EDIT : If i declare all the variables in my interface they should be public by default, which opposes encapsulation and i know that i can go with abstract class instead of interface. But, in my question the main point is focused on Spring dependency injection, how can i get an object of the implementation class to access its own methods, i have used #Qualifier but it doesn't seem to work.
I have an interface with the name Account and its two implementation classes SavingsAccount and SalaryAccount. A customer class linking account details with customer with the help of an account object defined in customer class. Please go through the example for a better understanding.
My question is how to access the methods of SavingsAccount in Customer class with the spring bean of the SavingsAccount which is autowired.
If I hardcode something like SavingsAccount sa = new Savingsccount(); in the customer class i can achieve accessing the methods of the SavingsAccount class. In spring i have to do that using #Autowiredand #Qualifier annotation.
#Autowired
#Qualifier("savingsAccount")
private Account account;
But i cannot access the methods of the savingsAccount with the spring bean account.getAccNum();, like it is done with sa.getAccNum;
The complete example:
public interface Account {
public double showFunds();}
#Component
public class SalaryAccount implements Account {
private long accNum;
private double funds;
// getter setter methods
#Override
public double showFunds() {
return funds;
}
}
#Component
public class SavingsAccount implements Account {
private long accNum;
private double funds;
// getter setter methods
#Override
public double showFunds() {
return funds;
}
}
I have used autowired annotation along with qualifier, so that i get a spring bean of the implementation class specified in the #Qualifier
#Component
public class Customer {
private int id;
private String name;
private long mobile;
#Autowired
#Qualifier("savingsAccount")
private Account account;
public Customer() {
super();
}
public Customer(int id, String name, long mobile, Account account) {
super();
this.id = id;
this.name = name;
this.mobile = mobile;
this.account = account;
}
//getters and setters
public void customerAccountProfile(){
// here if i want to access members and variables of Savingsaccount with the bean account. How?
account.getAccNum() ;// throws error
account.showFunds();// works fine beacause it is defined in the interface Account
System.out.println("do nothing");
}
}
You need to add all variables and methods in interface for access as below:
public interface Account {
long accNum;
double showFunds();
}
I couldn't see any proper answer yet, so I am posting this so it might help people in the future.
Solution: I think in the #Component on the SavingsAccount class, you can use the bean name. Like here, you can use #Component("savingsAccount"). And then pass it with the #Qualifier annotation in the Customer class.
Related
I want to be able to create a new account for my application. I have an account class which represents one entity and another class that represents personal information of the account. In order to create the new account and have it be in the database I want to add some information into the account table and some information into the PersonalInfo table as detailed in the classes below. How do I do this with a CrudRespository interface. As I understand it, the crudrepository can interact with one table in the database. In my example that would be Accounts. This is fine because most of my checking and communicating will be with the accounts table. But for when I am creating a new account I need to add the data that will be given into two tables. Do I have to make manual queries and add it as a method in there?
#Entity
#Component
public class Account {
#Id
private int accountNum;
private String accountType;
private int accountBalance;
private String accountStatus;
#Entity
#Component
public class PersonalInfo {
#Id
private int accountNum;
private String firstName;
private String lastName;
private String SSN;
private String streetName;
private String city;
private String state;
private String zipcode;
#RepositoryRestResource(collectionResourceRel="accounts",path="accounts")
public interface AccountsDB extends CrudRepository<Account, Integer>{
}
Just create a repository for PersonalInfo and invoke two save() methods (of the two different repositories respectively) with the two created entities respectively.
Just make sure to set the identical ids (accountNum) for these two entities.
