原因：org.hibernate.MappingException：未知实体：新对象。 JTA

Question

由于我的数据库很复杂，我决定创建一个新对象，我只在其中指定我将使用的参数。为了进行自定义查询，我在 JPA 中使用了 createNativeQuery。关注我的对象。

POJO:

@Entity
public class Unidade implements Serializable{

    private static final long serialVersionUID = 1L;


    private Long numeroUnidade;
    private String nomeUnidade;
    private Integer numeroCep;
    private Integer numeroCepComplemento;
    private String tipoServico;
    private Integer status;
    private String logradouro;
    private String bairro;
    private String cidade;
    private String uf;
    private Integer statusVinculo; 

   //getter and setters

目前我正在使用以下 select 到 return 的简单整数并且没有显示任何问题。

@PersistenceContext(unitName="projetoAtividade")
private EntityManager em;

public void selectTeste(){
        int quantidade = (Integer) em.createNativeQuery("select count(*) from abc.nattbu29_escritorio;").getSingleResult();
        System.out.println("========== QUANTIDADE DE REGISTRO NA TABELA ===========");
        System.out.println(quantidade);
    }

虽然之前的查询工作正常，但是当提交一个复杂的查询并放置相应的别名时，也使用 createNativeQuery 并放置对象 return，hibernate 抛出异常（org.hibernate.MappingException：未知） .

记得我创建了一个新的 class 将 @Entity 指定。

查询：

@SuppressWarnings("unchecked")
        public List<UnidadeTeste> buscarUnidadeGenerica(int numeroCeps, int numeroComplementoCeps){
            List<UnidadeTeste> lsita = new ArrayList<UnidadeTeste>();
            String query = "SELECT "
                    +"TBZ21.NATURAL as numeroPv, "
                    +"TBZ21.UNIDADE as nomePv, "
                    +"TBZ21.SITUACAO as statusVinculacao, "
                    +"TBJ55.TPO_ATUACAO_U44 as tipoServAZW, "
                    +"TBJ55.SITUACAO_ATUAL as status, "
                    +"TBH18.LOCALIDADE as cidade, "
                    +"'' as bairro, "
                    +"'' as logradouro, "
                    +"TBH18.CEP as numeroCep, "
                    +"TBH18.CEP_COMPLEMENTO as numeroCepComplemento, "
                    +"TBH18.FK_AZWVWL22_UNISG as uf "
                    +"FROM AZW.AZWTBJ55_RGATUACAO TBJ55 "
                    +"LEFT JOIN AZW.AZWTBH18_LOCALIDADE TBH18 ON "
                    +"TBH18.LOCALIDADEE = TBJ55.LOCALIDADEE "
                    +"LEFT JOIN AZW.AZWTBZ21_UNIDADE TBZ21 ON "
                    +"TBJ55.NATURAL = TBZ21.NATURAL "
                    +"WHERE TBH18.CEP = ?1 "
                        +"AND TBH18.CEP_COMPLEMENTO = ?2; "

            lsita = em.createNativeQuery(query, Unidade.class)
                    .setParameter(1, numeroCeps)
                    .setParameter(2, numeroComplementoCeps)
                    .getResultList();
            System.out.println(lsita.size()+"<<<<=====================");
            return lsita;
        }

错误：

Caused by: org.hibernate.MappingException: Unknown entity: br.com.project.modelo.UnidadeTeste
    at org.hibernate.internal.SessionFactoryImpl.getEntityPersister(SessionFactoryImpl.java:1031) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.loader.custom.sql.SQLQueryReturnProcessor.getSQLLoadable(SQLQueryReturnProcessor.java:336) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.loader.custom.sql.SQLQueryReturnProcessor.processRootReturn(SQLQueryReturnProcessor.java:377) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.loader.custom.sql.SQLQueryReturnProcessor.processReturn(SQLQueryReturnProcessor.java:356) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.loader.custom.sql.SQLQueryReturnProcessor.process(SQLQueryReturnProcessor.java:172) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.loader.custom.sql.SQLCustomQuery.<init>(SQLCustomQuery.java:87) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.engine.query.spi.NativeSQLQueryPlan.<init>(NativeSQLQueryPlan.java:67) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.engine.query.spi.QueryPlanCache.getNativeSQLQueryPlan(QueryPlanCache.java:155) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.internal.AbstractSessionImpl.getNativeSQLQueryPlan(AbstractSessionImpl.java:218) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.internal.AbstractSessionImpl.list(AbstractSessionImpl.java:224) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.internal.SQLQueryImpl.list(SQLQueryImpl.java:156) [hibernate-core-4.0.1.Final.jar:4.0.1.Final]
    at org.hibernate.ejb.QueryImpl.getResultList(QueryImpl.java:252) [hibernate-entitymanager-4.0.1.Final.jar:4.0.1.Final]
    ... 84 more

在我通过 Internet 搜索时，几乎所有论坛主题的解决方案答案都是一致的：“将 @Entity 放在 class 声明之前”，但这对我不起作用。

我 "spent" 一整天都在尝试寻找有效的解决方案，但不幸的是，我没有成功。有人可以说一种解决我的问题的新方法吗？我在 2 个不同的项目中进行了测试，遇到了同样的问题。

