Select one column using Spring Data JPA - spring

Question

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!"}

Answer 1

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.

Answer 2

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
}

Answer 3

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]

Answer 4

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.

Answer 5

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

Answer 6

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));
}
}