PersistenceContext;
get EntityManager from this way:
#PersistenceContext
private EntityManager entityManager;
UserObject userObject = getUserByRoll(entityManager,"1001");
Now you have data in this userObject.
Note:
query.getSingleResult() return a object array. You have to maintain the column position and data type with query column position.
select id,name from users where roll_no = 1001
query return a array and it's [0] --> id and 1 -> name.
More info visit this thread .
There is no special support for this. But what you can do is create a custom method with a String parameter and in your implementation get the EntityManager injected and execute it.
Possibly helpful links:
https://docs.spring.io/spring-data/jpa/docs/current/reference/html/#repositories.custom-implementations
How to access entity manager with spring boot and spring data
Note: I would reconsider if what you are trying to do is a good idea because it bleeds implementation details of the repository into the rest of the application.
if you want to add custom query you should add #Param
#Query("from employee where name=:name")
employee findByName(#Param("name)String name);
}
this query will select unique record with match name.this will work
Thank you #ilya. Is there an alternative approach to achieve this task using Spring Data JPA? Without #Query annotation?
I just want to act on this part. yes there is a way you can go about it without using the #query annotation. what you need is to define a derived query from your interface that implements the JPA repository instance.
then from your repository instance you will be exposed to all the methods that allow CRUD operations on your database such as
interface UserRepository extends CrudRepository<User, Long> {
long deleteByLastname(String lastname);
List<User> removeByLastname(String lastname);
}
with these methods spring data will understand what you are trying to archieve and implement them accordingly.
Also put in mind that the basic CRUD operations are provided from the base class definition and you do not need to re define them. for instance this is the JPARepository class as defined by spring so extending it gives you all the methods.
public interface CrudRepository<T, ID extends Serializable>
extends Repository<T, ID> {
<S extends T> S save(S entity);
Optional<T> findById(ID primaryKey);
Iterable<T> findAll();
long count();
void delete(T entity);
boolean existsById(ID primaryKey);
}
For more current information check out the documentation at https://docs.spring.io/spring-data/jpa/docs/current/reference/html/
Based on #jelies answer, I am using the following approach
You can create another interface for your custom methods (as example MyQueryCustom) and then implement it as follows.
public class MyQueryRepositoryImpl implements MyQueryRepositoryCustom {
#PersistenceContext
private EntityManager entityManager;
public int executeQuery(String query) {
return entityManager.createNativeQuery(query).executeUpdate();
}
}
This will execute a custom query.
I am trying to make a simple round-trip with a REST API that leads to storing an entity into the db and then returns the stored entity.
Going down works fine and the entity is stored and correctly returned to the REST Controller. However, when I return it, Jackson seems to serialize it incorrectly, as the "name" attribute is not included.
This is the entity:
#Entity
#Configurable
public class MyEntity extends IdentifiableEntity {
private String name;
protected MyEntity() {
};
public MyEntity(String name) {
this.name = name;
}
}
and the extended entity:
#Configurable
#Inheritance(strategy = InheritanceType.JOINED)
#Entity
public abstract class IdentifiableEntity {
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
#Column(name = "id")
private Long id;
#Version
#Column(name = "version")
private Integer version = 1;
public String toString() {
return ReflectionToStringBuilder.toString(this,
ToStringStyle.SHORT_PREFIX_STYLE);
}
public Long getId() {
return this.id;
}
public void setId(Long id) {
this.id = id;
}
public Integer getVersion() {
return this.version;
}
public void setVersion(Integer version) {
this.version = version;
}
}
The REST controller is:
#RestController
#RequestMapping("/service")
public class Service {
#RequestMapping(value = "/public/{name}", method = RequestMethod.GET)
public MyEntity storeEntityPublic(#PathVariable String name) {
System.out.println("Hello " + name
+ ", I am saving on the db. (PUBLIC)");
MyEntity saved = controller.saveEntity(name);
return saved;
}
}
Then my business logic:
#Service
public class LogicController {
#Autowired
private MyEntityRepository myEntityRepository;
public MyEntity saveEntity(String name) {
MyEntity cg = new MyEntity(name);
return myEntityRepository.save(cg);
}
}
I am using Spring repositories:
#Repository
public interface MyEntityRepository extends JpaSpecificationExecutor<MyEntity>,
JpaRepository<MyEntity, Long> {
}
The returned JSON is:
{"id":12,"version":1}
Where is my "name" attribute? Is is set in the variable being returned by the REST controller.
I found the trick: MyEntity needs to have a public get for the property that has to be shown. A good reason to use a DTO pattern.
In response to your "I don't want to have my Entity "dirty"" comment: Jackson allows the use of so-called Mixins. They allow you to define annotations for your class outside the class itself. In your case it could look like this:
public abstract class MyEntityMixin {
#JsonProperty
private String name;
}
You may keep it as a field and annotate the field with #JsonProperty if you like.
Im very new to Hibernate so this will probably a easy task for you guys.
As the Topic says I'm trying to reference the same entity in multiple Lists. But when I try to do so I get an exception saying: "Duplicate entry '5' for key 'military_id'".
I googled but could not find a solution to my problem.
I have an Entity called MilitaryUnitData like this:
#Entity
public class MilitaryUnitData implements IMovable{
private long Id;
//snip
#Id
#GeneratedValue(strategy=GenerationType.TABLE)
public long getId() {
return Id;
}
public void setId(long id) {
Id = id;
}
//snip
}
and a class City where I want to store my units in.
#Entity
public class CityData {
private Collection<MilitaryUnitData> military = new ArrayList<MilitaryUnitData>();
private String name;
//snip
#Id
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
#OneToMany
#Column(nullable=false)
public Collection<MilitaryUnitData> getMilitary() {
return military;
}
public void setMilitary(Collection<MilitaryUnitData> military) {
this.military = military;
}
//snip
}
The problem occurs when I want to put a Unit into 2 cities at the same time.
How do I have to change the mapping to achive this?
Thx in advance.
I'm trying to reference the same entity in multiple Lists
After looking at your code, I think you mean, that the same MilitaryUnitData is used in several CityData?
IF this is correct, than the realtion ship is a M:N relation ship, and you need to use a #ManyToMany instead of an #OneToMany.