Question

使用QUeryDSL加入时遇到问题。我试图收到一份＆＃34; Clube＆＃34;与＃34; Federacao＆＃34;。

有一个ManyToOne关系

有人可以帮助我吗？

感谢。

关系是在下面的类之间。

@Entity
@Table(name = "federacoes")
@SequenceGenerator(name = "idgen", sequenceName = "federacoes_id_federacao_seq", allocationSize = 1)
@AttributeOverride(name = "id", column = @Column(name = "id_federacao"))
public class Federacao extends AbstractEntity{

private String sigla;
private String uf;
private String nome;
...
}

@Entity
@Table(name = "clubes")
@SequenceGenerator(name = "idgen", sequenceName = "clubes_id_clube_seq", allocationSize = 1)
@AttributeOverride(name = "id", column = @Column(name = "id_clube"))
public class Clube extends AbstractEntity{

private Federacao federacao;
...

@ManyToOne
@JoinColumn(name = "id_federacao")
@Column(name = "federacao")
public Federacao getFederacao() {
    return federacao;
}
...
}

我使用QueryDSL并在扩展QueryDslRepositorySupport的存储库类中执行此操作：

...
@PersistenceContext
private EntityManager em;

private static final QClube qClube = QClube.clube;
private static final QFederacao qFederacao = QFederacao.federacao;

@Override
public List<Clube> findAll(FilterClubeDTO filterClubeDTO) {

JPAQuery query = new JPAQuery(em);
    return query.from(qClube)
        .innerJoin(qFederacao).on(qClube.federacao.eq(qFederacao))
    .orderBy(qClube.id.desc())
    .list(qClube);

}

我收到此例外：

2016-04-12 12:32:38.485 ERROR 2853 --- [ qtp36627152-15] o.h.hql.internal.ast.ErrorCounter        :  Path expected for join!


org.springframework.web.util.NestedServletException: Request processing failed; nested exception is org.springframework.dao.InvalidDataAccessApiUsageException: org.hibernate.hql.internal.ast.QuerySyntaxException: Path expected for join! [select clube
from br.com.cbfm.core.models.Clube clube
inner join Federacao federacao with clube.federacao = federacao
order by clube.id desc]; nested exception is java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: Path expected for join! [select clube
from br.com.cbfm.core.models.Clube clube
inner join Federacao federacao with clube.federacao = federacao
order by clube.id desc]

Answer 1

这应该按预期工作：

JPAQuery query = new JPAQuery(em);
    return query.from(qClube)
        .innerJoin(qClube.federacao, qFederacao)
    .orderBy(qClube.id.desc())
    .list(qClube);

Answer 2

Meiko提供的查询是对的。我已下载了您的项目，并发现了与模型关系相关的一些问题。

请勿使用以下代码

@ManyToOne
@JoinColumn(name = "id_federacao")
@Column(name = "federacao")
public Federacao getFederacao() {
    return federacao;
}

但是使用，

@ManyToOne
@JoinColumn(name = "federacao")
private Federacao federacao;

使用名称＆＃34; federacao＆＃34;相反＆＃34; id_federacao＆＃34;因为您要加入的列是clubes.federacao而不是clubes.id_federacao。
您不能将@Column与@ManyToOne一起使用。

这应该解决你的问题。我也修改了一些小东西，但这不会影响代码的工作。

Answer 3

问题在于我的映射。我刚刚解决了这个问题：

Class Clube

@ManyToOne
@JoinColumn(name="id_federacao", nullable=false)
private Federacao federacao;

和类Federacao

@OneToMany(mappedBy="federacao")
private Set<Clube> clubes;

我现在有另一个问题，我现在就发布了。

谢谢大家。

加入关系ManyToOne时出错。使用QueryDSL

3 个答案: