%Like% Query in spring JpaRepository

Question

I would like to write a like query in JpaRepository but it is not returning anything :

LIKE \'%place%\'-its not working.

LIKE

Answer 1

when call funtion, I use: findByPlaceContaining("%" + place);

or: findByPlaceContaining(place + "%");

or: findByPlaceContaining("%" + place + "%");

Answer 2

You dont actually need the @Query annotation at all.

You can just use the following

    @Repository("registerUserRepository")
    public interface RegisterUserRepository extends JpaRepository<Registration,Long>{

    List<Registration> findByPlaceIgnoreCaseContaining(String place);

    }

Answer 3

I use this:

@Query("Select c from Registration c where lower(c.place) like lower(concat('%', concat(:place, '%')))")

lower() is like toLowerCase in String, so the result isn't case sensitive.

Answer 4

Found solution without @Query (actually I tried which one which is "accepted". However, it didn't work).

Have to return Page<Entity> instead of List<Entity>:

public interface EmployeeRepository 
                          extends PagingAndSortingRepository<Employee, Integer> {
    Page<Employee> findAllByNameIgnoreCaseStartsWith(String name, Pageable pageable);
}

IgnoreCase part was critical for achieving this!

Answer 5

The spring data JPA query needs the "%" chars as well as a space char following like in your query, as in

@Query("Select c from Registration c where c.place like %:place%").

Cf. http://docs.spring.io/spring-data/jpa/docs/current/reference/html.

You may want to get rid of the @Queryannotation alltogether, as it seems to resemble the standard query (automatically implemented by the spring data proxies); i.e. using the single line

List<Registration> findByPlaceContaining(String place);

is sufficient.

Answer 6

Try this.

@Query("Select c from Registration c where c.place like '%'||:place||'%'")