异常QuerySyntaxException。。。。 is not mapped

org.hibernate.hql.ast.QuerySyntaxException: user is not mapped [from user where name='123' and password='123']
at org.hibernate.hql.ast.util.SessionFactoryHelper.requireClassPersister(SessionFactoryHelper.java:158)
at org.hibernate.hql.ast.tree.FromElementFactory.addFromElement(FromElementFactory.java:87)
at org.hibernate.hql.ast.tree.FromClause.addFromElement(FromClause.java:70)
at org.hibernate.hql.ast.HqlSqlWalker.createFromElement(HqlSqlWalker.java:255)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.fromElement(HqlSqlBaseWalker.java:3056)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.fromElementList(HqlSqlBaseWalker.java:2945)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.fromClause(HqlSqlBaseWalker.java:688)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:544)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:281)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:229)
at org.hibernate.hql.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:228)
at org.hibernate.hql.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:160)
at org.hibernate.hql.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:111)
at org.hibernate.engine.query.HQLQueryPlan.<init>(HQLQueryPlan.java:77)
at org.hibernate.engine.query.HQLQueryPlan.<init>(HQLQueryPlan.java:56)
at org.hibernate.engine.query.QueryPlanCache.getHQLQueryPlan(QueryPlanCache.java:72)
at org.hibernate.impl.AbstractSessionImpl.getHQLQueryPlan(AbstractSessionImpl.java:133)
at org.hibernate.impl.AbstractSessionImpl.createQuery(AbstractSessionImpl.java:112)
at org.hibernate.impl.SessionImpl.createQuery(SessionImpl.java:1623)
at hibernate.yan.LoginServlet.doPost(LoginServlet.java:51)
at hibernate.yan.LoginServlet.doGet(LoginServlet.java:24)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:617)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:717)
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:233)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:191)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:127)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:102)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:109)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:298)
at org.apache.coyote.http11.Http11Processor.process(Http11Processor.java:852)
at org.apache.coyote.http11.Http11Protocol$Http11ConnectionHandler.process(Http11Protocol.java:588)
at org.apache.tomcat.util.net.JIoEndpoint$Worker.run(JIoEndpoint.java:489)
at java.lang.Thread.run(Thread.java:619)

try{
    System.out.println("3");
    Configuration conf=new Configuration().configure();
    SessionFactory sf=conf.buildSessionFactory();
    session=sf.openSession();
    tx=session.beginTransaction();
    System.out.println("1");
    Query query=session.createQuery("from user where name='"+name+"' and password='"+password+"'");
    System.out.println("2");
    List list=(List) query.list();
    tx.commit();
       Iterator it=list.iterator();
       if(it.hasNext())
       {
        User user=(User) it.next();
        System.out.println(user.getName()+" "+user.getPassword()+" "+user.getPhonetype());
        out.print("successfully login!");
       }
      
   
   }catch(Exception e)
   {
    e.printStackTrace();
   }





解决方法，把user改为User，因为User.java中定义了User类和数据库中user表相对应。


hql是写的是PO对象，不是table名

故改为配置文件中的类名即可。