. lang。ClassCastException:[Ljava.lang.Object;不能将其转换为com.xl.entity.Users。

Here is the code:

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;
    }

Error：

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)

(Translations courtesy Google Translate)

(翻译由谷歌翻译)

4 个解决方案

#1

If you need the Users object, edit your query to be like

如果您需要用户对象，请编辑您的查询。

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

Otherwise the result will be an array of objects so you need to either use a transformer or simply cast it to object[]:

否则，结果将是一个对象数组，因此您需要使用transformer或简单地将其转换为对象[]:

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];

#2

You can create another constructor for Users as public Users(String name, String password) and change query to

您可以为用户创建另一个构造函数，作为公共用户(字符串名称、字符串密码)和更改查询。

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

#3

You need a ResultTransformer backed on root entity in this manner

您需要以这种方式在根实体上支持的ResultTransformer。

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

Using a projection (the select list) will produce a result of Object[] type: the ResultTransformer will create a User object and maps raw Object[] to correct User's property

使用投影(选择列表)将产生对象[]类型的结果:ResultTransformer将创建一个用户对象并映射原始对象[]来纠正用户的属性。

#4

-1

Since query.list() returns List, and here you are casting the content of the list to type User, Now, as exception says Object can not be cast to User. Please make sure, are you getting the content in form of user.

由于query.list()返回列表，这里您正在将列表的内容转换为类型用户，现在，作为异常表示对象不能被转换为用户。请确保你的内容是以用户的形式出现的。

You can make a check:

你可以做一个检查:

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

#1