Question

Hibernate报错:could not resolve property

代码如下
public static List type(String username){ //返回用户类型
Configuration config=new Configuration().configure();
SessionFactory sf=config.buildSessionFactory();
Session session=sf.openSession();
Transaction tx=session.beginTransaction();

String str=new String();
str="select login.cloginType from Logininfo as login where login.cloginId='"+username+"'";
Query query=session.createQuery(str);
List list=query.list();

tx.commit();
tx=null;
session.close();
return list;
}
Hibernate配置
<?xml version="1.0" encoding="utf-8"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">
<!--
Mapping file autogenerated by MyEclipse Persistence Tools
-->
<hibernate-mapping>
<class name="com.library.pojo.Logininfo" table="logininfo" catalog="library">
<id name="cloginId" type="java.lang.String">
<column name="cLoginID" length="20" />
<generator class="assigned" />
</id>
<property name="cloginPwd" type="java.lang.String">
<column name="cLoginPwd" length="20" not-null="true" />
</property>
<property name="iloginType" type="java.lang.Integer">
<column name="iLoginType" not-null="true" />
</property>
</class>
</hibernate-mapping>

运行是出现could not resolve property请问是哪错了。谢谢

Answer 1

看看com.library.pojo.Logininfo类中的属性名和类型跟映射文件中的是否一致，还有列名在数据库中是否存在

Answer 2

用a.user.id。
例如你那句话写成：
1List<Article> articles=(List<Article>)hibernateTemplate.find("from Ariticle a where a.user.id='"+id+"'");

Answer 3

你那貌似many to one 的主外键关系，可能是外没有找到主