Why do I search and can't use nickname in the hibernate?

cnmyth · **Joined:** Wed Aug 30, 2006 9:06 pm **Posts:** 1

session = HibernateSessionFactory.currentSession();
Query query = session.createQuery("from Who where name='aa'");
这样才行

session = HibernateSessionFactory.currentSession();
Query query = session.createQuery("from Who a where a.name='aa'");
要是这样就出错
org.hibernate.QueryException: could not resolve property: name of: Who [ from Who a where a.name='aa']
at org.hibernate.persister.entity.AbstractPropertyMapping.throwPropertyException(AbstractPropertyMapping.java:43)
at org.hibernate.persister.entity.AbstractPropertyMapping.toType(AbstractPropertyMapping.java:37)
at org.hibernate.persister.entity.BasicEntityPersister.toType(BasicEntityPersister.java:1094)
at org.hibernate.hql.ast.FromElementType.getPropertyType(FromElementType.java:273)
at org.hibernate.hql.ast.FromElement.getPropertyType(FromElement.java:349)
at org.hibernate.hql.ast.DotNode.getDataType(DotNode.java:474)
at org.hibernate.hql.ast.DotNode.prepareLhs(DotNode.java:208)
at org.hibernate.hql.ast.DotNode.resolve(DotNode.java:166)
at org.hibernate.hql.ast.FromReferenceNode.resolve(FromReferenceNode.java:87)
at org.hibernate.hql.ast.FromReferenceNode.resolve(FromReferenceNode.java:83)
at org.hibernate.hql.ast.HqlSqlWalker.resolve(HqlSqlWalker.java:463)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.expr(HqlSqlBaseWalker.java:863)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.exprOrSubquery(HqlSqlBaseWalker.java:3713)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.comparisonExpr(HqlSqlBaseWalker.java:3190)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.logicalExpr(HqlSqlBaseWalker.java:1405)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.whereClause(HqlSqlBaseWalker.java:599)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.query(HqlSqlBaseWalker.java:404)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.selectStatement(HqlSqlBaseWalker.java:201)
at org.hibernate.hql.antlr.HqlSqlBaseWalker.statement(HqlSqlBaseWalker.java:151)
at org.hibernate.hql.ast.QueryTranslatorImpl.analyze(QueryTranslatorImpl.java:189)
at org.hibernate.hql.ast.QueryTranslatorImpl.doCompile(QueryTranslatorImpl.java:130)
at org.hibernate.hql.ast.QueryTranslatorImpl.compile(QueryTranslatorImpl.java:83)
at org.hibernate.impl.SessionFactoryImpl.getQuery(SessionFactoryImpl.java:427)
at org.hibernate.impl.SessionImpl.getQueries(SessionImpl.java:884)
at org.hibernate.impl.SessionImpl.list(SessionImpl.java:834)
at org.hibernate.impl.QueryImpl.list(QueryImpl.java:74)
at DAOFactory.query(DAOFactory.java:26)
at DAOFactory.main(DAOFactory.java:47)
Exception in thread "main"

是不是我配置出错啊，我是用myeclipse自动生成的
<?xml version="1.0"?>
<!DOCTYPE hibernate-mapping PUBLIC "-//Hibernate/Hibernate Mapping DTD 3.0//EN"
"http://hibernate.sourceforge.net/hibernate-mapping-3.0.dtd">

<hibernate-mapping>
<class name="Who" table="who" schema="dbo" catalog="spdb">
<composite-id name="id" class="WhoId">
<key-property name="name" type="string">
<column name="name" length="50" />
</key-property>
<key-property name="password" type="string">
<column name="password" length="50" />
</key-property>
</composite-id>
</class>
</hibernate-mapping>

gskempes · **Joined:** Mon Nov 22, 2004 11:21 am **Posts:** 42

Hi cnmyth,

the property "name" is part of your composite id, therefore you have to qualify it using "id":

Code:

session = HibernateSessionFactory.currentSession();
Query query = session.createQuery("from Who where id.name='aa'"); 

or:

Code:

session = HibernateSessionFactory.currentSession();
Query query = session.createQuery("from Who a where a.id.name='aa'"); 

Regards,
Georg

P.S.: Please rate this posting if it helps you solving the problem