[FIXED] How to select and order an EmbeddedID in JPA

Issue

I have one table with a composite primary key:

public class IdmTenantBoard implements Serializable {
private static final long serialVersionUID = 1L;

@EmbeddedId
private IdmTenantBoardPK id;

@Column(name="board_display_name")
private String boardDisplayName;

@Column(name="board_name")
private String boardName;

The embedded class looks like this:

public class IdmTenantBoardPK implements Serializable {
//default serial version id, required for serializable classes.
private static final long serialVersionUID = 1L;

@Column(name="board_id")
private int boardId;

@Column(name="tenant_id")
private String tenantId;

The method I am using to call query is as follows:

public String getMaxBoardId(){
      TypedQuery<String> query = entityManager.createQuery("select o from IdmTenantBoard o order by o.IdmTenantBoardPK.boardId desc",String.class);
      query.setMaxResults(1);
      List<String> lResults = query.getResultList();
      String maxBoardId=null;
      if((null!=lResults)&&(!lResults.isEmpty())){
          maxBoardId=lResults.get(0);
      }
      return maxBoardId;
}

What do I need to get the boardId sorted in Descending order? I am getting the following error:

Caused by: org.hibernate.QueryException: could not resolve property: IdmTenantBoardPK of: com.newrubric.idm.common.entities.IdmTenantBoard [select o from com.newrubric.idm.common.entities.IdmTenantBoard o order by o.IdmTenantBoardPK.boardId desc]

Solution

In following IdmTenantBoardPK is type of the embedded id and name of the property is id:

@EmbeddedId
private IdmTenantBoardPK id;

That's why o.IdmTenantBoardPK.boardId cannot be used in queries. Instead of it o.id.boardId should be used.

Answered By - Mikko Maunu