Or, you could create a service to do it for you, like so:
public interface AccountAndPersonalInfoService {
void save(Account account, PersonalInfo personalInfo);
}
#Service
public class AccountAndPersonalInfoServiceImpl implements AccountAndPersonalInfoService {
#Autowired
private AccountsDB accountsDB;
#Autowired
private PersonalInfoDB personalInfoDB;
#Override
void save(Account account, PersonalInfo personalInfo) {
if (account.getAccountNum() == personalInfo.getAccountNum()) {
accountsDB.save(account);
personalInfoDB.save(personalInfo);
} else throw new IllegalArgumentException("The ids of the entities do not match.");
}
}
I am creating a simple one to many relationship using JPA and Spring Data Rest. But i am getting this error
org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'designationRepository' defined in com.example.relational.BootRelations.repository.DesignationRepository defined in #EnableJpaRepositories declared on JpaRepositoriesRegistrar.EnableJpaRepositoriesConfiguration: Cannot resolve reference to bean 'jpaMappingContext' while setting bean property 'mappingContext'; nested exception is org.springframework.beans.factory.BeanCreationException: Error creating bean with name 'jpaMappingContext': Invocation of init method failed; nested exception is org.hibernate.AnnotationException: No identifier specified for entity: com.example.relational.BootRelations.models.Designation
That is because I have #Entity classes like Designation and Employee, the code follows like:
#Entity
public class Designation {
private String Designation;
private float Salary;
#OneToMany(targetEntity=Employee.class, mappedBy="designation", fetch=FetchType.LAZY, cascade=CascadeType.ALL)
#JoinColumn(name="Emp_Id")
private Set<Employee> employee;
// Getter & Setter and Constructors
}
Now Employee class:
#Entity
#Table(name="Employee")
public class Employee {
#Id
#GeneratedValue(strategy=GenerationType.IDENTITY)
#Column(name="Emp_Id")
private int Emp_Id;
#Column(name="Emp_Name")
private String Emp_Name;
#Column(name="Emp_Email")
private String Emp_Email;
#OneToOne(targetEntity=Designation.class, mappedBy="employee", fetch=FetchType.LAZY, cascade=CascadeType.ALL)
private Designation designation;
// // Getter & Setter and Constructors
}
The repositories are:
public interface DesignationRepository extends JpaRepository<Designation, Integer> {}
public interface EmployeeRepository extends JpaRepository<Employee, Integer> {}
Services:
#Service
public class DesignationServiceImpl implements DesignationService {
private DesignationRepository designationRepository;
#Autowired
public DesignationServiceImpl(DesignationRepository designationRepository) {
this.designationRepository = designationRepository;
}
#Override
public List<Designation> findAllDesignation() {
return designationRepository.findAll();
}
}
public class EmployeeServiceImpl implements EmployeeService {
private EmployeeRepository employeeRepository;
#Autowired
public EmployeeServiceImpl(#Qualifier("employeeJPAImpl") EmployeeRepository employeeRepository) {
this.employeeRepository = employeeRepository;
}
}
#Service
public class DesignationServiceImpl implements DesignationService {
private DesignationRepository designationRepository;
#Autowired
public DesignationServiceImpl(#Qualifier("DesignationRepository") DesignationRepository designationRepository) {
this.designationRepository = designationRepository;
}
#Override
public List<Designation> findAllDesignation() {
return designationRepository.findAll();
}
}
As I had two entities I had two controllers:
#RestController
#RequestMapping("/api")
public class DesignationController {
private DesignationService designationService;
#Autowired
public DesignationController(DesignationService designationService) {
this.designationService = designationService;
}
#GetMapping("/designation")
public List<Designation> getAllDesignation(){
return designationService.findAllDesignation();
}
}
and
#RestController
#RequestMapping("/api")
public class EmployeeController {
private EmployeeService employeeService;
#Autowired
public EmployeeController(EmployeeService employeeService) {
this.employeeService = employeeService;
}
// Mappings
}
Can anyone help me solve this issue? Until I had one entity it was working fine; as soon as I added another one it failed to start, with the error. Please help me fix the issue.
In Employee class make this changes.
#Entity
#Table
public class Employee {
#Id
#GeneratedValue(strategy=GenerationType.IDENTITY)
#Column
private Long empId;
#Column
private String Name;
#Column
private String Email;
#OneToMany(targetEntity=Designation.class, fetch=FetchType.LAZY, cascade=CascadeType.ALL)
private Designation designation;
// // Getter & Setter and Constructors
}
In the Designation class, make this changes..
#Entity
public class Designation {
#Id
private Long id;
private String role;
private float salary;
#ManyToOne(targetEntity=Employee.class, fetch=FetchType.LAZY, cascade=CascadeType.ALL)
#JoinColumn
private Set<Employee> employee;
// Getter & Setter and Constructors
}
This will result in a table structure like this.
Employee
EMP_ID
NAME
EMAIL
Designation
ID
ROLE
SALARY
EMPLOYEE_ID
In OneToMany and ManyToOne association, we must keep a note that, which class is holding the Many side. That side is known as the owning side. As that class will contain the foreign key. Here, our designation class is containing the Many side. So, it will create a foreign key automatically.
If we give the #JoinColumn in the Employee class. Still we will get the Designation class as the Owning class.
You can refer this article, for more details.
Rather than this, let me tell you some of your major mistakes.
Do not give the variables name same as your class name.
By writing variables name as empId in java file, it will result EMP_ID in database.
If a column name or a table name is same your variable name, you can avoid re-mentioning it.
means, if my entity class name is Employee. And I want my database Table name should also be EMPLOYEE. I can avoid specifying #Table(name="EMPLOYEE").
I hope it would be helpful !!
#ConfigurationProperties("person")
public class PersonConfiguration {
private List<Details> details = new ArrayList<>();
//getters and setters and tostring
public static class Details {
private String key;
private String company;
private String family;
//getters and setters and tostring
}
}
public class Person {
public String key;
public SomeClass obj;
//constructor
//getter, setter and tostring
}
#Component
public class PersonFactory {
#Autowired
PersonConfiguration personConfig;
public Person getPerson(String key) {
//if key exists in PersonConfiguration
//create SomeClass based on family and compnay details for corresponding key
//create Person bean using key and SomeClass.
}
}
How can i create multiple beans of Person in the above class in lazy manner only when requested based on the different keys in the request.
Basically have a configuration and factory class. Based on an incoming request need to check if it exists in the configuration and if it does, need to create an object of some type using factory class.
Does anyone have any idea how to get a single column using Spring Data JPA? I created a repository like below in my Spring Boot project, but always get the {"cause":null,"message":"PersistentEntity must not be null!"} error when accessing the Restful URL.
#RepositoryRestResource(collectionResourceRel = "users", path = "users")
public interface UsersRepository extends CrudRepository<Users, Integer> {
#Query("SELECT u.userName FROM Users u")
public List<String> getUserName();
}
Then if I access the Restful URL like ../users/search/getUserName, I get the error:
{"cause":null,"message":"PersistentEntity must not be null!"}
Create a Projection interface
public interface UserNameOnly {
String getUserName();
}
Then in your repository interface return that type instead of the user type
public interface UserRepository<User> extends JpaRepository<User,String> {
List<UsernameOnly> findNamesByUserNameNotNull();
}
The get method in the projection interface must match a get method of the defined type on the JPA repository, in this case User.
The "findBySomePropertyOnTheObjectThatIsNotNull" allows you to get a List of the entities (as opposed to an Iterable) based on some criteria, which for a findAll can simply be if the unique identifier (or any other NonNull field) is not null.
Concept is : In your entity class create a constructor with only required instant variables. And use that constructor in the repository method shown below.
Lets say you have a interface Repository like below
Repository implementation:
public interface UserRepository<User> extends JpaRepository<User,String>
{
#Query(value = "select new com.org.User(usr.userId) from User usr where usr.name(:name)")
List<User> findUserIdAlone(#Param("name") String user);
}
In Controller
#RestController
public class UserController
{
#Autowired
private UserRepository<User> userRepository;
#Res
public ResponseEntity<User> getUser(#PathVariable("usrname") String userName)
{
User resultUser = usrRepository.findUserIdAlone(userName);
return ResponseEntity.ok(resultUser);
}
}
public class User
{
private String userId,userName;
public User(String userId)
{
this.userId=userId;
}
// setter and getters goes here
}
This Works for me.
public interface UserDataRepository extends JpaRepository<UserData, Long> {
#Query(value = "SELECT emp_name FROM user_data", nativeQuery = true)
public List<Object[]> findEmp_name();
}
System.out.println("data"+ userDataRepository.findEmp_name());
The above line gave me this result :
data[abhijeet, abhijeet1, abhijeet2, abhijeet3, abhijeet4, abhijeet5]
If you want to only return a single column you should look at Projections and Excerpts which will allow you to filter specific columns and other things that are usefule.
