I have been reading a lot about using enums as parameters in queries. I have some queries in my project that use the value from these enums as parameters.
For example:
public enum YesNo {
Y, N
}
Query:
select ent
from
Entity ent
where
ent.id = :id
and ent.deleted = project.path.example.YesNo.N
Entity:
#Entity
public class Entity{
Long id;
#Enumerated(EnumType.STRING)
YesNo deleted;
}
The above works correctly as expected.
However, when I have the following:
interface Commons{
interface MostCommonTypesofAnimals {
long DOG = 1L;
long CAT = 2L;
}
}
Query
select a
from
Animal a
where
a.id = :id
and a.type = project.path.example.Commons.MostCommonTypesofAnimals.DOG
Entity
#Entity
public class Animal{
Long id;
Type type;
}
#Entity
public class Type{
public Long id;
}
It does not work telling me that the path is incorrect even though it is actually correct.
Is there any work around? Or interface values cannot be mapped? Can anyone provide me an example that works? I could not find anything similar.
Please note that this is just an example to illustrate the situation., those are not the real names that I am using or anything.
For using enum while using hibernate / jpa (based on your tags), you should use annotation in your Pojo class.
#Enumerated(EnumType.ORDINAL)
In your example, something like:
#Entity
#Table(name = "tableName")
public class entityName {
#Enumerated(EnumType.ORDINAL)
private YesNo yesNoEnum;
}
The annotation can go here or in the getter, as you prefer.
You can find more info here
ps: for yes or no I suggest you using a boolean value, not an enum
Related
I'm trying to use querydsl to build a query which joins two tables. However, a slight discrepancy between the corresponding Java class data types seems to prevent me from directly comparing the two columns values in my query. The data type within the object corresponding to table A is java.util.UUID while the data type of the object corresponding to table B is String.
I have something like the following:
#Table(name = "TABLE_A")
public class TableA {
#Column(name = "uuid")
#Type(type = "uuid-char")
private UUID uuid;
}
#Table(name = "TABLE_B")
public class TableB {
#Column(name = "uuid")
private String uuid;
}
#Service
public class QueryService {
private final JPQLQueryFactory queryFactory;
public UUID getData(UUID input) {
return queryFactory.select(QTableA.tableA.uuid)
.from(QTableA.tableA)
.innerJoin(QTableB.tableB)
.on(QTableB.tableB.uuid.eq(QTableA.tableA.uuid.toString()))
.where(QTableA.tableA.uuid.eq(input))
.fetchOne();
}
}
The above code does not return anything. However, the below code seems to work:
#Service
public class QueryService {
private final JPQLQueryFactory queryFactory;
public UUID getData(UUID input) {
return queryFactory.select(QTableA.tableA.uuid)
.from(QTableA.tableA)
.innerJoin(QTableB.tableB)
.on(QTableA.tableA.uuid.eq(input)
.and(QTableB.tableB.uuid.eq(input.toString()))
.where(QTableA.tableA.uuid.eq(input))
.fetchOne();
}
}
I don't understand why directly comparing the columns doesn't work, but comparing them to a common variable does work. Would QTableA.tableA.uuid.toString() not call the proper toString() method?
Please try:
.on(QTableB.tableB.uuid.toString().eq(QTableA.tableA.uuid.toString()))
Apply toString on QTableB.tableB.uuid
It's looking like a direct comparison is not possible with the current version. Querydsl's type checking is too strict for a direct comparison to be done on these kinds of differing data types.
i have following entities
A.class
#Entity
public class A {
List<B> listOfB;
//getters and setters
}
B.class
#Entity
public class B {
private long id;
private List<C> listOfC;
//getters and setters
}
C.class
#Entity
public class C {
long getId();
}
In some cases C.class is unnecessary. So I tried to ignore it at the query with projection;
AView.class
public interface AView {
List<BView> getB();
}
BView.class
public interface BView {
long getId();
}
The JSON Response looks fine, but the Hibernate Query log is showing the SELECT of class C.
I read a custom query at solution. But the Problem is that the classes have too many fields.
Is it posible to ignore the select of c in the query without writing a custom Query?
This seemed like a FetchType related issue to me.
If you are providing the relations with spring data jpa annotations, You might consider using FetchType.LAZY.
