%Like% Query in spring JpaRepository
JavaJpaSpring Data-JpaJava Problem Overview
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<Registration,Long> {
@Query("Select c from Registration c where c.place like :place")
List<Registration> findByPlaceContaining(@Param("place")String place);
}
Java Solutions
Solution 1 - Java
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 @Query
annotation 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.
Solution 2 - Java
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);
}
Solution 3 - Java
For your case, you can directly use JPA methods. That code 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:
-
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 :)
Solution 4 - Java
You can also implement the like queries using Spring Data JPA supported keyword "Containing".
List<Registration> findByPlaceContaining(String place);
Solution 5 - Java
Try this.
@Query("Select c from Registration c where c.place like '%'||:place||'%'")
Solution 6 - Java
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);
Solution 7 - Java
when call funtion, I use:
findByPlaceContaining("%" + place);
or:
findByPlaceContaining(place + "%");
or:
findByPlaceContaining("%" + place + "%");
Solution 8 - Java
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.
Solution 9 - Java
answer exactly will be
-->` @Query("select u from Category u where u.categoryName like %:input%") ListfindAllByInput(@Param("input") String input);
Solution 10 - Java
We can use native query
@Query(nativeQuery = true, value ="Select * from Registration as c where c.place like %:place%")
List<Registration> findByPlaceContaining(@Param("place")String place);
Solution 11 - Java
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!
Solution 12 - Java
You can just simply say 'Like' keyword after parameters..
List<Employee> findAllByNameLike(String name);
Solution 13 - Java
Us like this
@Query("from CasFhgDeviceView where deviceGroupName like concat(concat('%CSGW%', :usid), '%') ")
Solution 14 - Java
There can be various approaches. As mentioned in answer of many, if possible you can use JPA predefined template query.
List<Registration> findByPlaceContainingIgnoreCase(String place);
Also, you can append '%' in java layer before calling the above method.
If complex query, then you can normally use @Query one
@Query("Select r from Registration r where r.place like '%' || :place || '%'")
For readability, you can use below one
@Query("Select r from Registration r where r.place like CONCAT('%', :place, '%'")