%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

You can also implement the like queries using Spring Data JPA supported keyword "Containing".

List<Registration> findByPlaceContaining(String place);

Answer 2

You can have one alternative of using placeholders as:

@Query("Select c from Registration c where c.place LIKE  %?1%")
List<Registration> findPlaceContainingKeywordAnywhere(String place);

Answer 3

For your case, you can directly use JPA methods. That is like bellow :

Containing: select ... like %:place%

List<Registration> findByPlaceContainingIgnoreCase(String place);

here, IgnoreCase will help you to search item with ignoring the case.

Using @Query in JPQL :

@Query("Select registration from Registration registration where 
registration.place LIKE  %?1%")
List<Registration> findByPlaceContainingIgnoreCase(String place);

Here are some related methods:

1. Like findByPlaceLike

   … where x.place like ?1

2. StartingWith findByPlaceStartingWith

   … where x.place like ?1 (parameter bound with appended %)

3. EndingWith findByPlaceEndingWith

   … where x.place like ?1 (parameter bound with prepended %)

4. Containing findByPlaceContaining

   … where x.place like ?1 (parameter bound wrapped in %)

More info , view this link , this link and this

Hope this will help you :)

Answer 4

answer exactly will be

-->` @Query("select u from Category u where u.categoryName like %:input%")
     List findAllByInput(@Param("input") String input);