178
votes

I am using Spring JPA to perform all database operations. However I don't know how to select specific columns from a table in Spring JPA?

For example:
SELECT projectId, projectName FROM projects

15
The idea behind JPA not looking for specific fields is that is cost (efficiency wise) the same to bring one column or all columns from one row of the table. - Desorder
@Desorder -- the cost is not always the same. It's probably not a big deal for simpler, primitive sort of datatypes but the reason I ended up on this page is because I noticed a simple "list documents" query was running slow. That entity has a BLOB column (needs it for file upload/storage) & I suspect it is slow because it is loading the BLOBs into memory even though they're not required for listing the docs. - jm0
@jm0 As far as you remember, how many tables had BLOB columns? - Desorder
@Desorder it was just one table but I was doing a "list" function (multirow -- list all docs created by a given id). The only reason I noticed this issue was because this simple list query was taking several seconds, whereas more complex queries on other tables were happening almost instantly. Once I realized, I knew it would suffer more and more as rows are added because Spring JPA is loading every BLOB into memory even tho they are not used. I found a decent solution for Spring data (posted below) but I think I have an even better one that is pure JPA annotation, I will post tmrw if it works - jm0

15 Answers

91
votes

You can set nativeQuery = true in the @Query annotation from a Repository class like this:

public static final String FIND_PROJECTS = "SELECT projectId, projectName FROM projects";

@Query(value = FIND_PROJECTS, nativeQuery = true)
public List<Object[]> findProjects();

Note that you will have to do the mapping yourself though. It's probably easier to just use the regular mapped lookup like this unless you really only need those two values:

public List<Project> findAll()

It's probably worth looking at the Spring data docs as well.

229
votes

You can use projections from Spring Data JPA (doc). In your case, create interface:

interface ProjectIdAndName{
    String getId();
    String getName();
}

and add following method to your repository

List<ProjectIdAndName> findAll();
155
votes

I don't like the syntax particularly (it looks a little bit hacky...) but this is the most elegant solution I was able to find (it uses a custom JPQL query in the JPA repository class):

@Query("select new com.foo.bar.entity.Document(d.docId, d.filename) from Document d where d.filterCol = ?1")
List<Document> findDocumentsForListing(String filterValue);

Then of course, you just have to provide a constructor for Document that accepts docId & filename as constructor args.

28
votes

In my situation, I only need the json result, and this works for me:

public interface SchoolRepository extends JpaRepository<School,Integer> {
    @Query("select s.id, s.name from School s")
    List<Object> getSchoolIdAndName();
}

in Controller:

@Autowired
private SchoolRepository schoolRepository;

@ResponseBody
@RequestMapping("getschoolidandname.do")
public List<Object> getSchool() {
    List<Object> schools = schoolRepository.getSchoolIdAndName();
    return schools;
}
15
votes

In my case i created a separate entity class without the fields that are not required (only with the fields that are required).

Map the entity to the same table. Now when all the columns are required i use the old entity, when only some columns are required, i use the lite entity.

e.g.

@Entity
@Table(name = "user")
Class User{
         @Column(name = "id", unique=true, nullable=false)
         int id;
         @Column(name = "name", nullable=false)
         String name;
         @Column(name = "address", nullable=false)
         Address address;
}

You can create something like :

@Entity
@Table(name = "user")
Class UserLite{
         @Column(name = "id", unique=true, nullable=false)
         int id;
         @Column(name = "name", nullable=false)
         String name;
}

This works when you know the columns to fetch (and this is not going to change).

won't work if you need to dynamically decide the columns.

10
votes

With the newer Spring versions One can do as follows:

If not using native query this can done as below:

public interface ProjectMini {
    String getProjectId();
    String getProjectName();
}

public interface ProjectRepository extends JpaRepository<Project, String> { 
    @Query("SELECT p FROM Project p")
    List<ProjectMini> findAllProjectsMini();
}

Using native query the same can be done as below:

public interface ProjectRepository extends JpaRepository<Project, String> { 
    @Query(value = "SELECT projectId, projectName FROM project", nativeQuery = true)
    List<ProjectMini> findAllProjectsMini();
}

For detail check the docs

9
votes

I guess the easy way may be is using QueryDSL, that comes with the Spring-Data.

