How Spring Data JPA Works Inside

Question

I went through the Spring Data JPA Tutorial. I am confused about how this structure works internally. Let me specify a specific scenario.

Code has been defined

/**
 * Custom finder
 */
public List<Location> getLocationByStateName(String name) {
    @SuppressWarnings("unchecked")
    List<Location> locs = entityManager
        .createQuery("select l from Location l where l.state like :state")
        .setParameter("state", name + "%").getResultList(); // note
        return locs;
}

It was simply replaced by the following interface

@Repository
public interface LocationJPARepository extends JpaRepository<Location, Long> {
    List<Location> findByStateLike(String stateName);
}

And the corresponding test case worked fine

@Test
public void testFindWithLike() throws Exception {
   List<Location> locs = locationRepository.getLocationByStateName("New");
   assertEquals(4, locs.size());
}

New test case

@Test
public void testFindWithLike() throws Exception {
   List<Location> locs = locationJPARepository.findByStateLike("New");
   assertEquals(4, locs.size());
}

My question

• As the framework knows, am I looking for an exact match using = or a partial match using an SQL statement (can't it be a method name?)
• If for some reason this decides that I am looking for a partial match, then there are still auxiliary parameters ... for example, the name% or% name or% name% ...
• Just like this solves the matter, how is it important? (I can be case insensitive using SQL, as with toUpper (), i.e. by comparing everything in uppercase)
• ( ) EXACT SQL , ?

, . , .

Answer 1

. .

, , , findByFirstnameIgnoreCase, , UPPER(x.firstame) = UPPER(?1).

, findByProperty, , , LIKE, findByFirstnameLike, where x.firstname like ?1.

, . @Query , .