休眠拋出QuerySyntaxException給出正確的類名稱

Question

1. 我有Product類;

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

2. 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(); 
       } 
   } 
   

3. 產品DAO class;

   public class ProductDao extends BaseDao<Product> { }

4. 一種產品，JAX-RS服務;

   @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; 
       } 
   } 
   

   當我運行應用程序並調用端點時，我得到一個很好的QuerySyntaxException;

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

的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

爲了避免衝突，指定實現類的類和實體管理器。例如：

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(); 
} 

然後實現類：

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

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

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

Answer 2

試圖把全名，而不是simpleClassName。

如下

「FROM」 + type.getName（）