Using to your question the answer can be

JPAQuery query = new JPAQuery(entityManager);
List<Tuple> result = query.from(projects).list(project.projectId, project.projectName);
for (Tuple row : result) {
 System.out.println("project ID " + row.get(project.projectId));
 System.out.println("project Name " + row.get(project.projectName)); 
}}

The entity manager can be Autowired and you always will work with object and clases without use *QL language.

As you can see in the link the last choice seems, almost for me, more elegant, that is, using DTO for store the result. Apply to your example that will be:

JPAQuery query = new JPAQuery(entityManager);
QProject project = QProject.project;
List<ProjectDTO> dtos = query.from(project).list(new QProjectDTO(project.projectId, project.projectName));

Defining ProjectDTO as:

class ProjectDTO {

 private long id;
 private String name;
 @QueryProjection
 public ProjectDTO(long projectId, String projectName){
   this.id = projectId;
   this.name = projectName;
 }
 public String getProjectId(){ ... }
 public String getProjectName(){....}
}
8
votes

In my opinion this is great solution:

interface PersonRepository extends Repository<Person, UUID> {

    <T> Collection<T> findByLastname(String lastname, Class<T> type);
}

and using it like so

void someMethod(PersonRepository people) {

  Collection<Person> aggregates =
    people.findByLastname("Matthews", Person.class);

  Collection<NamesOnly> aggregates =
    people.findByLastname("Matthews", NamesOnly.class);
}
6
votes

Using Spring Data JPA there is a provision to select specific columns from database

---- In DAOImpl ----

@Override
    @Transactional
    public List<Employee> getAllEmployee() throws Exception {
    LOGGER.info("Inside getAllEmployee");
    List<Employee> empList = empRepo.getNameAndCityOnly();
    return empList;
    }

---- In Repo ----

public interface EmployeeRepository extends CrudRepository<Employee,Integer> {
    @Query("select e.name, e.city from Employee e" )
    List<Employee> getNameAndCityOnly();
}

It worked 100% in my case. Thanks.

3
votes

You can use JPQL:

TypedQuery <Object[]> query = em.createQuery(
  "SELECT p.projectId, p.projectName FROM projects AS p", Object[].class);

List<Object[]> results = query.getResultList();

or you can use native sql query.

Query query = em.createNativeQuery("sql statement");
List<Object[]> results = query.getResultList();
3
votes

You can apply the below code in your repository interface class.

entityname means your database table name like projects. And List means Project is Entity class in your Projects.

@Query(value="select p from #{#entityName} p where p.id=:projectId and p.projectName=:projectName")

List<Project> findAll(@Param("projectId") int projectId, @Param("projectName") String projectName);
2
votes

It is possible to specify null as field value in native sql.

@Query(value = "select p.id, p.uid, p.title, null as documentation, p.ptype " +
            " from projects p " +
            "where p.uid = (:uid)" +
            "  and p.ptype = 'P'", nativeQuery = true)
Project findInfoByUid(@Param("uid") String uid);
0
votes

Using Native Query:

Query query = entityManager.createNativeQuery("SELECT projectId, projectName FROM projects");
List result = query.getResultList();
0
votes

You can use the answer suggested by @jombie, and:

  • place the interface in a separate file, outside the entity class;
  • use native query or not (the choice depended on your needs);
  • don't override findAll() method for this purpose but use name of your choice;
  • remember to return a List parametrized with your new interface (e.g. List<SmallProject>).
0
votes
public static final String FIND_PROJECTS = "select ac_year_id,ac_year from tbl_au_academic_year where ac_year_id=?1";

    @Query(value = FIND_PROJECTS, nativeQuery = true)
    public  List<Object[]> findByAcYearId(Integer ac_year_id);

this works for me