Hibernate抛出一个QuerySyntaxException与给定的正确的类名 [英] Hibernate throwing a QuerySyntaxException with correct class name given

问题描述

1. 我有一个Product类；

@Entity
public class Product {
    .
    .
    public Product() { }
    .
    .
}

通用DAO；

A generic DAO;

public class GenericDao<T> {

    private Class<T> type;

    @Inject
    protected EntityManager entityManager;

    public GenericDao() { }


    public List<T> list() {
        return entityManager.createQuery("FROM " + type.getSimpleName(), type).getResultList();
    }
}

产品DAO类；

A Product DAO class;

public class ProductDao extends BaseDao<Product> { }

产品JAX-RS服务；

A product JAX-RS service;

@Path("/product")
public class ProductService {

    @Inject
    private ProductDao productDao;

    @GET
    @Path("/getProducts")
    @Produces(MediaType.APPLICATION_JSON)
    public List<Product> getProducts() {
        List<Product> response = productDao.list();
        return response;
    }
}

当我运行应用程序并调用端点时，我得到了一个不错的QuerySyntaxException;

When I run the app and make a call to the endpoint I get a nice QuerySyntaxException;

org.jboss.resteasy.spi.UnhandledException: java.lang.IllegalArgumentException: org.hibernate.hql.internal.ast.QuerySyntaxException: Product is not mapped [FROM Product]

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="mainconfig">
        <provider>org.hibernate.jpa.HibernatePersistenceProvider</provider>
       <properties>
            <property name="javax.persistence.jdbc.driver" value="com.mysql.jdbc.Driver" />
            <property name="javax.persistence.jdbc.url" value="jdbc:mysql://localhost:3306/awsapp" />
            <property name="hibernate.dialect" value="org.hibernate.dialect.MySQLDialect" />
            <property name="hibernate.hbm2ddl.auto" value="create" />
            <property name="javax.persistence.jdbc.user" value="${conf.jdbc.user}" />
            <property name="javax.persistence.jdbc.password" value="${conf.jdbc.password}" />
            <property name="hibernate.show_sql" value="true" />

        </properties>
    </persistence-unit>

</persistence>

推荐答案

为避免冲突，请在实现类中指定您的类和实体管理器.例子:

To avoid conflicts, specify your Class and entity manager in implementation classes. Example :

public abstract class GenericDao<T> {
    private Class<T> clazz;

    public GenericDao(Class<T> clazz) {
        this.clazz = clazz;
    }

    public T getById(Long key) {
        return getEntityManager().find(clazz, key);
    }
    protected abstract EntityManager getEntityManager();
}

然后是您的实现类:

public class ProductDao extends BaseDao<Product> { 
    @PersistenceContext
    private EntityManager em;

    public ProductDao(){
        super(Product.class);
    }

        /**
     * {@inheritDoc}
     */
    @Override
    protected EntityManager getEntityManager() {
        return em;
    }
}

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