If you need list all of the users, try select userName from Users, if you need one user use "where" look at spring data JPA http://docs.spring.io/spring-data/jpa/docs/current/reference/html/ , try change CrudRepository to JpaRepository
It is possible to provide custom implementations of methods in a Spring Data JPA repository, which enables complete control on queries and return types. The approach is as follows:
Define an interface with the desired method signatures.
Implement the interface to achieve the desired behavior.
Have the Repository extend both JpaRepository and the custom interface.
Here is a working example that uses JpaRepository, assuming a user_table with two columns, user_id and user_name.
UserEntity class in model package:
#Entity
#Table(name = "user_table")
public class UserEntity {
#Id
#GeneratedValue(strategy=GenerationType.AUTO)
#Column(name = "user_id")
private Long userId;
#Column(name = "user_name")
private String userName;
protected UserEntity() {}
public UserEntity(String userName) {
this.userName = userName;
// standard getters and setters
}
Define interface for the custom repository in the repository package:
public interface UserCustomRepository {
List<String> findUserNames();
}
Provide implementation class for the custom interface in the repository package:
public class UserCustomRepositoryImpl implements UserCustomRepository {
// Spring auto configures a DataSource and JdbcTemplate
// based on the application.properties file. We can use
// autowiring to get a reference to it.
JdbcTemplate jdbcTemplate;
#Autowired
public void setJdbcTemplate(JdbcTemplate jdbcTemplate) {
this.jdbcTemplate = jdbcTemplate;
}
// Now our custom implementation can use the JdbcTemplate
// to perform JPQL queries and return basic datatypes.
#Override
public List<String> findUserNames() throws DataAccessException {
String sql = "SELECT user_name FROM user_table";
return jdbcTemplate.queryForList(sql, String.class);
}
}
Finally, we just need to have the UserRepository extend both JpaRepository and the custom interface we just implemented.
public interface UserRepository extends JpaRepository<UserEntity, Long>, UserCustomRepository {}
Simple test class with junit 5 (assuming the database is initially empty):
#SpringBootTest
class UserRepositoryTest {
private static final String JANE = "Jane";
private static final String JOE = "Joe";
#Autowired
UserRepository repo;
#Test
void shouldFindUserNames() {
UserEntity jane = new UserEntity(JANE);
UserEntity joe = new UserEntity(JOE);
repo.saveAndFlush(jane);
repo.saveAndFlush(joe);
List<UserEntity> users = repo.findAll();
assertEquals(2, users.size());
List<String> names = repo.findUserNames();
assertEquals(2, names.size());
assertTrue(names.contains(JANE));
assertTrue(names.contains(JOE));
}
}
Can I use Method Injection Lookup -- with a entity class?.I use Spring+JPA+Hibernate. This allow to inject a prototype bean into a singleton bean.Is this also possible with entity beans?A is prototype scoped bean.I want to put A(#Entity) into a class B (ex. DAO) with scope=singleton.Thanks
#Entity
public class A(){
private String name;
private String surname;
...//get and set
}//A
public interface DAO{
public void method();
}//DAO
public class DAOImpl implements DAO{
private A object_a;
public void method(){
//In this method I use everytime a new istance of A
}//method
}//DAOImpl
You can use #Embedded to include your sub bean, and use in your sql.
#Entity
public class User(){
private String name;
#Embedded
private Address address;
#Bean(scope=DefaultScopes.PROTOTYPE)
public User() {
}
...//get and set
}
#Entity
public class Address(){
private String name;
...//get and set
}
public interface UserRepository extends JpaRepository<User, Long> {
#Query(value = "select u from users u where u.address.name = :addressName")
List<Blog> findUserByAddress(#Param("addressName") String addressName);
}