Question

我无法使此查询有效：

Query query = eManager.createQuery("select c FROM News c WHERE c.NEWSID = :id",News.class);
        return (News)query.setParameter("id", newsId).getSingleResult();

我得到了这个例外：

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].

为什么会这样？：id和命名参数相同

修改我的实体类

@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

Answer 1

这是因为News实体没有名为NEWSID的持久属性。持久属性的名称在JPQL查询中区分大小写，并且应该使用与它们在实体中出现的完全相同的大小写。

因为实体具有名为newsId的持久属性，所以也应该在查询中使用，而不是NEWSID：

select c FROM News c WHERE c.newsId = :id

Answer 2

实体具有名为newsId的持久属性。但在查询中，您使用了NEWSID。试试这个

select c FROM News c WHERE c.newsId = :id

Answer 3

我的实体是：

@Entity
@Table(name = "TBL_PERSON_INFO")
public class Person implements Serializable {
    @Id
    @Column(name = "ID", nullable = false)
    private Integer id;

    @Column(name = "USER_ID", nullable = false)
    private Integer user_id;
    .
    .
    .
}

我的查询是（JPQL）：

String queryName = "from Person p where p.user_id = :user_id";

所以我这样使用它：

        javax.persistence.Query query = em.createQuery(queryName);
        query.setParameter("user_id", userId);

        try {
            obj = query.getSingleResult();
        }
        catch (javax.persistence.NoResultException nre) {
            logger.error("javax.persistence.NoResultException: " + nre.getMessage());
        }
        catch (javax.persistence.NonUniqueResultException nure) {
            logger.error("javax.persistence.NonUniqueResultException: " + nure.getMessage());
        }
        catch (Exception e) {
            e.printStackTrace();
        }

        if (obj == null) {
            System.out.println("obj is null!");
            return null;
        }

        Person person = (Person) obj;

这是工作; - ）

JPQL：无法将状态字段路径解析为有效类型

3 个答案: