i am developing a RESTFUL API using spring-boot for a future Angular Front-End. i am having this problem with creating my entities into postgres tables.
connectivity with the database was checked and all is fine. using mvn clean install & mvn spring-boot run commands generate normal tomcat deployment without any errors. however no tables are create
here's my code:
entity:
#Entity
#Table(name = "questions")
public class Question {
#Id
#GeneratedValue(generator = "question_generator")
#SequenceGenerator(
name = "question_generator",
sequenceName = "question_sequence",
initialValue = 1000
)
private Long id;
#NotBlank
#Size(min = 3, max = 100)
private String title;
#Column(columnDefinition = "text")
private String description;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getTitle() {
return title;
}
public void setTitle(String title) {
this.title = title;
}
public String getDescription() {
return description;
}
public void setDescription(String description) {
this.description = description;
}
public Question() {
}
}
application.propreties:
# ===============================
# DATABASE CONNECTION
# ===============================
spring.datasource.driver-class-name=org.postgresql.Driver
spring.datasource.url=jdbc:postgresql://localhost:5432/postgres
spring.datasource.username=postgres
spring.datasource.password=postgres
# ===============================
# JPA / HIBERNATE
# ===============================
spring.jpa.show-sql=true
spring.jpa.hibernate.ddl-auto=update
spring.jpa.properties.hibernate.dialect=org.hibernate.dialect.PostgreSQLDialect
# Fix Postgres JPA Error:
# Method org.postgresql.jdbc.PgConnection.createClob() is not yet implemented.
spring.jpa.properties.hibernate.temp.use_jdbc_metadata_defaults=false
and here's my repo:
import model.Question;
import org.springframework.data.jpa.repository.JpaRepository;
import org.springframework.stereotype.Repository;
#Repository
public interface QuestionRepository extends JpaRepository<Question, Long> {
}
i was able to resolve this issue by changing the entity package and made it visible to the app. now it works perfectly fine.
the main package name is: com.testAppBlaBla
entities's package should be com.testAppBlaBla.model
otherwise the entity won't be generated.
Try to change spring.jpa.hibernate.ddl-auto property to create value.
Related
I have a spring boot application where I need to update a migratedCustomer db table based on userId and phoneNumber.
Since I have to use for loop in the service layer for every update, it is creating a
new transaction and performance is hampered.
how could I make sure only one transaction is created and hence to improve the performance. code is like below
#Entity
#Table(name = "MigratedCustomer")
public class MigratedCustomer {
#Id
#GeneratedValue(strategy = GenerationType.IDENTITY)
private Long id;
private String userId;
private String phoneNumber;
#Temporal(TemporalType.TIMESTAMP)
private Date createdTimestamp;
private int batchNumber;
private String comment;
}
public class MigratedCustomerService {
#Autowired
private UserRepository userRepository;
public void updateMsisdn(List<MigratedCustomer> savedCustomers) {
for (MigratedCustomer savedCustomer : savedCustomers) {
userRepository.updateStatus(savedCustomer.getUserId(),
savedCustomer.getPhoneNumber());
}
}
}
public interface MsisdnRepository extends JpaRepository<Msisdn, Long> {
#Modifying
#Query(value = "UPDATE Msisdn SET status=INACTIVE where userId=:userId and phoneNumber=:phoneNumber",
nativeQuery = true)
void updateStatus(#Param("userId") String userId, #Param("phoneNumber") String phoneNumber);
}
i'm new to Springboot. I'm trying to implement a simple REST api using :
-Springboot, JPA & rest along with hibernate
I have a 2 tables database, Notebook that contains 1 to many notes
I already setup the 2 tables and relationships. I also created a NotebookRepository and NoteRepository to get basic CRUD operations via the springboot rest. The Database connection and relationships are functionning
but i don't know how to add a new note (it has a notebook_id foreign key which msut NOT be NULL) and everytime i tryto post something along these lines
{
"title:"abc",
"text":"whatever",
"notebook":{
"id":2
}
}
i get this error :
Caused by: java.sql.SQLIntegrityConstraintViolationException: Column 'notebook_id' cannot be null
#Entity
