I can't make this query work:
Query query = eManager.createQuery("select c FROM News c WHERE c.NEWSID = :id",News.class);
return (News)query.setParameter("id", newsId).getSingleResult();
and I got this exception:
Exception Description: Problem compiling [select c FROM News c WHERE c.NEWSID = :id].
[27, 35] The state field path 'c.NEWSID' cannot be resolved to a valid type.] with root cause
Local Exception Stack:
Exception [EclipseLink-0] (Eclipse Persistence Services - 2.5.0.v20130507-3faac2b): org.eclipse.persistence.exceptions.JPQLException
Exception Description: Problem compiling [select c FROM News c WHERE c.NEWSID = :id].
Why does it happen? :id and named parameter are identical
EDIT: my entity class
@Entity
@Table(name="NEWS")
public class News implements Serializable{
@Id
@SequenceGenerator(name = "news_seq_gen", sequenceName = "news_seq")
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "news_seq_gen")
private int newsId;
private String newsTitle;
private String newsBrief;
private String newsContent;
private Date newsDate;
@Transient
private boolean selected=false;
//constructor and getters and setters
My entity is:
my query is (JPQL):
so I use it like this:
It's work ;-)
That happens because
News
entity does not have persistent attribute namedNEWSID
. Names of the persistent attributes are case sensitive in JPQL queries and those should be written with exactly same case as they appear in entity.Because entity have persistent attribute named
newsId
, that should also be used in query instead ofNEWSID
:entity have persistent attribute named newsId.but in query you have used NEWSID . try with this