Picketlink with custom model and long Id - jpa

I have a existing Model and want to use it with Picketlink. But I am using Long as #Id field. But Picketlink expect this to be a String field. I have found some hints to use another entity which maps to the corresponding entity of my model. But actually I don't now how to do it.
I have a base class, which all entities derive from:
#MappedSuperclass
public abstract class AbstractEntity implements Serializable, Cloneable {
#Id
#Identifier
#Column(name = "SID")
private Long sid;
#Column(name = "INSERT_TIME")
private Date insertTime;
#Column(name = "UPDATE_TIME")
private Date updateTime;
// getters and setters
}
And a derived realm entity:
#Entity
#IdentityManaged(Realm.class)
public class RealmEntity extends AbstractEntity {
#AttributeValue
private String name;
#PartitionClass
private String typeName;
#ConfigurationName
private String configurationName;
#AttributeValue
private boolean enforceSSL;
#AttributeValue
private int numberFailedLoginAttempts;
// getters and setters
}
And the mapping class for Picketlink looks as follows:
#IdentityPartition(supportedTypes = {
Application.class,
User.class,
Role.class
})
public class Realm extends AbstractPartition {
#AttributeProperty
private boolean enforceSSL;
#AttributeProperty
private int numberFailedLoginAttempts;
private Realm() {
this(null);
}
public Realm(String name) {
super(name);
}
}
The PartitionManager is defined as follows:
builder
.named("default.config")
.stores()
.jpa()
.supportType(User.class, Role.class, Application.class, Realm.class)
.supportGlobalRelationship(Grant.class, ApplicationAccess.class)
.mappedEntity(App.class, AppUserRole.class, AppRole.class, AppUser.class, UserEntity.class, RelationshipIdentityTypeEntity.class, RealmEntity.class)
.addContextInitializer((context, store) -> {
if (store instanceof JPAIdentityStore) {
if (!context.isParameterSet(JPAIdentityStore.INVOCATION_CTX_ENTITY_MANAGER)) {
context.setParameter(JPAIdentityStore.INVOCATION_CTX_ENTITY_MANAGER, entityManager);
}
}
});
When I try to create a new Realm Hibernate throws an error while trying to load the Realm because the #Id is defined as Long but the #Identifier of the Picketlink model is a String.
this.shsRealm = new Realm(REALM_SHS_NAME);
this.shsRealm.setEnforceSSL(true);
this.shsRealm.setNumberFailedLoginAttempts(3);
this.partitionManager.add(this.shsRealm);
java.lang.IllegalArgumentException: Provided id of the wrong type for class de.logsolut.common.picketlink.model.RealmEntity. Expected: class java.lang.Long, got class java.lang.String
How can I map the JPA model correctly to Picketlink?

Related

Build method in JPA with composite key

This is my entity with composite key:
#Entity
#Table(name = "course_ratings")
public class CourseRating {
#EmbeddedId
private CourseRatingKey id;
}
Where CourseRatingKey looks like this:
#Embeddable
public class CourseRatingKey implements Serializable {
#Column(name = "user_id")
private int userId;
#Column(name = "course_id")
private int courseId;
public CourseRatingKey() {
}
// getters, setters, equals(), hashCode()
}
And my JPA repository
#Repository
public interface CourseRatingRepository extends JpaRepository<CourseRating, CourseRatingKey> {
}
I am trying to build method that will return list of all CourseRating with given courseId property of CourseRatingKey. Below method doesn't work because JPA doesn't recognize it:
repository.findAllByIdCourseId(int id);
How can I build my method name to achieve my goal?
I have solved my problem by declaring this method in repository. I'm a bit confused as the other methods work without declaring.
#Repository
public interface CourseRatingRepository extends JpaRepository<CourseRating, CourseRatingKey> {
List<CourseRating> findAllByIdCourseId(Integer id);
}

JPA composite Foreign Key part of composite Primary Key unable to find ID

We have tables,
'Lin_Code_Groups' with fields,
Project_ID (PK),
CG_ID(PK),
CG_Name
Corresponding entity class,
public class Lin_Code_Groups implements Serializable {
#EmbeddedId
private LinCodeGroupPK pk;
private String CG_name;
#Embeddable
public static class LinCodeGroupPK implements Serializable {
private Integer Project_ID;
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer CG_ID;
}
#OneToMany(mappedBy = "lin_Code_Groups")
private List<Lin_CG_Params> lin_CG_Params;
}
table Lin_CG_Params with fields,
Project_ID (PK)..FK to Lin_Code_Groups,
CG_ID(PK)...FK to Lin_Code_Groups,
Param_name(PK),
Param_value
Corresponding entity class,
public class Lin_CG_Params implements Serializable {
#EmbeddedId
private LinCodeGroupParamPK pk;
private String Param_value;
#Embeddable
public static class LinCodeGroupParamPK implements Serializable {
private String Param_name;
private LinCodeGroupPK linCodeGroupPK;
}
#MapsId("linCodeGroupPK")
#ManyToOne
#JoinColumns( {
#JoinColumn(name = "Project_ID",referencedColumnName= "Project_ID"),
#JoinColumn(name = "CG_ID",referencedColumnName= "CG_ID")
})
private Lin_Code_Groups lin_Code_Groups;
}
in controller class, i am using JPA's .Save method to save the data in to the tables.
#PostMapping(value = {"/hello"}, consumes = "application/json", produces = "application/json")
public ResponseEntity<Object> saveNewCodeGroupsDetails(#RequestBody Lin_Code_Groups objLin_Code_Groups ) {
respository.save(objLin_Code_Groups);
}
but getting an error 'Unable to find Lin_CG_Params with id Lin_CG_Params.LinCodeGroupParamPK'
Can anyone is please help ?

how to store PostgreSQL jsonb using SpringBoot + JPA?

