状态字段路径无法解析为有效类型。 JPA查询

Question

我正在尝试进行jpa查询，但我得到了一个例外：

  

状态字段“n.id”路径无法解析为有效类型。

我的查询是：

select distinct n from News n 
left join n.commentsList as c 
left join n.tagSet as nt 
left join n.author as a 

group by n.id, n.title, n.shortText, n.fullText, n.creationDate,
n.modificationDate, n.author.authorId, n.version 
order by n.modificationDate desc, count(c.news) desc

我的实体：

@Entity
@Table(name = "News")
public final class News implements Serializable, IEntity {

/**
 * For deserialization with no exception after modification.
 */
private static final long serialVersionUID = 3773281197317274020L;

@Id
@SequenceGenerator(name = "NEWS_SEQ_GEN", sequenceName = "NEWS_SEQ")
@GeneratedValue(strategy = GenerationType.SEQUENCE, generator = "NEWS_SEQ_GEN")
@Column(name = "NEWS_ID", precision = 0)
private Long newsId; // Primary key

@Column(name = "TITLE")
private String title;

@Column(name = "SHORT_TEXT")
private String shortText;

@Column(name = "FULL_TEXT")
private String fullText;

@Temporal(TemporalType.DATE)
@Column(name = "CREATION_DATE")
private Date creationDate;

@Temporal(TemporalType.DATE)
@Column(name = "MODIFICATION_DATE")
private Date modificationDate;

@OneToMany(fetch = FetchType.EAGER, mappedBy = "news")
private List<Comment> commentsList;

@ManyToMany(fetch = FetchType.EAGER)
@JoinTable(name = "NEWS_TAG", joinColumns = { @JoinColumn(name = "NEWS_ID") }, inverseJoinColumns = { @JoinColumn(name = "TAG_ID") })
private Set<Tag> tagSet;

@ManyToOne(fetch = FetchType.EAGER)
@JoinTable(name = "NEWS_AUTHOR", joinColumns = { @JoinColumn(name = "NEWS_ID") }, inverseJoinColumns = { @JoinColumn(name = "AUTHOR_ID") })
private Author author;

@Version
private Long version;

我认为别名和group by子句中存在错误。

Answer 1

您的ID被称为

private Long newsId; // Primary key

因此，这是您应在查询中使用的值。换句话说：

group by n.newsId, ...

而不是n.Id

我认为您需要调整查询，并在n.newsId子句中使用group by。

select n, count(c) from News n 
left join n.commentsList as c 
left join n.tagSet as nt 
left join n.author as a 

group by n.id
order by n.modificationDate desc, count(c) desc

我认为这可以确保您可以获得评论的计数（虽然我不确定）。