Like : #ManyToMany(fetch = FetchType.LAZY)
I am building an API to return two fields as such:
{
currentPoints: 325,
badgeName: "Some Badge"
}
However, I am having trouble using hibernate in order populate those two fields. I made two attempts and both are throwing errors. Both of these errors can be found in their respective Repository file. In the 2nd attempt, I am using native=true and am able to get it to work using a SELECT *. However, I am trying to only populate and return two fields of the entity.
One solution I thought about is using the 2nd approach with a SELECT * and creating another package named response with CurrentInfoResponse class and just returning that class. However, I wanted to see if there was a way to avoid this using the current model that I have.
Possible Solution:
#Getter
#AllArgsConstructor
public class CurrentInfoResponse{
private Integer currentPoints;
private String badgeName
}
Package Structure:
Controller.java:
#GetMapping("/current-badge/{userId}")
public CurrentBadgeInfoModel getCurrentBadge(#PathVariable Integer userId){
return currentBadgeInfoService.getCurrentBadge(userId);
}
ServiceImpl.java:
#Override
public CurrentBadgeInfoModel getCurrentBadge(Integer userId){
return currentBadgeInfoRepository.getCurrentBadge(userId);
}
CurrentBadgeInfoModel.java:
#Getter
#Entity
#Table(name = "user_current_badge_info")
public class CurrentBadgeInfoModel {
#Id
#Column(name = "user_current_info_id")
private Integer userCurrentBadgeInfo;
#Column(name = "user_id")
private Integer userId;
#Column(name = "current_points")
private Integer currentPoints;
#ManyToOne
#JoinColumn(name = "badge_id")
private BadgeModel badgeModel;
}
BadgeModel.java
#Getter
#Entity
#Table(name = "badge_info")
public class BadgeModel {
#Id
#JoinColumn(name= "badge_id")
private Integer badgeId;
#Column(name = "badge_name")
private String badgeName;
}
Repository.java - ATTEMPT 1:
#Repository
public interface CurrentBadgeInfoRepository extends JpaRepository<CurrentBadgeInfoModel, Integer> {
#Query("SELECT cbim.currentPoints, cbim.badgeModel.badgeName FROM CurrentBadgeInfoModel cbim JOIN
cbim.badgeModel WHERE cbim.userId=?1")
CurrentBadgeInfoModel getCurrentBadge(Integer userId);
}
//Error: No converter found capable of converting from type [java.lang.Integer] to type [com.timelogger.model.CurrentBadgeInfoModel]
Repository.java - ATTEMPT 2:
#Repository
public interface CurrentBadgeInfoRepository extends JpaRepository<CurrentBadgeInfoModel, Integer> {
#Query(value = "SELECT current_points, badge_name FROM user_current_badge_info ucbi JOIN badge_info bi ON ucbi.badge_id=bi.badge_id WHERE user_id=?1", nativeQuery = true)
CurrentBadgeInfoModel getCurrentBadge(Integer userId);
}
//Error: Column 'user_current_info_id' not found
Using the SELECT clause of HQL should help you here.
If you don't have that constructor, you can add it
#Query("SELECT new CurrentBadgeInfoModel(cbim.currentPoints, cbim.badgeModel.badgeName) FROM CurrentBadgeInfoModel cbim JOIN
cbim.badgeModel WHERE cbim.userId=?1")
Notice the usage of new CurrentBadgeInfoModel(cbim.currentPoints, cbim.badgeModel.badgeName)
I think this is a perfect use case for Blaze-Persistence Entity Views.
I created the library to allow easy mapping between JPA models and custom interface or abstract class defined models, something like Spring Data Projections on steroids. The idea is that you define your target structure(domain model) the way you like and map attributes(getters) via JPQL expressions to the entity model.
A DTO model for your use case could look like the following with Blaze-Persistence Entity-Views:
#EntityView(CurrentBadgeInfoModel.class)
public interface CurrentInfoResponse {
Integer getCurrentPoints();
#Mapping("badgeModel.badgeName")
String getBadgeName();
}
The Spring Data integration allows you to use it almost like Spring Data Projections: https://persistence.blazebit.com/documentation/entity-view/manual/en_US/index.html#spring-data-features
CurrentInfoResponse findByUserId(Integer userId);
The best part is, it will only fetch the state that is actually necessary!