#Table(name="notebook")
public class NoteBook {
#Id
#GeneratedValue(strategy=GenerationType.IDENTITY)
#Column(name="id")
private int id;
#Column(name="name")
private String name;
#OneToMany(mappedBy="notebook", cascade=CascadeType.ALL)
List<Note> notes;
public NoteBook() {
}
public NoteBook(String name) {
this.name = name;
}
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 List<Note> getNotes() {
return notes;
}
public void setNotes(List<Note> notes) {
this.notes = notes;
}
public void addNote(Note note) {
if(notes == null) {
notes = new ArrayList<>();
}
note.setNotebook(this);
notes.add(note);
}
#Entity
#Table(name="note")
public class Note {
#Id
#GeneratedValue(strategy=GenerationType.IDENTITY)
#Column(name="id")
private int id;
#Column(name="title")
private String title;
#Column(name="text")
private String text;
#ManyToOne(cascade={CascadeType.MERGE, CascadeType.DETACH, CascadeType.PERSIST, CascadeType.REFRESH})
#JoinColumn(name="notebook_id")
private NoteBook notebook;
public Note() {
}
public Note(String title, String text) {
this.title = title;
this.text = text;
}
#RepositoryRestResource(collectionResourceRel = "note", path = "notes")
public interface NoteRepository extends JpaRepository<Note, Integer>{
//No code...
}
#RepositoryRestResource(collectionResourceRel = "notebook", path = "notebooks")
public interface NotebookRepository extends JpaRepository<NoteBook, Integer>{
}
The problem is that the class Note doesn't have a constructor with NoteBook parameter to pass the created NoteBook object to, so the solution is to add this constructor:
public Note(String title, String text, NoteBook noteBook) {
this.title = title;
this.text = text;
this.noteBook = noteBook;
}
and it's enough to send the JSON object as you do, but just be aware of case-sensitivity:
{ "title:"abc", "text":"whatever", "noteBook":{ "id":2 } }
I think you need to add referencedColumnName = "id" for JoinColumn annotation for notebook field in Note class.
Maybe you have problem with IDENTITY generation type. See this problem with null pointer
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;
}
I'm working with Spring Data which is great stuff, but sometimes I need to get more data from database than my model can handle. For example I have model like below.
#Entity
#Table(name = "email")
public class Mail implements Serializable {
#Getter
#Setter
#GeneratedValue(strategy = GenerationType.IDENTITY)
#Id
private Long id;
#Getter
#Setter
private String text;
}
An I my query will be more complex than usual. I want to get my model and in addition number of similar entities, using group by.
#Query(value = "SELECT m, COUNT(m) as countValue FROM Mail m GROUP BY m.text")
List<Mail> findAllNewsletters();
How I should handle something like that? My model does't contain countValue so I will get List<Object[]>
How to deal with that situation, keep my code clean, easiness
of using this.
Step 1: Create a container class to hold the output from your query.
class MailOccurence {
private final Mail mail;
private final Long recurrence;
public MailOccurence(final Mail mail, final Long recurrence) {
this.mail = mail;
this.recurrence = recurrence;
}
public Mail getMail() { return mail; }
public Long getRecurrence() { return recurrence; }
}
Step 2: Populate and return instances of the container class from the query.
Query(value = "SELECT new MailOccurence(m, COUNT(m)) FROM Mail m GROUP BY m.text")
List<MailGroup> findAllNewsletters();
For full details, see the JPA specification.
You can go for a DTO like following
public class MailEntry {
private Long id;
private String text;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getText() {
return text;
}
public void setText(String text) {
this.text = text;
}
}
and inside your business logic you can take the advantage of spring template something like following
#Autowired
JdbcTemplate jdbcTemplate;
private static final String SQL = "SELECT m, COUNT(m) as countValue FROM Mail m GROUP BY m.text";
public List<MailEntry> getMailEntries() {
List<MailEntry> mailEntryList = jdbcTemplate.query(SQL, new RowMapper<MailEntry>() {
public MailEntry mapRow(ResultSet rs, int rowNum) throws SQLException {
MailEntry mailEntry = new MailEntry();
mailEntry.setId(rs.getInt(1));
mailEntry.setText(rs.getString(2));
return mailEntry;
}
});
return mailEntryList;
}
Hope this help.
