CODEWITHSUNDEEP
javahibernatejpaspring-dataspring-data-jpa
m.gavr
m.gavr

Reputation: 83

spring-data-jpa repository use Collection as @Param in Query

I'm having an issue performing a custom query through the use of a spring data jpa repository.

@Query annitated method doesn't specify parametr of a List<> type.

here are entity classes:

@Entity
@Table(name = "\"user\"")
public class User {
  @Id
  @GeneratedValue(generator = "increment")
  private long id;

  @Column(unique = true, nullable = false)
  private String mail;

  @ManyToMany(targetEntity = Interest.class, mappedBy = "users")
  private List<Interest> interests = new ArrayList<Interest>();
  ...
  ... setters and getters.

@Entity
@Table(name = "interest")
public class Interest {
  @Id
  @Column(name = "interest_name", nullable = false, unique = true)
  private String name;

  @ManyToMany(targetEntity = User.class, fetch = FetchType.LAZY)
  private List<User> users = new ArrayList<User>();
  ...
  ... setters and getters.

here is query:

@Repository
public interface UserRepository extends JpaRepository<User, Long> {
  @Query("select distinct u from User u where u <> :currentUser and
         u.interests in :currentInterests")
  List<User> getUsersWithSameInterests(@Param("currentUser") User user,
         @Param("currentInterests") List<Interest> interests);
}

i used this like:

@Autowired
private UserRepository userRepository;
  @Override
  public List<User> getUsersWithSameInterests(User user) {
    return userRepository.getUsersWithSameInterests(user, user.getInterests());
  }

but got

org.springframework.dao.InvalidDataAccessResourceUsageException: could not extract ResultSet; SQL [n/a]; nested exception is org.hibernate.exception.SQLGrammarException: could not extract ResultSet
.
.
.
Caused by: java.sql.SQLException: No value specified for parameter 2
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:957)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:896)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:885)
at com.mysql.jdbc.SQLError.createSQLException(SQLError.java:860)
at com.mysql.jdbc.PreparedStatement.checkAllParametersSet(PreparedStatement.java:2205)
at com.mysql.jdbc.PreparedStatement.fillSendPacket(PreparedStatement.java:2185)
at com.mysql.jdbc.PreparedStatement.fillSendPacket(PreparedStatement.java:2115)
at com.mysql.jdbc.PreparedStatement.executeQuery(PreparedStatement.java:1936)
at org.apache.commons.dbcp2.DelegatingPreparedStatement.executeQuery(DelegatingPreparedStatement.java:83)
at org.apache.commons.dbcp2.DelegatingPreparedStatement.executeQuery(DelegatingPreparedStatement.java:83)
at org.hibernate.engine.jdbc.internal.ResultSetReturnImpl.extract(ResultSetReturnImpl.java:70)
... 73 more

No value specified for parameter 2, although 2nd parametr has valid value

Upvotes: 4

Views: 4555

Answers (1)

rahulnikhare
rahulnikhare

Reputation: 1486

   Change the code as per below:

   Old Code:

   `@Repository
   public interface UserRepository extends JpaRepository<User, Long> {
   @Query("select distinct u from User u where u <> :currentUser and
         u.interests in :currentInterests")
   List<User> getUsersWithSameInterests(@Param("currentUser") User user,
         @Param("currentInterests") List<Interest> interests);
   }`

   Updated Code:

   `@Repository
   public interface UserRepository extends JpaRepository<User, Long> {
   @Query("select distinct u from User u where u.user :currentUser 
           and u.interests in :currentInterests")
   List<User> getUsersWithSameInterests(@Param("currentUser") User user,
         @Param("currentInterests") List<Interest> interests);
  }`

Upvotes: 1

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