Question

我的项目是java ee web应用程序，我使用hibernate，因为我想搜索值为item_id的值，所以，我编写了搜索结果的代码，但搜索值没有显示，所以我怎么解决这个问题请帮助我。< / p>

        if(submitType != null && item_id != null)
     {


     try {
        AssetsObject assets = new AssetsObject();

        assets.setItem_id(item_id);

        AssetsObject assets1 =assetsProvider.getAssets(assets);






                item_name= assets1.getItem_name();

                hardware = assets1.getHardware();

                software = assets1.getSoftware();

                vender = assets1.getVender();

                venderTel = assets1.getVenderTel();

                venderaddress= assets1.getVenderaddress();


                return INPUT;


                }
         catch (Exception e)
         {
        addActionError("Not vailed item No or its not in the database");
                        //return INPUT;
        }

我的implimenter类查询是这个我认为这个类的问题，你能解决这个问题吗？

公共类AssetsProviderImpl实现AssetsProvider { protected static Logger log = Logger.getLogger（AssetsProviderImpl.class .getName（））;

        private HibernateProvider hibernateProvider;

        public AssetsProviderImpl(HibernateProvider hibernateProvider) throws ProviderException
        {
            this.hibernateProvider = hibernateProvider;

            log.info("AssetsProviderImpl created");
        }   

        public AssetsObject insertAssets(AssetsObject assets) throws ProviderException
        {


            return (AssetsObject) hibernateProvider.save(assets);


        }

        public AssetsObject getAssets(AssetsObject assets) throws ProviderException
        {
            try
            {
                Session session = hibernateProvider.openSession();          

                try
                {               

                    Query query = (Query) session.createQuery("from org.efs.openreports.objects.AssetsObject as assets "
                                + "where assets.item_id = ?");
                    List list1 = (List) ((org.hibernate.Query) query).list();

                    AssetsObject as = (AssetsObject) list1; 

                    return as;
                }
                catch (HibernateException he)
                {               
                    throw he;
                }
                finally
                {
                    session.close();
                }
            }
            catch (HibernateException he)
            {
                throw new ProviderException(he);
            }
        }

＆GT;

Answer 1

您不会将getAssets(assets)方法的结果分配给任何变量。我看不出任何人如何显示结果。我猜代码应该是：

AssetsObject assets1 = assetsProvider.getAssets(assets);

不要把它弄糟，但是Java EE和Hibernate是复杂的野兽，看着你的代码，看起来你并没有真正掌握一些基本的Java东西，如调用方法和分配结果。您也不尊重Java命名约定。你不应该花些时间先学习基本的东西吗？

如何检索搜索结果？

1 个答案: