Hibernate 抛出 QuerySyntaxException 并给出正确的 class 名称

Question

我有一个Product class;

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

一个通用的 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();
    }
}

一个产品 DAO class;

public class ProductDao extends BaseDao<Product> { }

产品 JAX-RS 服务；

@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;
    }
}

当我运行应用程序调用端点时，我得到一个很好的 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>

Answer 1

为避免冲突，请在实施 classes 中指定您的 Class 和实体管理器。示例：

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();
}

然后你的实现 class :

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

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

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

Answer 2

尝试使用全名而不是 simpleClassName。

如下

"FROM " + type.getName()

Hibernate 抛出 QuerySyntaxException 并给出正确的 class 名称

Hibernate throwing a QuerySyntaxException with correct class name given

java

hibernate

resteasy