I'll try to formulate the question more simple:
#Entity
public class One implements Serializable {
...
#Id
#GeneratedValue
private Long id;
#OneToMany
#OrderBy("name ASC")
private List<Many> many;
...
First I populate the List with some Many-Entities and persist the One-Entity. Second I retrieve (em.find) the One-Entity expecting the List in ascending order by Many#name, but it's not ordered by name. The List is ordered by id. Complete code see below if necessary.
Original post some days ago:
I'm using a current Netbeans Glassfish bundle.
Product Version: NetBeans IDE 8.0 (Build 201403101706)
Updates: NetBeans IDE is updated to version NetBeans 8.0 Patch 2
Java: 1.7.0_51; Java HotSpot(TM) 64-Bit Server VM 24.51-b03
Runtime: Java(TM) SE Runtime Environment 1.7.0_51-b13
System: Mac OS X version 10.9.3 running on x86_64; UTF-8; de_DE (nb)
The JPA #OrderBy annotation is completely ignored.
#Entity
public class One implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue
private Long id;
#OneToMany
#OrderBy("name ASC")
private List<Many> many;
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public List<Many> getMany() {
return many;
}
public void setMany(List<Many> many) {
this.many = many;
}
}
The many Entity
#Entity
public class Many implements Serializable {
private static final long serialVersionUID = 1L;
#Id
#GeneratedValue
private Long id;
private String name;
public Many() {
}
public Many(String name) {
this.name = name;
}
public Long getId() {
return id;
}
public void setId(Long id) {
this.id = id;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
The service class (EJB)
#Stateless
public class Service {
#PersistenceContext(unitName = "cwPU")
private EntityManager em;
public One createOne() {
return em.merge(new One());
}
public Many createMany(String name) {
return em.merge(new Many(name));
}
public One add(Long oneId, Long manyId) {
One one = em.find(One.class, oneId);
Many many = em.find(Many.class, manyId);
one.getMany().add(many);
return one;
}
public One find(Long id) {
One one = em.find(One.class, id);
return one;
}
}
The main class
public class Main {
public static void main(String[] args) throws NamingException {
EJBContainer container = EJBContainer.createEJBContainer();
Context ctx = container.getContext();
Service service = (Service) ctx.lookup("java:global/classes/Service");
One one = service.createOne();
Many many = service.createMany("the-first");
service.add(one.getId(), many.getId());
many = service.createMany("a-second");
one = service.add(one.getId(), many.getId());
one = service.find(one.getId());
System.out.println("-------------------------------------------------");
for (Many m : one.getMany()) {
System.out.println(m.getName());
}
container.close();
}
}
The output:
the-first
a-second
No matter what I write to the #OrderBy annotation (name ASC, name DESC, id ASC, id DESC), the output is always the same ascending order by the id.
Any idea what I'm missing?
The #Orderby annotation doesn't actually work that way. According to the javadoc, the annotation "Specifies the ordering of the elements of a collection ...at the point when the collection is retrieved."
So the annotation affects the result of the query (find), but does not dictate the order in the collection you store the result set into.
The solution is calling em.refresh (at the right place) as stated from Chris and WPrecht. I had to do this in a separate EJB method.
This did not work:
public One find(Long id) {
em.refresh(em.find(One.class, id)); // did not work
One one = em.find(One.class, id);
return one;
}
Adding a separate refresh method
public void refresh(Long id) {
em.refresh(em.find(One.class, id));
}
and calling it in the main program
...
service.refresh(one.getId());
one = service.find(one.getId());
...
works!
Probably I have to do more reading to understand caching.