[FIXED] JPQL Like Case Insensitive

Issue

I want to search data in User table by name case insensitive.

@Repository
public interface UserRepository extends JpaRepository<User, Long> {

  @Query("select u from User u where lower(u.name) like %lower(?1)%")
  public List<User> findByNameFree(String name);

}

I got an error: unexpected token: %. Where should I place '%'?

Solution

You can use the concat operator:

@Query("select u from User u where lower(u.name) like lower(concat('%', ?1,'%'))")
public List<User> findByNameFree(String name);

or with a named parameter:

@Query("select u from User u where lower(u.name) like lower(concat('%', :nameToFind,'%'))")
public List<User> findByNameFree(@Param("nameToFind") String name);

(Tested with Spring Boot 1.4.3)

Answered By - Wim Deblauwe
Answer Checked By - Gilberto Lyons (JavaFixing Admin)