java.lang.ClassCastException：[Ljava.lang.Object; 无法转换为com.xl.entity.Users

Question

这是代码：

public Users login(String username) {
        Users user=null;
        try {
            String hql="select user.name,user.password from Users user where user.name=:name";
            Query query = session.createQuery(hql);
            query.setString("name", username);
            user=(Users) query.list().get(0);
        } catch (HibernateException e) {
            e.printStackTrace();
        }
        return user;
    }

错误：

java.lang.ClassCastException: [Ljava.lang.Object; cannot be cast to com.xl.entity.Users
    at com.xl.impl.HouseDaoImpl.login(HouseDaoImpl.java:51)
    at com.xl.biz.HouseBiz.login(HouseBiz.java:25)
    at com.xl.Servlet.UserServlet.doGet(UserServlet.java:25)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:690)
    at javax.servlet.http.HttpServlet.service(HttpServlet.java:803)
    at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:290)
    at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
    at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:230)
    at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:175)
    at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:128)
    at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:104)
    at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
    at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:261)
    at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:844)
    at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:581)
    at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:447)
    at java.lang.Thread.run(Thread.java:619)

（翻译由Google翻译提供）

Answer 1

如果需要Users对象，请编辑查询，使其类似于

String hql="from Users user where user.name=:name";

否则结果将是对象数组，因此您需要使用转换器或将其强制转换为object []：

String hql="select user.password from Users user where user.name=:name";
Query query = session.createQuery(hql);
query.setString("name", username);
Object[] result =(Object[]) query.list().get(0);
String password = result[0];

Answer 2

您可以作为public Users(String name, String password)为Users创建另一个构造函数，并将查询更改为

String hql="select new yourpath.Users(user.name,user.password) from Users user
where user.name=:name";

Answer 3

您需要以这种方式支持根实体的ResultTransformer

Query query = session.createQuery(hql);
query.setResultTransformer(RootEntityResultTransformer.INSTANCE);
query.list()

使用投影（选择列表）将产生Object[]类型的结果： ResultTransformer将创建一个User对象并映射原始Object[]以更正User的属性

Answer 4

由于query.list()返回List ，并且您在此处将List的内容强制转换为User类型，因此，由于异常， Object can not be cast to User强制转换Object can not be cast to User 。 请确保以用户形式获取内容。

您可以进行检查：

 if(query.list() instanceOf User){
        // can cast to user.
 }else{
        // can not cast to User.
 }