我正在尝试获取一个不同的 id 返回列表，这将是很长的……但我正在取回这个 DistinctResultList ……如何处理这个问题，以便他们获得他们期望的结果……这就是我正在做...
@NamedQuery(name="getProvidersByResourceIds", query = "SELECT DISTINCT p.resourceId FROM Provider p WHERE p.resourceId `in :resourceIds")`
除此之外，我尝试通过这样做来获取 resourceIds'...
List<Long> provIDs = (List<Long>) emf.createNamedQuery("getProvidersByResourceIds").setParameter("resourceIds", values).getResultList();
但就像我说的，我不断返回 DistinctResultList... 查看调试器，我可以看到返回的值。我怎样才能把它翻译成有用的东西？
javax.ejb.EJBException: See nested exception; nested exception is: java.lang.IllegalArgumentException: Parameter "Parameter<long>('resourceIds')" declared in "SELECT p FROM Provider p WHERE p.resourceId = :resourceIds" is set to value of "org.apache.openjpa.kernel.DistinctResultList@35fb35fb" of type "org.apache.openjpa.kernel.DistinctResultList", but this parameter is bound to a field of type "long".
java.lang.IllegalArgumentException: Parameter "Parameter<long>('resourceIds')" declared in "SELECT p FROM Provider p WHERE p.resourceId = :resourceIds" is set to value of "org.apache.openjpa.kernel.DistinctResultList@35fb35fb" of type "org.apache.openjpa.kernel.DistinctResultList", but this parameter is bound to a field of type "long"