问题补充：

伙计们，我尝试在我的 class 中创建一个 @SqlResultSetMapping，编码如下，但是现在，休眠器抛出了另一个异常：

 ERROR [org.jboss.ejb3.invocation] (http--127.0.0.1-8888-2) JBAS014134: EJB Invocation failed on component AtividadeEJB for method public void br.com.project.jboss.modelo.AtividadeEJB.testeChamada(): javax.ejb.EJBException: javax.persistence.PersistenceException: org.hibernate.MappingException: Unknown SqlResultSetMapping [UnidadeResult]
    at org.jboss.as.ejb3.tx.CMTTxInterceptor.handleExceptionInOurTx(CMTTxInterceptor.java:166) [jboss-as-ejb3-7.1.1.Final.jar:7.1.1.Final]
    at org.jboss.as.ejb3.tx.CMTTxInterceptor.invokeInOurTx(CMTTxInterceptor.java:230) [jboss-as-ejb3-7.1.1.Final.jar:7.1.1.Final]
    at org.jboss.as.ejb3.tx.CMTTxInterceptor.required(CMTTxInterceptor.java:304) [jboss-as-ejb3-7.1.1.Final.jar:7.1.1.Final]
    at org.jboss.as.ejb3.tx.CMTTxInterceptor.processInvocation(CMTTxInterceptor.java:190) [jboss-as-ejb3-7.1.1.Final.jar:7.1.1.Final]
    at org.jboss.invocation.InterceptorContext.proceed(InterceptorContext.java:288) [jboss-invocation-1.1.1.Final.jar:1.1.1.Final]

我的新 POJO：

@Entity
@SqlResultSetMapping(name="UnidadeResult", 
entities={ 
    @EntityResult(entityClass=UnidadeTeste.class, fields={
        @FieldResult(name="tx_descricao", column="nomePv"),
        @FieldResult(name="nu_idntr_pessoa", column="id")})},
columns={
    @ColumnResult(name="nomePv")}
)


public class UnidadeTeste implements Serializable{

    private static final long serialVersionUID = 1L;

    @Id
    @GeneratedValue
    private Long id;

    private String nomePv;

    public String getNomePv() {
        return nomePv;
    }

    public void setNomePv(String nomePv) {
        this.nomePv = nomePv;
    }

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

我的 EJB：

@SuppressWarnings("unchecked")
    public void testeChamada(){
        List<Object> lista = em.createNativeQuery("SELECT nu_idntr_pessoa as id, email as nomePv FROM pessoa;", "UnidadeResult").getResultList();
        System.out.println("===============>>"+lista.size());
    }

Answer 1

如果你想return一个对象的本地查询结果我建议使用SqlResultSetMapping。

否则，将您的 return 类型设置为：

List<Object[]> results = em.createNativeQuery(query)
                .setParameter(1, numeroCeps)
                .setParameter(2, numeroComplementoCeps)
                .getResultList();

然后遍历结果列表，并将 Object 数组的每个成员转换为其数据类型。

Answer 2

List<UnidadeTeste> lsita = new ArrayList<UnidadeTeste>();

lsita = em.createNativeQuery(查询,Unidade.class)

此处存在类型不匹配。异常很明显：未知实体：br.com.project.modelo.UnidadeTeste

您的实体是 Unidade，您必须 return List<Unidade> 而不是 UnidadeTeste。

否则试试SqlResultSetMapping

Answer 3

我的解决方案是：

本主题中：准确描述了我的问题。但是之前看书还是有点看不懂，解决了我的问题。

即使为我的查询结果的接收者创建一个简单的实体，也有必要将我的 class 映射到 persistence.xml。这是我的错误！

所以，我现在的项目：

persistence.xml:

<?xml version="1.0" encoding="UTF-8"?>
<persistence version="2.0"
    xmlns="http://java.sun.com/xml/ns/persistence" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
    xsi:schemaLocation="http://java.sun.com/xml/ns/persistence http://java.sun.com/xml/ns/persistence/persistence_2_0.xsd">

    <persistence-unit name="projetoAtividade" transaction-type="JTA">
        <provider>org.hibernate.ejb.HibernatePersistence</provider>
        <jta-data-source>java:/projetoAtividade</jta-data-source>

        //another classes

        <class>br.com.project.modelo.UnidadeTeste</class>

        //properties of connection

UnidadeTeste.class:

package br.com.project.modelo;

import javax.persistence.Entity;
import javax.persistence.Id;

@Entity
public class UnidadeTeste {

    @Id
    private Long id;

    private String description;

    public Long getId() {
        return id;
    }

    public void setId(Long id) {
        this.id = id;
    }

    public String getDescription() {
        return description;
    }

    public void setDescription(String description) {
        this.description = description;
    }

}

EJB:

@SuppressWarnings("unchecked")
    public void testeChamada(){
        List<UnidadeTeste> lista = em.createNativeQuery("SELECT nu_idntr_pessoa AS id, email AS description FROM Pessoa", UnidadeTeste.class).getResultList();
        System.out.println("===============>>"+lista.size());
    }

请记住，您需要为您在 class.

中创建的属性上的同名列设置别名

原因：org.hibernate.MappingException：未知实体：新对象。 JTA

Caused by: org.hibernate.MappingException: Unknown entity: New Object. JTA

java

jpa

jta