条件API联接：java.lang.IllegalArgumentException：无法解析属性

Question

我试图通过jpa标准api进行一个简单的连接操作,但我得到了一个错误：

java.lang.IllegalArgumentException：无法针对路径解析属性[Company]（位于org.hibernate.ejb.Criteria.path.AbstractPathimpl.java.Unknownatribute（AbstractPathimpl.java:120)（位于org.hibernate.ejb.Criteria.path.AbstractPathimpl.LocateAttribute（AbstractPathimpl.java:229)（位于org.hibernate.ejb.Criteria.path.AbstractFromimpl.join（WorkMethod.InvokeExplory(FrameworkMethod.Java:47)在org.junit.Internal.Runners.Statements.InvokeMethod.Evalue(InvokeMethod.Java:17)在org.junit.Runners.ParentRunner.Runleaf（ParentRunner.Java:325)在.Java：211）在com.intellij.rt.execution.junit.junitstarter.main(junitstarter.Java:67)在sun.reflect.nativeMethodAccessorInvoke0（本机方法）在sun.reflect.nativeMethodAccessorInvoke.Java:62）在com.intellij.rt.execution.application.appmain.main(appmain.Java:134）

方法类似于：

    public List findAllWithCompaniesByCriteria() {

    EntityManager em = EMgrUtil.createEntityManager();

    CriteriaBuilder builder = em.getCriteriaBuilder();
    CriteriaQuery<Object[]> query = builder.createQuery(Object[].class);

    Metamodel m = em.getMetamodel();

    EntityType<Services> Services_ = m.entity(Services.class);
    EntityType<Companies> Companies_ = m.entity(Companies.class);

    /*Root<Services> services = query.from(Services.class);
    Join<Services, Companies> companies = query.join(Companies_);*/

    Root<Services> s = query.from(Services.class);
    Join<Services, Companies> c = s.join("Companies", JoinType.INNER);
    query.multiselect(s.get("avatar"), c.get("name"));

    List<Object[]> list = em.createQuery(query).getResultList();

    return list;

}

我们必须连接的表的名称是“zaks_companies”，这两种情况的错误都是一样的。

@Entity
@Table(name = "zaks_services")
public class Services {

@Id
@GeneratedValue(strategy=GenerationType.IDENTITY)
@Column(name = "id")
private Integer id;

@Column(name = "title", nullable = false)
@Size(min = 0, max = 255)
private String title;

//@Column(name = "descr", length = 65535, columnDefinition = "Text")
//@Column(name = "descr", length = 4294967295, columnDefinition = "Longtext")
@Column(name = "descr", nullable = false, length = 16777215, columnDefinition = "Mediumtext")
private String descr;

@Column(name = "avatar", nullable = false)
private String avatar;

@Column(name = "date_add", nullable = false, insertable = false, updatable = false, columnDefinition = "Datetime DEFAULT CURRENT_TIMESTAMP")
@Temporal(TemporalType.TIMESTAMP)
private Date date_add;

@OneToOne
@JoinColumn(name = "company_id")
private Companies companies;

...getters/setters
...default constructor

和公司模型：

@Entity
@Table(name = "zaks_companies")
public class Companies {

@Id
@GeneratedValue(strategy = GenerationType.IDENTITY)
@Column(name = "id")
private Integer id;

@OneToOne(mappedBy = "companies")
private Services services;

@Column(name = "name", nullable = false)
@Size(min = 0, max = 255)
private String name;

...getters/setters
...default constructor

我的一本手册在这里：https://www.youtube.com/watch?v=j-F4jvljpgq在16:50

Answer 1

这是因为持久属性的名称区分大小写。在本例中，持久属性的名称为company:

private Companies companies;

但query尝试使用companies:

Join<Services, Companies> c = s.join("Companies", JoinType.INNER);

在join中，应使用companies，如下所示：

Join<Services, Companies> c = s.join("companies", JoinType.INNER);