标准 API 连接：java.lang.IllegalArgumentException：无法解析属性

Question

我正在尝试通过 jpa 标准 api 进行简单的连接操作，但出现错误：

java.lang.IllegalArgumentException：无法针对路径解析属性 [Companies] 在 org.hibernate.ejb.criteria.path.Abs​​tractPathImpl.unknownAttribute(AbstractPathImpl.java:120) 在 org.hibernate.ejb.criteria.path.Abs​​tractPathImpl.locateAttribute(AbstractPathImpl.java:229) 在 org.hibernate.ejb.criteria.path.Abs​​tractFromImpl.join(AbstractFromImpl.java:411) 在 com.maven_test.models.jpa.dao.ServicesDAO.findAllWithCompaniesByCriteria（ServicesDAO.java:106） 在 com.maven_test.models.jpa.ServicesFindByIdByCriteria.test2(ServicesFindByIdByCriteria.java:44) 在 sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) 在 sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62) 在 sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43) 在 org.junit.runners.model.FrameworkMethod$1.runReflectiveCall(FrameworkMethod.java:50) 在 org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java:12) 在 org.junit.runners.model.FrameworkMethod.invokeExplosively（FrameworkMethod.java:47） 在 org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:17) 在 org.junit.runners.ParentRunner.runLeaf（ParentRunner.java:325） 在 org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:78) 在 org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:57) 在 org.junit.runners.ParentRunner$3.run(ParentRunner.java:290) 在 org.junit.runners.ParentRunner$1.schedule(ParentRunner.java:71) 在 org.junit.runners.ParentRunner.runChildren(ParentRunner.java:288) 在 org.junit.runners.ParentRunner.access$000(ParentRunner.java:58) 在 org.junit.runners.ParentRunner$2.evaluate(ParentRunner.java:268) 在 org.junit.runners.ParentRunner.run(ParentRunner.java:363) 在 org.junit.runner.JUnitCore.run(JUnitCore.java:137) 在 com.intellij.junit4.JUnit4IdeaTestRunner.startRunnerWithArgs(JUnit4IdeaTestRunner.java:74) 在 com.intellij.rt.execution.junit.JUnitStarter.prepareStreamsAndStart(JUnitStarter.java:211) 在 com.intellij.rt.execution.junit.JUnitStarter.main(JUnitStarter.java:67) 在 sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) 在 sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62) 在 com.intellij.rt.execution.application.AppMain.main(AppMain.java:134)

方法如下：

    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;

}

我们要加入的表名为“zaks_companies”，两种情况都报错。

@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

和公司模式：

@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 16:50

Answer 1

这是因为持久属性的名称区分大小写。在这种情况下，持久属性的名称是companies：

private Companies companies;

但查询尝试使用Companies:

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

在连接中应使用companies，如下所示：

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