Question

我正在尝试通过jpa标准api进行简单的连接操作，但我收到错误：

java.lang.IllegalArgumentException：无法针对路径解析属性[Companies] at org.hibernate.ejb.criteria.path.AbstractPathImpl.unknownAttribute（AbstractPathImpl.java:120）在org.hibernate.ejb.criteria.path.AbstractPathImpl.locateAttribute（AbstractPathImpl.java:229）在org.hibernate.ejb.criteria.path.AbstractFromImpl.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） at sun.reflect.NativeMethodAccessorImpl.invoke0（Native Method） at sun.reflect.NativeMethodAccessorImpl.invoke（NativeMethodAccessorImpl.java:62） at 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） at sun.reflect.NativeMethodAccessorImpl.invoke0（Native Method） at 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;

}

我们必须加入的表名是“zaks_companies”，两种情况的错误相同。

@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

和公司模特：

@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于16:50

Answer 1

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

private Companies companies;

但查询尝试使用Companies：

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

在加入时，应使用companies，如下所示：

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

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

1 个答案: