CODEWITHSUNDEEP
springjpaspring-data-jpaspring-data
Roman Sterlin
Roman Sterlin

Reputation: 1617

Spring data @Query insert, DBC style parameters (?) are not supported for JPA queries

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?

Upvotes: 1

Views: 2608

Answers (2)

Diego Camilo Silva Tellez
Diego Camilo Silva Tellez

Reputation: 33

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);

}

Upvotes: 0

Pivoter
Pivoter

Reputation: 419

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.

Upvotes: 1

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