Question

我正在使用hibernate和REST服务。查看 REST SERVICE 下面的代码，在这种情况下如何使用HQL语句（Hibernate查询）提取content_id？

例如我的尝试......

<query cache-mode="get" cacheable="true" flush-mode="auto"
name="Languages.getLanguageById" read-only="true">
select languageId  as id,displayName as name from Languages where id=<query-param name="languageId" type="java.lang.Long"/>
</query>

// REST服务

@Path("{resource}/getLanguageById")

@GET
@Produces(MediaType.TEXT_XML)  
public String getLanguageById(@PathParam("resource") String id, 
                              @QueryParam("content-id")String content_id)
{

            ........
            .......

}

// Hibernate XML

<hibernate-mapping>
 <class catalog="CONTENT" name="test.db.com.Languages" table="languages">
  <id name="languageId" type="java.lang.Long">
   <column name="languageId"/>
   <generator class="identity"/>
  </id>
  <property generated="never" lazy="false" name="language" type="string">
   <column length="50" name="language" not-null="true"/>
  </property>
  <property generated="never" lazy="false" name="displayName" type="string">
   <column length="50" name="displayName" not-null="true"/>
  </property>
 </class>
</hibernate-mapping>

Answer 1

没关系我找到了替代解决方案

public static List listObject1(String queryname,Long content_id) {
        String query = "select languageId  as id,displayName as name from Languages where id =:query_id";
        Transaction tx = null;    

        Session session = SessionFactoryUtil.getInstance().openSession();
        Query myquery = session.createQuery(query);
        myquery.setParameter("query_id",content_id);

......

.......more code
    }

如何从REST路径中提取查询参数？

1 个答案: