I am trying to make a custom insert query in my interface that exnteds JpaRepository
public interface CustomerCouponDAO extends JpaRepository<CustomerCoupons, Integer>{
@Query("INSERT into customer_coupons (customerId, couponId) values (?,?)")
public void insert(Integer custid, Integer coup);
}
but when I get the exception:
Caused by: java.lang.IllegalArgumentException: JDBC style parameters (?) are not supported for JPA queries.
any ideas on how to make that insert query?
Use PersistentContext.
Interface:
@Repository
public interface CustomerCouponDAOExtend {
public void insert(Integer custid, Integer coup);
}
Implementation:
public class CustomerCouponDAOExtendImpl {
@PersistenceContext
private EntityManager em;
@Transactional
public void insert(Integer custid, Integer coup) {
CustomerCoupons custCoupons = new CustomerCoupons();
custCoupons.setCustid(custid);
custCoupons.setCoup(coup);
em.merge(custCoupons);
}
}
Also you can use persist, but in this case your necessary add
CustomerCoupons custCoupons = em.find(CustomerCoupons.class, coup); to avoid problems if the row is already in DB.
Extend your own interface:
@Repository
public interface CustomerCouponDAO
extends JpaRepository<CustomerCoupons, Integer>,CustomerCouponDAOExtend{
}
UPDATE:
Observe the naming convention that Spring finds implementation: if extend repository has name CustomerCouponDAOExtend then the implementation should be called CustomerCouponDAOExtendImpl.
@RepositoryRestResource annotation, but not @Repository. Remove custom method from CustomerCouponDAO and this should work correctly. And remember mark both interfaces as @Repository. Complemented the answer.it's too late to answer anycase: in JPA for custom method you must set the number of parameter in the query annotation:
public interface CustomerCouponDAO extends JpaRepository<CustomerCoupons, Integer>{
@Query("INSERT into customer_coupons (customerId, couponId) values (?1,?2)")
public void insert(Integer custid, Integer coup);
}
