org.hibernate.MappingException：否JDBC类型的方言映射：2002 [英] org.hibernate.MappingException: No Dialect mapping for JDBC type: 2002

问题描述

当我尝试执行JPA nativeQuery以获取几何字段类型时，出现 org.hibernate.MappingException：No对于JDBC类型：2002 的Dialect映射。 / p>

我使用Oracle和 org.hibernatespatial.oracle.OracleSpatial10gDialect 。

几何字段映射为：

  @Column（name =geometry）
 @Type（type =org.hibernatespatial.GeometryUserType）
 private几何几何; 
 
 // ... 
 
 List< Object> listFeatures = new LinkedList< Object>（）; 
 
 Query query = entityManager.createNativeQuery（
SELECT+ slots +，geometry FROM edtem_features feature，edtem_dades dada WHERE+ 
feature。+ tematic.getIdGeomField ）+= dada。+ tematic.getIdDataField（）+ 
AND dada.capesid =+ tematic.getCapa（）。getId（）+ 
AND feature.geometriesid =+ tematic。 。getGeometria（）的getId（））; 
 listFeatures.addAll（query.getResultList（））; 
  

这是我在spring + struts2上的hibernate配置

 < bean id =entityManagerFactory
 class =org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean> 
< property name =dataSourceref =dataSource/> 
 
< property name =jpaVendorAdapter> 
< bean class =org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter> 
< property name =databasePlatformvalue =org.hibernatespatial.oracle.OracleSpatial10gDialect/> 
< property name =showSqlvalue =true/> 
< / bean> 
< / property> 
< property name =jpaProperties> 
<道具> 
< prop key =hibernate.dialect> org.hibernatespatial.oracle.OracleSpatial10gDialect< / prop> 
< prop key =hibernate.default_schema> my_schema< / prop> 
< /道具> 
< / property> 
< / bean> 
  

这怎么解决？或者如何强制几何的类型得到这个工作？

解决方案

您可以尝试使用以下映射定义： b>

  @Column（name =geometry，columnDefinition =Geometry，nullable = true）
 private几何几何; 
  

而不是：

  @Column（name =geometry）
 @Type（type =org.hibernatespatial.GeometryUserType）
 private几何几何; 
  

I'm getting org.hibernate.MappingException: No Dialect mapping for JDBC type: 2002 when I try to do a JPA nativeQuery to get a geometry field type.

I'm using Oracle and org.hibernatespatial.oracle.OracleSpatial10gDialect.

The geometry field is mapped as:

@Column(name="geometry")  
@Type(type = "org.hibernatespatial.GeometryUserType")  
private Geometry geometry;

// ...

List<Object> listFeatures = new LinkedList<Object>();

Query query = entityManager.createNativeQuery(
   "SELECT "+ slots +" , geometry FROM  edtem_features feature, edtem_dades dada WHERE" +
   " feature."+ tematic.getIdGeomField() +" = dada."+ tematic.getIdDataField()+ 
   " AND dada.capesid= "+ tematic.getCapa().getId() +
   " AND feature.geometriesid= "+ tematic.getGeometria().getId());
listFeatures.addAll(query.getResultList());

This is my hibernate configuration over spring+struts2

<bean id="entityManagerFactory"
            class="org.springframework.orm.jpa.LocalContainerEntityManagerFactoryBean">
    <property name="dataSource" ref="dataSource" />

    <property name="jpaVendorAdapter">
        <bean class="org.springframework.orm.jpa.vendor.HibernateJpaVendorAdapter">
            <property name="databasePlatform" value="org.hibernatespatial.oracle.OracleSpatial10gDialect" />
            <property name="showSql" value="true" />
        </bean>
    </property>
    <property name="jpaProperties">
        <props>
            <prop key="hibernate.dialect">org.hibernatespatial.oracle.OracleSpatial10gDialect</prop>
            <prop key="hibernate.default_schema">my_schema</prop>
        </props>
    </property>
</bean>

How can this be solved? Or how to force the type of the geometry to get this working?

解决方案

Could you try with the following mapping definition:

@Column(name = "geometry", columnDefinition="Geometry", nullable = true) 
private Geometry geometry;

Instead of:

@Column(name="geometry")  
@Type(type = "org.hibernatespatial.GeometryUserType")  
private Geometry geometry;

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