Spring JPA @Query with LIKE

Try to use the following approach (it works for me):

@Query("SELECT u.username FROM User u WHERE u.username LIKE CONCAT('%',:username,'%')")
List<String> findUsersWithPartOfName(@Param("username") String username);

Notice: The table name in JPQL must start with a capital letter.

Using Query creation from method names, check table 4 where they explain some keywords.

1. Using Like: select ... like :username

    List<User> findByUsernameLike(String username);
   

2. StartingWith: select ... like :username%

    List<User> findByUsernameStartingWith(String username);
   

3. EndingWith: select ... like %:username

    List<User> findByUsernameEndingWith(String username);
   

4. Containing: select ... like %:username%

    List<User> findByUsernameContaining(String username);
   

Notice that the answer that you are looking for is number 4. You don't have to use @Query

Another way: instead CONCAT function we can use double pipe: :lastname || '%'

@Query("select c from Customer c where c.lastName LIKE :lastname||'%'")
List<Customer> findCustomByLastName( @Param("lastname") String lastName);

You can put anywhere, prefix, suffix or both

:lastname ||'%'  
'%' || :lastname  
'%' || :lastname || '%'  

Easy to use following (no need use CONCAT or ||):

@Query("from Service s where s.category.typeAsString like :parent%")
List<Service> findAll(@Param("parent") String parent);

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