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  place  works perfectly   Here is my code      Repository  registerUserRepository   public interface RegisterUserRepository extendsJpaRepository lt Registration Long gt          Query  Select c from Registration c where c place like  place        List lt Registration gt  findByPlaceContaining  Param  place  String place

User · Answer

answer exactly will be    --    Query  select u from Category u where u categoryName like   input         List findAllByInput  Param  input   String input

User · Answer

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 lt Registration gt  findByPlaceContaining String place     is sufficient

User · Answer

when call funtion  I use  findByPlaceContaining       place    or  findByPlaceContaining place          or  findByPlaceContaining       place

User · Answer

You can also implement the like queries using Spring Data JPA supported keyword  Containing     List lt Registration gt  findByPlaceContaining String place

User · Answer

For your case  you can directly use JPA methods  That code is like bellow   Containing  select     like   place  List lt Registration gt  findByPlaceContainingIgnoreCase String place    here  IgnoreCase will help you to search item with ignoring the case  Using  Query in JPQL    Query  quot Select registration from Registration registration where  registration place LIKE    1  quot   List lt Registration gt  findByPlaceContainingIgnoreCase String place    Here are some related methods   Like findByPlaceLike     where x place like  1  StartingWith findByPlaceStartingWith     where x place like  1  parameter bound with appended     EndingWith findByPlaceEndingWith     where x place like  1  parameter bound with prepended     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

User · Answer

Try this    Query  Select c from Registration c where c place like       place

User · Answer

You can have one alternative of using placeholders as    Query  Select c from Registration c where c place LIKE    1    List lt Registration gt  findPlaceContainingKeywordAnywhere String place

User · Answer

You dont actually need the  Query annotation at all  You can just use the following      Repository  quot registerUserRepository quot       public interface RegisterUserRepository extends JpaRepository lt Registration Long gt            List lt Registration gt  findByPlaceIgnoreCaseContaining String place

User · Answer

Found solution without  Query  actually I tried which one which is  accepted   However  it didn t work     Have to return Page lt Entity gt  instead of List lt Entity gt    public interface EmployeeRepository                            extends PagingAndSortingRepository lt Employee  Integer gt        Page lt Employee gt  findAllByNameIgnoreCaseStartsWith String name  Pageable pageable       IgnoreCase part was critical for achieving this

User · Answer

I use this   Query  quot Select c from Registration c where lower c place  like lower concat      concat  place         quot    lower   is like toLowerCase in String  so the result isn t case sensitive

[java] %Like% Query in spring JpaRepository

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