I'm working on a migration software that will consume unknown data from REST services.
I already think about use MongoDB but I decide to not use it and use PostgreSQL.
After read this I'm trying to implement it in my SpringBoot app using Spring JPA but I don't know to map jsonb in my entity.
Tried this but understood nothing!
Here is where I am:
#Repository
#Transactional
public interface DnitRepository extends JpaRepository<Dnit, Long> {
#Query(value = "insert into dnit(id,data) VALUES (:id,:data)", nativeQuery = true)
void insertdata( #Param("id")Integer id,#Param("data") String data );
}
and ...
#RestController
public class TestController {
#Autowired
DnitRepository dnitRepository;
#RequestMapping(value = "/dnit", method = RequestMethod.GET)
public String testBig() {
dnitRepository.insertdata(2, someJsonDataAsString );
}
}
and the table:
CREATE TABLE public.dnit
(
id integer NOT NULL,
data jsonb,
CONSTRAINT dnit_pkey PRIMARY KEY (id)
)
How can I do this?
Note: I don't want/need an Entity to work on. My JSON will always be String but I need jsonb to query the DB
Tried this but understood nothing!
To fully work with jsonb in Spring Data JPA (Hibernate) project with Vlad Mihalcea's hibernate-types lib you should just do the following:
1) Add this lib to your project:
<dependency>
<groupId>com.vladmihalcea</groupId>
<artifactId>hibernate-types-52</artifactId>
<version>2.2.2</version>
</dependency>
2) Then use its types in your entities, for example:
#Data
#NoArgsConstructor
#Entity
#Table(name = "parents")
#TypeDef(name = "jsonb", typeClass = JsonBinaryType.class)
public class Parent implements Serializable {
#Id
#GeneratedValue(strategy = SEQUENCE)
private Integer id;
#Column(length = 32, nullable = false)
private String name;
#Type(type = "jsonb")
#Column(columnDefinition = "jsonb")
private List<Child> children;
#Type(type = "jsonb")
#Column(columnDefinition = "jsonb")
private Bio bio;
public Parent(String name, List children, Bio bio) {
this.name = name;
this.children = children;
this.bio = bio;
}
}
#Data
#NoArgsConstructor
#AllArgsConstructor
public class Child implements Serializable {
private String name;
}
#Data
#NoArgsConstructor
#AllArgsConstructor
public class Bio implements Serializable {
private String text;
}
Then you will be able to use, for example, a simple JpaRepository to work with your objects:
public interface ParentRepo extends JpaRepository<Parent, Integer> {
}
parentRepo.save(new Parent(
"parent1",
asList(new Child("child1"), new Child("child2")),
new Bio("bio1")
)
);
Parent result = parentRepo.findById(1);
List<Child> children = result.getChildren();
Bio bio = result.getBio();
You are making things overly complex by adding Spring Data JPA just to execute a simple insert statement. You aren't using any of the JPA features. Instead do the following
Replace spring-boot-starter-data-jpa with spring-boot-starter-jdbc
Remove your DnitRepository interface
Inject JdbcTemplate where you where injecting DnitRepository
Replace dnitRepository.insertdata(2, someJsonDataAsString ); with jdbcTemplate.executeUpdate("insert into dnit(id, data) VALUES (?,to_json(?))", id, data);
You were already using plain SQL (in a very convoluted way), if you need plain SQL (and don't have need for JPA) then just use SQL.
Ofcourse instead of directly injecting the JdbcTemplate into your controller you probably want to hide that logic/complexity in a repository or service.
There are already several answers and I am pretty sure they work for several cases. I don't wanted to use any more dependencies I don't know, so I look for another solution.
The important parts are the AttributeConverter it maps the jsonb from the db to your object and the other way around. So you have to annotate the property of the jsonb column in your entity with #Convert and link your AttributeConverter and add #Column(columnDefinition = "jsonb") as well, so JPA knows what type this is in the DB. This should already make it possible to start the spring boot application. But you will have issues, whenever you try to save() with the JpaRepository. I received the message:
PSQLException: ERROR: column "myColumn" is of type jsonb but
expression is of type character varying.
Hint: You will need to rewrite or cast the expression.
This happens because postgres takes the types a little to serious.
You can fix this by a change in your conifg:
datasource.hikari.data-source-properties: stringtype=unspecified
datasource.tomcat.connection-properties: stringtype=unspecified
Afterwards it worked for me like a charm, and here is a minimal example.
I use JpaRepositories:
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.stereotype.Repository;
#Repository
public interface MyEntityRepository extends JpaRepository<MyEntity, Integer> {
}
The Entity:
import javax.persistence.Column;
import javax.persistence.Convert;
public class MyEntity {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
protected Integer id;
#Convert(converter = MyConverter.class)
#Column(columnDefinition = "jsonb")
private MyJsonObject jsonContent;
}
The model for the json:
public class MyJsonObject {
protected String name;
protected int age;
}
The converter, I use Gson here, but you can map it however you like:
import javax.persistence.AttributeConverter;
import javax.persistence.Converter;
#Converter(autoApply = true)
public class MyConverter implements AttributeConverter<MyJsonObject, String> {
private final static Gson GSON = new Gson();
#Override
public String convertToDatabaseColumn(MyJsonObject mjo) {
return GSON.toJson(mjo);
}
#Override
public MyJsonObject convertToEntityAttribute(String dbData) {
return GSON.fromJson(dbData, MyJsonObject.class);
}
}
SQL:
create table my_entity
(
id serial primary key,
json_content jsonb
);
And my application.yml (application.properties)
datasource:
hikari:
data-source-properties: stringtype=unspecified
tomcat:
connection-properties: stringtype=unspecified
For this case, I use the above tailored converter class, you are free to add it in your library. It is working with the EclipseLink JPA Provider.
import com.fasterxml.jackson.core.JsonProcessingException;
import com.fasterxml.jackson.core.type.TypeReference;
import com.fasterxml.jackson.databind.ObjectMapper;
import org.apache.log4j.Logger;
import org.postgresql.util.PGobject;
import javax.persistence.AttributeConverter;
import javax.persistence.Converter;
import java.io.IOException;
import java.sql.SQLException;
import java.util.Map;
#Converter
public final class PgJsonbToMapConverter implements AttributeConverter<Map<String, ? extends Object>, PGobject> {
private static final Logger LOGGER = Logger.getLogger(PgJsonbToMapConverter.class);
private static final ObjectMapper MAPPER = new ObjectMapper();
#Override
public PGobject convertToDatabaseColumn(Map<String, ? extends Object> map) {
PGobject po = new PGobject();
po.setType("jsonb");
try {
po.setValue(map == null ? null : MAPPER.writeValueAsString(map));
} catch (SQLException | JsonProcessingException ex) {
LOGGER.error("Cannot convert JsonObject to PGobject.");
throw new IllegalStateException(ex);
}
return po;
}
#Override
public Map<String, ? extends Object> convertToEntityAttribute(PGobject dbData) {
if (dbData == null || dbData.getValue() == null) {
return null;
}
try {
return MAPPER.readValue(dbData.getValue(), new TypeReference<Map<String, Object>>() {
});
} catch (IOException ex) {
LOGGER.error("Cannot convert JsonObject to PGobject.");
return null;
}
}
}
Usage example, for an entity named Customer.
#Entity
#Table(schema = "web", name = "customer")
public class Customer implements Serializable {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Integer id;
#Convert(converter = PgJsonbToMapConverter.class)
private Map<String, String> info;
public Customer() {
this.id = null;
this.info = null;
}
// Getters and setter omitted.
If you're using R2DBC you can use dependency io.r2dbc:r2dbc-postgresql, and use type io.r2dbc.postgresql.codec.Json in your member attributes of an entity class, e.g.:
public class Rule {
#Id
private String client_id;
private String username;
private String password;
private Json publish_acl;
private Json subscribe_acl;
}