I'm using the javax.persistence package to map my Java classes.
I have entities like these:
public class UserEntity extends IdEntity {
}
which extends a mapped superclass named IdEntity:
#MappedSuperclass
public class IdEntity extends VersionEntity {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
// Getters and setters below...
}
The IdEntity super class extends another mapped super class named VersionEntity to make all entities inherit version properties:
#MappedSuperclass
public abstract class VersionEntity {
#Version
private Integer version;
// Getters and setters below...
}
Why?
Because now I can make generic queries on the IdEntity class for all entities, and it will look like this: (example)
CriteriaBuilder builder = JPA.em().getCriteriaBuilder();
CriteriaQuery<IdEntity> criteria = builder.createQuery(IdEntity.class);
Now to the problem.
Some of my entities will have timestamps like created_at and deleted_at. But not all entities.
I could provide these properties in my entity classes like this:
public class UserEntity extends IdEntity {
#Basic(optional = false)
#Column(name = "updated_at")
#Temporal(TemporalType.TIMESTAMP)
private Date updatedAt;
}
But as I have a lot of entities, this will make me put a lot of redundant code in all entities that should have timestamps. I wish there was some way I could make the relevant classes inherit these fields in some way.
One possible solution is to create a parallell IdEntity superclass, maybe named IdAndTimeStampEntity and make those entities that should have timestamps inherit from this new superclass instead, but hey that's not fair to my colleague-developers because now they have to know which super class to choose from when writing generic queries:
CriteriaBuilder builder = JPA.em().getCriteriaBuilder();
CriteriaQuery<???> criteria = builder.createQuery(???); // Hmm which entity should I choose IdEntity or IdAndTimeStampEntity ?? *Annoyed*
And the generic entity queries become not so generic..
My question: How can I make all of my entities inherit id and
version fields, but only a sub part of all entities inherit
timestamp fields, but keep my queries to a single type of entities?
Update #1
Question from Bolzano: "can you add the code which you specify the path(holds table info) for entities ?"
Here is a working example of querying a UserEntity which is a IdEntity
CriteriaBuilder builder = JPA.em().getCriteriaBuilder();
CriteriaQuery<IdEntity> criteria = builder.createQuery(IdEntity.class);
Root<IdEntity> from = criteria.from(IdEntity.class);
criteria.select(from);
Path<Integer> idPath = from.get(UserEntity_.id); //generated meta model
criteria.where(builder.in(idPath).value(id));
TypedQuery<IdEntity> query = JPA.em().createQuery(criteria);
return query.getSingleResult();
I would pick a solution that didn't enforce a class-based object model like you've outlined. What happens when you don't need optimistic concurrency checking and no timestamps, or timestamps but no OCC, or the next semi-common piece of functionality you want to add? The permutations will become unmanageable.
I would add these common interactions as interfaces, and I would enhance your reusable find by id with generics to return the actual class you care about to the caller instead of the base superclass.
Note: I wrote this code in Stack Overflow. It may need some tweaking to compile.
#MappedSuperclass
public abstract class Persistable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
// getter/setter
}
public interface Versioned {
Integer getVersion();
}
public interface Timestamped {
Date getCreated();
Date getLastUpdated();
}
#Embeddable
public class TimestampedEntity {
#Column(name = "create_date")
#Temporal
private Date created;
#Column
#Temporal
private Date lastUpdated;
// getters/setters
}
#Entity
public class UserEntity extends Persistable implements Versioned, Timestamped {
#Version
private Integer version;
#Embedded
private TimestampedEntity timestamps;
/*
* interface-defined getters. getTimestamps() doesn't need to
* be exposed separately.
*/
}
public class <CriteriaHelperUtil> {
public <T extends Persistable> T getEntity(Class<T> clazz, Integer id, SingularAttribute idField) {
CriteriaBuilder builder = JPA.em().getCriteriaBuilder();
CriteriaQuery<T> criteria = builder.createQuery(clazz);
Root<T> from = criteria.from(clazz);
criteria.select(from);
Path<Integer> idPath = from.get(idField);
criteria.where(builder.in(idPath).value(id));
TypedQuery<T> query = JPA.em().createQuery(criteria);
return query.getSingleResult();
}
}
Basic Usage:
private UserEntity ue = CriteriaHelperUtil.getEntity(UserEntity.class, 1, UserEntity_.id);
ue.getId();
ue.getVersion();
ue.getCreated();
// FooEntity implements Persistable, Timestamped
private FooEntity fe = CriteriaHelperUtil.getEntity(FooEntity.class, 10, FooEntity_.id);
fe.getId();
fe.getCreated();
fe.getVersion(); // Compile Error!
#MappedSuperclass
public class IdEntity{
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
#Version
private Integer version;
}
#MappedSuperclass
public class IdAndTimeStampEntity extends IdEntity{
Date created;
}
#Entity
public class UserEntity extends IdAndTimeStampEntity{
String name;
}
#Entity
public class FooEntity extends IdEntity{...
Pros of this solution:
In simple and clear way uses OOP without need to embed duplicate code implementing intefaces in every subclass. (Every class is also interface)
Optimistic locking version column is mostly used approach. And should be part of base class. Except read only entities like codetables.
Usage:
public <T extends IdEntity> T persist(T entity) {
if (entity instanceof IdAndTimeStampEntity) {
((IdAndTimeStampEntity) entity).setCreated(new Date());
}
if (!em.contains(entity) && entity.getId() != null) {
return em.merge(entity);
} else {
em.persist(entity);
return entity;
}
}
I wish there was some way I could make the relevant classes inherit these fields in some way.
You could make a custom annotation #Timed and use an annotation processor to add the timestamp field and annotations, either by using a bytecode manipulation framework or creating a delegating subclass. Or, for example if you use Lombok, create a Lombok annotation.
That way, your team members only have to remember to use the #Timed annotation when you have entities with timestamps. Whether you like such approach or not is up to you.
I have two hibernate classes: a base class, and an extended class that has additional fields. (These fields are mapped by other tables.)
For example, I have:
#Entity
#Table(name="Book")
public class A {
private String ID;
private String Name;
// ...
}
#Entity
#Table(name="Book")
public class B extends A {
public String node_ID;
// ...
}
public class Node {
public String ID; // maps to B.node_ID
// ...
}
How do I map this in Hibernate? The hibernate documentation states three types of inheritence configurations: one table per class, one table with a type column, and a join table -- none of which apply here.
The reason I need to do this is because class A is from generic framework that's reused over multiple projects, and class B (and Node) are extensions specific to one project -- they won't be used again. In the future, I may have perhaps a class C with a house_ID or some other field.
Edit: If I try the above pseudo-code configuration (two entities mapped to the same table) I get an error that the DTYPE column doesn't exist. The HQL has a "where DTYPE="A" appended.
This is possible by mapping the #DiscriminatorColumn and #DiscriminatorValue to the same values for both classes; this can be from any column you use that has the same data regardless of which type (not sure if it works with null values).
The classes should look like so:
#Entity
#Table(name="Book")
#Inheritance(strategy=InheritanceType.SINGLE_TABLE)
#DiscriminatorColumn(name="published")
#DiscriminatorValue(value="true")
public class A {
private String ID;
private String Name;
// ...
}
#Entity
#Table(name="Book")
#DiscriminatorValue(value="true")
public class B extends A {
public String node_ID;
// ...
}
For anyone who got here like me and does not want to have the dtype column but instead want to use the same table for more than one entity as is I would recommend using this
Basically you can create a Base like this
#MappedSuperclass
public abstract class BaseBook<T extends BaseBook> {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Column(name = "id", nullable = false)
private Long id;
... any other variables, getters + setters
}
#Entity
#Table(name= "book")
public class BookA extends BaseBook<BookA>{
//Default class no need to specify any variables or getters/setters
}
#Entity
#Table(name= "book")
public class BookB extends BaseBook<BookB>{
#Column(name = "other_field")
private String otherFieldInTableButNotMapedInBase
... Any other fields, getter/setter
}
From the above we have created base super class which does not have any entity or table mapping. We then create BookA to be default with the Entity + Table mapping. From there we can create other Entities all extending from BaseBook but pointing to one table