Question

代码：

public void getDetails() {
try {
Session session = sessionFactory.openSession();
Transaction transaction = session.beginTransaction();
String hql = "select c.mobile, c.password FROM CrbtSubMasterDemo c where rownum<20";
Query query = session.createQuery(hql);
List<CrbtSubMasterDemo> itr = query.list();
session.getTransaction().commit();
for (CrbtSubMasterDemo pojo : itr) {//excepion line
System.out.println("[" + pojo.getMobile() + "]");
}
} catch (Exception e) {
e.printStackTrace();
}
}

CrbtSubMasterDemo 与数据库进行pojo映射。当我尝试运行它时，它会给出以下异常：

java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to com.telemune.demoPojo.CrbtSubMasterDemo
at com.telemune.demoHibernate.QueryTester.getDetails(QueryTester.java:57)
at com.telemune.demoHibernate.QueryTester.main(QueryTester.java:23)

问题是query.list（）返回pojo类的对象列表。那为什么是这个例外。我是Hibernate的新手，对不起，如果这是一个愚蠢的问题。

Answer 1

当你这样写：

String hql = "select c.mobile, c.password FROM CrbtSubMasterDemo c where rownum<20";

您的结果集不是List

的CrbtSubMasterDemo

尝试写：

String hql = "select FROM CrbtSubMasterDemo c where rownum<20";

另一种方法是定义CrbtSubMasterDemo的新构造函数，其中只传递两个字段c.mobile, c.password

所以您的查询变为：

String hql = "select new " + CrbtSubMasterDemo.class.getName() + "(c.mobile, c.password) FROM CrbtSubMasterDemo c where rownum<20";

如果您遵循此解决方案，请记住添加默认构造函数（不带参数），因此在您的pojo中您有：

public CrbtSubMasterDemo(String mobile, String password) {
    this.mobile = mobile;
    this.password = password
}

和

public CrbtSubMasterDemo() {
}

Answer 2

String hql = "select c.mobile, c.password FROM CrbtSubMasterDemo c where rownum<20";
Query query = session.createQuery(hql);

结果为List<Object[]>

List<Object[]> itr = query.list();

for (Object[] row : itr) {
  System.out.println(String.format("mobile:%s, password:%s", row[0], row[1]));
}

当然，如果mobile和password是字符串。您可以使用转换器将结果直接转换为CrbtSubMasterDemo。

Hibernate 3.2: Transformers for HQL and SQL

FluentHibernateResultTransformer

Answer 3

不要直接转换＆＃34; query.list（）;＆＃34;的结果。到CrbtSubMasterDemo列表。作为query.list（）返回对象列表。迭代收到的对象列表并逐个投射到列表中的CrbtSubMasterDemo列表

Answer 4

先生，很多时候用户都面临着这种要求。 Hibernate有ResultTransformer来转换对象中的hql / sql。

    public CrbtSubMasterDemo{
         private Stirng mobile;
         private String password;

          public CrbtSubMasterDemo(){
            --------------
         }

        #####after setting the transation set whichever columns you are selecting should be given as name of property of your object
        String hql = "select c.mobile as mobile, c.password as password FROM CrbtSubMasterDemo c where rownum<20";
        Query query = session.createQuery(hql);
        List<CrbtSubMasterDemo> itr = query.setResultTransformer(Transformers.aliasToBean(CrbtSubMasterDemo.class) ).list();
        ##No need to commit the transaction.
    }

它会将您的查询转换为CrbtSubMasterDemo

java.lang.ClassCastException：[Ljava.lang.Object;无法强制转换为className

4 个答案: