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Criteria API join:java.lang.IllegalArgumentException:无法解析属性

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我试图通过jpa标准api进行简单的连接操作,但是我收到一个错误:

java.lang.IllegalArgumentException:无法在org.hibernate.ejb.criteria.path.AbstractPathImpl的org.hibernate.ejb.criteria.path.AbstractPathImpl.unknownAttribute(AbstractPathImpl.java:120)中针对路径解析属性[Companies] . locateAttribute(AbstractPathImpl.java:229)atg.hibernate.ejb.criteria.path.AbstractFromImpl.join(AbstractFromImpl.java:411)at com.maven_test.models.jpa.dao.ServicesDAO.findAllWithCompaniesByCriteria(ServicesDAO.java:106)在sun.reflect的sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62)的sun.reflect.NativeMethodAccessorImpl.invoke0(本地方法)中的com.maven_test.models.jpa.ServicesFindByIdByCriteria.test2(ServicesFindByIdByCriteria.java:44) .deggatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)org.junit.runners.model.FrameworkMethod $ 1.runReflectiveCall(FrameworkMethod.java:50)org.junit.internal.runners.model.ReflectiveCallable.run(ReflectiveCallable.java: 12)在org.junit.runners.model.Fra megMethod.invokeExplosively(FrameworkMethod.java:47)org.junit.internal.runners.statements.InvokeMethod.evaluate(InvokeMethod.java:17)atg.junit.runners.ParentRunner.runLeaf(ParentRunner.java:325)org .junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:78)org.junit.runners.BlockJUnit4ClassRunner.runChild(BlockJUnit4ClassRunner.java:57)org.junit.runners.ParentRunner $ 3.run(ParentRunner.java:290) org.junit.runners.ParentRunner $ 1.schedule(ParentRunner.java:71)org.junit.runners.ParentRunner.runChildren(ParentRunner.java:288)org.junit.runners.ParentRunner.access $ 000(ParentRunner.java) :58)atg.junit.runners.ParentRunner $ 2.evaluate(ParentRunner.java:268)org.junit.runners.ParentRunner.run(ParentRunner.java:363)org.junit.runner.JUnitCore.run(JUnitCore) .java:137)at com.intellij.junit4.JUnit4IdeaTestRunner.startRunnerWithArgs(JUnit4IdeaTestRunner.java:74)at com.intellij.rt.execution.junit.JUnitStarter.prepareStreamsAndStart(JUnitStart) er.java:211)位于sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl . )的sun.reflect.NativeMethodAccessorImpl.invoke0(本地方法)的com.intellij.rt.execution.junit.JUnitStarter.main(JUnitStarter.java:67) . java:62)at 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;

}

我们必须加入的表的名称是“zaks_companies”,两种情况的错误都是相同的 .

@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

和公司型号:

@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

1 回答

  • 0

    那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);
    

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