There is no ID defined for this entity hierarchy

I am stuck with this error message, that appears every time I want to add a ManytoOne relationship with another entity class.
The class must use a consistent access type (either field or property). There is no ID defined for this entity hierarchy
This is my entity Transaction
#Entity
#Table(name = "CustomerTransaction")
public class CustomerTransaction implements Serializable {//this is the line with the error message
#Id
#GeneratedValue(strategy = GenerationType.AUTO)
private Long id;
#ManyToOne //This generates the problem
#JoinColumns({
#JoinColumn(name = "CUS_ID", referencedColumnName = "IDCUSTOMER") })
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
private long transactionID;
#Temporal(TemporalType.TIMESTAMP)
private Date buyDate;
public Date getBuyDate() {
return buyDate;
}
public void setBuyDate(Date buyDate) {
this.buyDate = buyDate;
}
public long getTransactionID() {
return transactionID;
}
public void setTransactionID(long transactionID) {
this.transactionID = transactionID;
}
public String getCarYear() {
return carYear;
}
public void setCarYear(String carYear) {
this.carYear = carYear;
}
public Date getTransactionDate() {
return transactionDate;
}
public void setTransactionDate(Date transactionDate) {
this.transactionDate = transactionDate;
}
private String carYear;
#Temporal(TemporalType.TIMESTAMP)
private Date transactionDate;
JPA annotation should all be placed either on fields or on accessor methods. You've placed the #Id and #GeneratedValue annotation on a field (private Long id), but #ManyToOne and #JoinColumns on a getter (public Long getId()). Move the latter on a field as well.
i had similar error but in the end, i realized #Id was referencing this package org.springframework.data.annotation.Id instead of javax.persistence.Id. i was using #MappedSuperClass approach so as soon as i corrected this, everything worked fine
You need to import #Id from "import javax.persistence.Id;"

Copy Entity ID at persist time

I want to copy the entity's UUID, generated at run time to another field.
The entity id is generated via the code described bellow:
package eclipselink.example;
public class UUIDSequence extends Sequence implements SessionCustomizer {
public UUIDSequence() {
super();
}
public UUIDSequence(String name) {
super(name);
}
#Override
public Object getGeneratedValue(Accessor accessor,
AbstractSession writeSession, String seqName) {
return UUID.randomUUID().toString().toUpperCase();
}
...
public void customize(Session session) throws Exception {
UUIDSequence sequence = new UUIDSequence("system-uuid");
session.getLogin().addSequence(sequence);
}
}
Persitence.xml:
property name="eclipselink.session.customizer" value="eclipselink.example.UUIDSequence"
The entity:
public abstract class MyEntity{
private String id;
private String idCopy;
#Id
#Basic(optional = false)
#GeneratedValue(generator="system-uuid")
#XmlElement(name = "ID")
public String getId() {
return id;
}
}
How can I instruct JPA (Eclipse-link) to copy the UUID generated at runtime to idCopy field as well?
I'm not 100% sure this will work (I don't know if EclipseLink calls the setter or assigns the field directly), but give this a try:
public abstract class MyEntity{
private String id;
private String idCopy;
#Id
#Basic(optional = false)
#GeneratedValue(generator="system-uuid")
#XmlElement(name = "ID")
public String getId() {
return id;
}
public void setId(String id) {
this.id = id;
this.idCopy = id;
// or
// this.setIdCopy(id);
}
}