Question

我有JPA加入的问题。正如数据库查询这样的工作：

SELECT *
FROM votingstats
INNER JOIN enginename ON enginename.id = votingstats.engine_name_id
WHERE engine_name LIKE 'zz'
  AND type_of_game_used_by_that_engine LIKE 'yy'
  AND vote_pack_name = 'xx'

然而，在Java JPA中，我遇到了一个问题，因为工作查询看起来像这样，并且工作正常：

 CriteriaQuery<VotingStats> criteria= 
 criteriaBuilder.createQuery(VotingStats.class);
    Root<VotingStats> currentElo = criteria.from(VotingStats.class);
    Join<VotingStats, EngineName> join = currentElo.join("engineNameId");
    CriteriaQuery<VotingStats> eloJoinQuerry = criteria
            .where(
                criteriaBuilder.and(
                        criteriaBuilder.equal(
                                join.get("engineName"), engineName),
                        criteriaBuilder.equal(
                                join.get("typeOfGameUsedByThatEngine"), typeOfGame.getTypeOfGame())
                )
            );

但是，在添加votePackName后，当我尝试引用VotingStats实体时它会中断：

  CriteriaQuery<VotingStats> criteria= 
  criteriaBuilder.createQuery(VotingStats.class);
    Root<VotingStats> currentElo = criteria.from(VotingStats.class);
    Join<VotingStats, EngineName> join = currentElo.join("engineNameId");
    CriteriaQuery<VotingStats> eloJoinQuerry = criteria
            .where(
                criteriaBuilder.and(
                        criteriaBuilder.equal(
                                join.get("engineName"), engineName),
                        criteriaBuilder.equal(
                                join.get("typeOfGameUsedByThatEngine"), typeOfGame.getTypeOfGame()),
                        criteriaBuilder.equal(
                                join.get("votePackName"), votePackName)
                )
            );

实体看起来像这样（但如果是普通连接，它们可以正常工作）：

@Entity
@Data
public class VotingStats {

@Id
@Column(name = "id", nullable = false)
@GeneratedValue(strategy=GenerationType.SEQUENCE, generator="votingstats_seq")
@SequenceGenerator(name="votingstats_seq", sequenceName="votingStats_id_seq", allocationSize=1)
@Access(AccessType.PROPERTY)
private Integer id;

@Basic
@Column(name = "vote_pack_name")
private String votePackName;

@Basic
@Column(name = "vote_amount")
private Integer voteAmount;

@ManyToOne(cascade = CascadeType.REFRESH)
@JoinColumn(name = "engine_name_id")
private EngineName engineNameId;
}

和第二个实体是：

@Entity
@Data
public class EngineName {

@Id
@Column(name = "id", nullable = false)
@GeneratedValue(strategy=GenerationType.SEQUENCE, generator="engineName_seq")
@SequenceGenerator(name="engineName_seq", sequenceName="enginename_id_seq", allocationSize=1)
@Access(AccessType.PROPERTY)
private Integer id;


@Basic
@Column(name = "engine_name")
private String engineName;


@Basic
@Column(name = "type_of_game_used_by_that_engine")
private String typeOfGameUsedByThatEngine;

}

这是我得到的错误：

java.lang.IllegalArgumentException：无法在此ManagedType上找到具有给定名称[votePackName]的Attribute [chess.database.entities.EngineName]

at org.hibernate.metamodel.internal.AbstractManagedType.checkNotNull（AbstractManagedType.java:128)
  在org.hibernate.metamodel.internal.AbstractManagedType.getAttribute（AbstractManagedType.java:113）
  at org.hibernate.query.criteria.internal.path.AbstractFromImpl.locateAttributeInternal（AbstractFromImpl.java:116)
  在org.hibernate.query.criteria.internal.path.AbstractPathImpl.locateAttribute（AbstractPathImpl.java:204）
  在org.hibernate.query.criteria.internal.path.AbstractPathImpl.get（AbstractPathImpl.java:177）
  at chess.database.dao.VotingStatsDAO.findByEngineName（VotingStatsDAO.java:32）

JPA将查询与每个查询联系起来

0 个答案: